Post by: xD_aQt on January 14, 2010, 10:01:33 am
Let f(x) =
Post by: the.watchman on January 14, 2010, 10:08:49 am
I think the reason for this is that the maximum value of the numerator is 21 (no matter what
For the minimum value, the maximum value of the denominator is
Sorry for my bad explaining!!!
Post by: xD_aQt on January 14, 2010, 10:12:17 am
The range isI guess that sounds about right! :) If someone could clarify?
I think the reason for this is that the maximum value of the numerator is 21 (no matter what) and the minimum value of the denominator is 7 (when
), so the maximum value of
is
For the minimum value, the maximum value of the denominator is, so
Sorry for my bad explaining!!!
Post by: the.watchman on January 14, 2010, 10:13:25 am
My bad!!!
Post by: xD_aQt on January 14, 2010, 10:15:41 am
Post by: xD_aQt on January 14, 2010, 10:16:10 am
Sorry, it's actuallyHaha, could you explain please?
My bad!!!
Post by: the.watchman on January 14, 2010, 10:17:43 am
Sorry, it's actuallyHaha, could you explain please?
My bad!!!
The maximum is defined at
Post by: the.watchman on January 14, 2010, 10:19:30 am
Ok, the 'central' point of the cosine graph is at
As the amplitude is 2, take the values 2 units from
SO the range is
Post by: xD_aQt on January 14, 2010, 10:21:58 am
pi is \pi (but it looks dodgy)Cheers .. umm so the graph starts from 6? and from there it moves up and down of 2 units? is that right?
Ok, the 'central' point of the cosine graph is at.
As the amplitude is 2, take the values 2 units from
SO the range is
Post by: the.watchman on January 14, 2010, 10:26:16 am
then the range is
Post by: xD_aQt on January 14, 2010, 10:30:08 am
It doesn't actually start frommmm .. think I get you :)or
,
then the range is
Post by: xD_aQt on January 14, 2010, 10:34:02 am
Post by: brightsky on January 14, 2010, 10:40:32 am
The graph is essentially a normal parabola moved 1 unit upwards. So the range (how much the graph covers in terms of y-values), starts at 1.
However, the graph is limited
Substitute those x-values into the equation:
So the range of the graph is
Post by: xD_aQt on January 14, 2010, 10:42:17 am
xD_aQt, for these kinds of questions, just imagine how the graph might look like.Haha, thanks I get you :)where
The graph is essentially a normal parabola moved 1 unit upwards. So the range (how much the graph covers in terms of y-values), starts at 1.
However, the graph is limited. Hence, we need to find the y-value when x = -2 and when x = 1.
Substitute those x-values into the equation:and
The biggest out of the numbers is 5, so the range of the graph is
Post by: xD_aQt on January 14, 2010, 10:46:00 am
The maximal domain of the function with equation f(x) =
Post by: simonhu81292 on January 14, 2010, 11:08:13 am
such as
f(x)= root x
the maximal domain is[0, infinity)
since the y values would be un-indentified if x values are lower than 0 ... ;D
for your question... under roots.. it must be greater than 0
3-x>0
3>x
therefore your domain (-infinity,3]
correct me if i am wrong .....the.watchman ;D
also sorry... i can't find the infinity symbol ;)
Post by: /0 on January 14, 2010, 11:17:33 am
If you haven't already, it's worth checking out the
If you put your cursor over other people's LaTeX, you should see the code that produces it. It wil let you type complicated equations in your posts.
For instance, putting your cursor over
Code: [Select]
(-\infty,3)This is a useful way to learn about new latex code :)Post by: xD_aQt on January 14, 2010, 11:18:33 am
so the maximal domain is basically the implied domain, the largest value of x you could in a functionWould you exclude the 3 then? so (-
such as
f(x)= root x
the maximal domain is[0, infinity)
since the y values would be un-indentified if x values are lower than 0 ... ;D
for your question... under roots.. it must be greater than 0
3-x>0
3>x
therefore your domain (-infinity,3]
correct me if i am wrong .....the.watchman ;D
also sorry... i can't find the infinity symbol ;)
Post by: the.watchman on January 14, 2010, 11:24:24 am
However, when the root is on the denominator of a function, it can't equal 0, so the endpoint is not included.
Post by: brightsky on January 14, 2010, 11:29:24 am
Post by: xD_aQt on January 14, 2010, 02:40:13 pm
In normal roots, the endpoint (when the root = 0) is included in the domain/range.
However, when the root is on the denominator of a function, it can't equal 0, so the endpoint is not included.
You exclude 3 in the domain becauseThanks Heaps! :)is an asymptote. The graph comes really close to
is undefined.
Post by: xD_aQt on January 14, 2010, 02:41:03 pm
Post by: simonhu81292 on January 14, 2010, 02:44:58 pm
;D
Post by: the.watchman on January 14, 2010, 02:48:17 pm
so the domain is
Post by: brightsky on January 14, 2010, 02:51:25 pm
so the domain isand the range is
+1
Post by: xD_aQt on January 14, 2010, 02:52:52 pm
Your going to ace methods this year :D
so the domain is
Post by: the.watchman on January 14, 2010, 02:53:43 pm
Post by: xD_aQt on January 14, 2010, 02:53:56 pm
The range of f(x) = esin(x) is
Post by: xD_aQt on January 14, 2010, 02:55:01 pm
I wish :DHave a little bit of confident ;D You've been only to help me out, so I know you can!
Post by: the.watchman on January 14, 2010, 02:59:13 pm
where
as min/max values for
when these are subbed into
so
Post by: xD_aQt on January 14, 2010, 03:02:10 pm
This is a composite function:Could you explain it in a different way, I'm a bit unsure how you solved the problem :-\
whereand
so
as min/max values forare
when these are subbed into
Post by: the.watchman on January 14, 2010, 03:03:35 pm
Post by: the.watchman on January 14, 2010, 03:06:47 pm
I'm just finding the range of
Post by: xD_aQt on January 14, 2010, 03:08:42 pm
Have you read the edit yet??? :):o Haha yeah :(
This is a composite function:as min/max values for
where
as min/max values for
when these are subbed into, you get
and
so
and so
Post by: xD_aQt on January 14, 2010, 03:10:39 pm
I typed e-1 and e1 into the calculator and I get values above 0 so would R+ be right?
Post by: the.watchman on January 14, 2010, 03:18:55 pm
I typed e-1 and e1 into the calculator and I get values above 0 so would R+ be right?
Nono, the approx range is
And in reference to
I think my explaining is dud (soz!)
Post by: xD_aQt on January 14, 2010, 03:21:33 pm
Nah your right! Its just my understanding, haha I'm a bit slow but I come around eventually :)
I typed e-1 and e1 into the calculator and I get values above 0 so would R+ be right?
Nono, the approx range is
And in reference tohas min/max values of
.
I think my explaining is dud (soz!)
Post by: xD_aQt on January 14, 2010, 03:22:14 pm
The least value of f(x) = 5 - 2 cos (x) is
Post by: the.watchman on January 14, 2010, 03:25:20 pm
General rule: For any
The min value of
It'll come in handy :)
Post by: the.watchman on January 14, 2010, 03:26:26 pm
So I'm typing left-handed only...
Post by: brightsky on January 14, 2010, 03:26:54 pm
Post by: the.watchman on January 14, 2010, 03:27:32 pm
xD_aQt,means, ALL real numbers above 0. :)
Ahhh, now that's a good explanation :D
Post by: xD_aQt on January 14, 2010, 03:28:11 pm
It'sdoes the general rule apply for both sin and cos?
General rule: For any,
The min value ofis
and the max value is
It'll come in handy :)
Post by: xD_aQt on January 14, 2010, 03:28:30 pm
Haha right, gotcha! Cheers so basically [0,xD_aQt,
Ahhh, now that's a good explanation :D
Post by: the.watchman on January 14, 2010, 03:29:35 pm
It'sdoes the general rule apply for both sin and cos?
General rule: For any
The min value of
It'll come in handy :)
Yup, but not tan :P
Post by: xD_aQt on January 14, 2010, 03:30:57 pm
Post by: xD_aQt on January 14, 2010, 03:32:13 pm
The implied domain of f =
Post by: the.watchman on January 14, 2010, 03:34:51 pm
This is at
So f is defined for the maximal domain
Post by: the.watchman on January 14, 2010, 03:36:18 pm
The implied domain, The maximal domain and The domain .. LOL they're all basically the domain, yeah? Are there any other domain?
If they give you a range restriction and ask you to find the domain, then it's not the (maximal/implied) domain you're after, it's the restricted domain (based on the range)
Post by: xD_aQt on January 14, 2010, 03:36:36 pm
First find whenDoes the sign change because when you square root? And its meant to include, right?,
This is atand
,
So f is defined for the maximal domain
Post by: the.watchman on January 14, 2010, 03:37:36 pm
First find whenDoes the sign change because when you square root?
This is at
So f is defined for the maximal domain
Sorry, I'm not sure what you mean.
Post by: xD_aQt on January 14, 2010, 03:38:11 pm
Okay, I'll keep that in mind.The implied domain, The maximal domain and The domain .. LOL they're all basically the domain, yeah? Are there any other domain?
If they give you a range restriction and ask you to find the domain, then it's not the (maximal/implied) domain you're after, it's the restricted domain (based on the range)
Post by: brightsky on January 14, 2010, 03:39:56 pm
When
Hope this explanation helps. :)
Post by: xD_aQt on January 14, 2010, 03:41:11 pm
Never mind, I think I understand how you got your answer now.First find whenDoes the sign change because when you square root?
This is at
So f is defined for the maximal domain
Sorry, I'm not sure what you mean.
Post by: xD_aQt on January 14, 2010, 03:41:40 pm
Imagine, the sine and cosine graphs. In the general form:Cheers, your explanation is just as good :), the amplitude would be
(the b changes nothing except the period). So the max and min values for y would be
and
respectively.
Whenis added to the function, the graph would be pushed upwards or downwards (depending on the value of c) c units. Hence, the max and min values of y becomes
and
.
Hope this explanation helps. :)
Post by: the.watchman on January 14, 2010, 03:42:10 pm
Imagine, the sine and cosine graphs. In the general form:
When
Hope this explanation helps. :)
Great, except the a's should be
Post by: brightsky on January 14, 2010, 03:42:46 pm
Post by: xD_aQt on January 14, 2010, 03:43:27 pm
It is less than because you cannot square root a negative number given that you are looking for real number solutions.Oh right!
Post by: the.watchman on January 14, 2010, 03:44:49 pm
It is less than because you cannot square root a negative number given that you are looking for real number solutions.
Yup!
Alternatively you could solve
Post by: brightsky on January 14, 2010, 03:45:35 pm
Oh yes, edited.Imagine, the sine and cosine graphs. In the general form:
When
Hope this explanation helps. :)
Great, except the a's should be, because if a is negative, then it goes the other way around. :P
Post by: xD_aQt on January 14, 2010, 03:45:43 pm
Great, except the a's should beMmm .. what does the | | mean ?
Post by: the.watchman on January 14, 2010, 03:47:13 pm
Great, except the a's should beMmm .. what does the | | ?
Modulus, absolute value etc.
The number of units from 0 to the number in question
YAY!!! Magical 200th post!!! :D
Post by: xD_aQt on January 14, 2010, 03:49:22 pm
Is it important to include the | | when dealing with problems like these?Great, except the a's should beMmm .. what does the | | ?
Modulus, absolute value etc.
The number of units from 0 to the number in question
YAY!!! Magical 200th post!!! :D
Post by: brightsky on January 14, 2010, 03:50:00 pm
Post by: xD_aQt on January 14, 2010, 03:50:11 pm
Haha 200 :) How many days?Great, except the a's should beMmm .. what does the | | ?
Modulus, absolute value etc.
The number of units from 0 to the number in question
YAY!!! Magical 200th post!!! :D
Post by: brightsky on January 14, 2010, 03:51:41 pm
Great, except the a's should beMmm .. what does the | | ?
Modulus, absolute value etc.
The number of units from 0 to the number in question
YAY!!! Magical 200th post!!! :D
Yep, so
Post by: the.watchman on January 14, 2010, 03:52:10 pm
Haha 200 :) How many days?Great, except the a's should beMmm .. what does the | | ?
Modulus, absolute value etc.
The number of units from 0 to the number in question
YAY!!! Magical 200th post!!! :D
HOLY @#$%!!!
11 days for the 200, and exactly 48 hours since the 100 milestone!!! I've been busy :D
Post by: xD_aQt on January 14, 2010, 03:52:43 pm
Haha, now I'm a little lost :(Great, except the a's should beMmm .. what does the | | ?
Modulus, absolute value etc.
The number of units from 0 to the number in question
YAY!!! Magical 200th post!!! :D
Yep, soand so on. It turns negative numbers into positive. But positives stay positive. :p
Post by: brightsky on January 14, 2010, 03:52:48 pm
Post by: xD_aQt on January 14, 2010, 03:53:15 pm
Busy helping, thanks heaps! :D BOTH!Haha 200 :) How many days?Great, except the a's should beMmm .. what does the | | ?
Modulus, absolute value etc.
The number of units from 0 to the number in question
YAY!!! Magical 200th post!!! :D
HOLY @#$%!!!
11 days for the 200, and exactly 48 hours since the 100 milestone!!! I've been busy :D
Post by: the.watchman on January 14, 2010, 03:54:48 pm
Hehe, you've been posting more than me, the.watchman! :)
And when I joined and saw your posts/day rate I was like WOW :P
Post by: brightsky on January 14, 2010, 03:55:04 pm
Post by: brightsky on January 14, 2010, 03:55:26 pm
Hehe, you've been posting more than me, the.watchman! :)
And when I joined and saw your posts/day rate I was like WOW :P
lolz, and now you've officially surpassed it. :p
Post by: TrueTears on January 14, 2010, 03:57:06 pm
Haven't seen anyone surpass him :P
Post by: the.watchman on January 14, 2010, 03:57:55 pm
over9000 had a post rate of 70 posts/day when he first joined.
Haven't seen anyone surpass him :P
WOW, that's amazing :D
EDIT: And kinda ridiculous... :P
Post by: xD_aQt on January 14, 2010, 03:58:12 pm
over9000 had a post rate of 70 posts/day when he first joined.LOL .. what a record, how to beat?!
Haven't seen anyone surpass him :P
Post by: brightsky on January 14, 2010, 04:00:08 pm
However, how about if a was negative. The only difference it makes to the graph is flip it, but it doesn't change any max and min calculations as the amplitude would still remain the same. This is why the amplitude is
Now, the error in mine was that if we make a negative, then
Post by: xD_aQt on January 14, 2010, 04:00:58 pm
Post by: brightsky on January 14, 2010, 04:01:05 pm
Post by: TrueTears on January 14, 2010, 04:04:09 pm
The domain of y =is given by
Solve by splitting into cases or sketch a diagram.
Whatever floats your boat.
Post by: the.watchman on January 14, 2010, 04:06:06 pm
I officially dictate that this thread is going WAYY to fast. :p
Yup!!!
The domain of y =
First, when is
When
Next, when is
When
That is
SO the domain is
Post by: the.watchman on January 14, 2010, 04:06:57 pm
The domain of y =
Solve by splitting into cases or sketch a diagram.
Whatever floats your boat.
AND when
Post by: TrueTears on January 14, 2010, 04:09:06 pm
no.The domain of y =
Solve by splitting into cases or sketch a diagram.
Whatever floats your boat.
AND when
Post by: brightsky on January 14, 2010, 04:09:31 pm
I officially dictate that this thread is going WAYY to fast. :p
Yup!!!The domain of y =
First, when isundefined???
When
Next, when is
When
That isor
SO the domain is
+1, but why did you need the "First, when is
Post by: the.watchman on January 14, 2010, 04:10:41 pm
I officially dictate that this thread is going WAYY to fast. :p
Yup!!!The domain of y =
First, when is
When
Next, when is
When
That is
SO the domain is
+1, but why did you need the "First, when is
Soz ...
Post by: xD_aQt on January 14, 2010, 04:12:39 pm
Post by: TrueTears on January 14, 2010, 04:15:11 pm
Post by: brightsky on January 14, 2010, 04:16:56 pm
Hence the range is R\{0} and the domain is R\ {9}.
Post by: brightsky on January 14, 2010, 04:17:36 pm
Sketch to trivialize the exercise.
TT, sketching graphs can often be the long and hard way of finding the domain and range. :p
Post by: TrueTears on January 14, 2010, 04:18:05 pm
Post by: brightsky on January 14, 2010, 04:19:19 pm
not for this one, don't underestimate the power of a sketch
LOLZ!
Post by: xD_aQt on January 14, 2010, 04:23:15 pm
Sketch to trivialize the exercise.
TT, sketching graphs can often be the long and hard way of finding the domain and range. :p
not for this one, don't underestimate the power of a sketchSketching would be my weakness - Hoping there is no sketch the graph this year.
Post by: TrueTears on January 14, 2010, 04:24:44 pm
Watch out 2010...
Post by: superflya on January 14, 2010, 04:26:15 pm
Not a single question on trig in the 2009 paper.
Watch out 2010...
yay :P
Post by: brightsky on January 14, 2010, 04:26:32 pm
Post by: TrueTears on January 14, 2010, 04:29:55 pm
Really? There's always got to be a question of some sort on the exam that involves sketching graphs.Ofcourse there's sketching graphs on the exam, was one on 2009 exam 2 :)
Post by: brightsky on January 14, 2010, 04:30:50 pm
Post by: TrueTears on January 14, 2010, 04:31:44 pm
Post by: xD_aQt on January 14, 2010, 04:33:43 pm
f (x) = 2 (1 -
2 (e-x - 1) for x
Post by: xD_aQt on January 14, 2010, 04:34:23 pm
TT, how are you online, yet offline? :pYou can appear offline, yet be online! :D
Post by: superflya on January 14, 2010, 04:36:18 pm
Post by: brightsky on January 14, 2010, 04:49:54 pm
Turning point at (0,2)
So implied domain = R, implied range =
But
So domain = [-3,0)
Post by: xD_aQt on January 14, 2010, 04:56:49 pm
Post by: xD_aQt on January 14, 2010, 05:22:40 pm
Post by: cipherpol on January 14, 2010, 05:27:43 pm
The range of y = 2 - e x-1 is
The graph of
Post by: xD_aQt on January 14, 2010, 05:31:20 pm
Post by: cipherpol on January 14, 2010, 05:33:45 pm
so what does the x-1 mean?
A translation of 1 unit in the positive direction of the x-axis, but since the domain of
Post by: xD_aQt on January 14, 2010, 05:35:24 pm
Thanks! :)so what does the x-1 mean?
A translation of 1 unit in the positive direction of the x-axis, but since the domain ofis
, it doesn't affect the range.
Post by: xD_aQt on January 14, 2010, 05:36:12 pm
Post by: cipherpol on January 14, 2010, 05:40:22 pm
The function f: R -> R, where f (x) = 2 0.1x + 0.1 - 3 has range
Exponential's have the same type of graph in general, so just visualize it as a normal exponential graph which has been translated 3 units down. Should be able to get the range from that :)
Post by: xD_aQt on January 14, 2010, 05:43:07 pm
I struggle with graphs, any pointers to help?The function f: R -> R, where f (x) = 2 0.1x + 0.1 - 3 has range
Exponential's have the same type of graph in general, so just visualize it as a normal exponential graph which has been translated 3 units down. Should be able to get the range from that :)
Post by: brightsky on January 14, 2010, 05:49:44 pm
Let
y-values are non-existent when
Hence, the range is
Post by: cipherpol on January 14, 2010, 05:50:32 pm
The basic exponential graph,I struggle with graphs, any pointers to help?The function f: R -> R, where f (x) = 2 0.1x + 0.1 - 3 has range
Exponential's have the same type of graph in general, so just visualize it as a normal exponential graph which has been translated 3 units down. Should be able to get the range from that :)
Post by: TrueTears on January 14, 2010, 05:52:36 pm
cihperpr0The basic exponential graph,I struggle with graphs, any pointers to help?The function f: R -> R, where f (x) = 2 0.1x + 0.1 - 3 has range
Exponential's have the same type of graph in general, so just visualize it as a normal exponential graph which has been translated 3 units down. Should be able to get the range from that :)where
has an asymptote at y=0. As x approaches
Post by: xD_aQt on January 14, 2010, 05:53:38 pm
y-values are non-existent when. You could probably work from that.
Mmm .. okay think I understandThe basic exponential graph,I struggle with graphs, any pointers to help?The function f: R -> R, where f (x) = 2 0.1x + 0.1 - 3 has range
Exponential's have the same type of graph in general, so just visualize it as a normal exponential graph which has been translated 3 units down. Should be able to get the range from that :)
Post by: QuantumJG on January 14, 2010, 06:07:28 pm
The function f: R -> R, where f (x) = 2 0.1x + 0.1 - 3 has range
(-3,infinity)
and btw,
If y = ax, then y = exln(a)
Post by: xD_aQt on January 14, 2010, 06:09:40 pm
Hmm .. how did you do that? :PThe function f: R -> R, where f (x) = 2 0.1x + 0.1 - 3 has range
(-3,infinity)
and btw,
If y = ax, then y = exln(a)
Post by: TrueTears on January 14, 2010, 06:13:24 pm
Let
Post by: xD_aQt on January 14, 2010, 06:21:53 pm
Post by: brightsky on January 14, 2010, 06:40:53 pm
In the general equation:
Knowing the sine graph, y-values can only go as far as the amplitude goes, hence the range of the sine graph would be [-a, a]. Hence, the range in this case is [-3, 3].
Post by: xD_aQt on January 14, 2010, 06:57:52 pm
Post by: TrueTears on January 14, 2010, 06:59:34 pm
Post by: superflya on January 14, 2010, 07:00:51 pm
Post by: xD_aQt on January 14, 2010, 07:03:33 pm
Post by: superflya on January 14, 2010, 07:06:55 pm
oh so since |a| determines the max and min then that tells us the range in my case right?
yep
Post by: xD_aQt on January 14, 2010, 07:12:57 pm
Cool :)oh so since |a| determines the max and min then that tells us the range in my case right?
yep
Post by: xD_aQt on January 14, 2010, 07:13:52 pm
Post by: TrueTears on January 14, 2010, 07:14:55 pm
Solve
:)
Post by: xD_aQt on January 14, 2010, 07:47:50 pm
Post by: brightsky on January 14, 2010, 07:56:02 pm
Split into two cases:
1.
2.
In Case 1:
Split the second inequality into two parts:
i.
ii.
In Case i,
Solving simultaneously using a number line,
Solving simultaneously with
In Case ii,
Solving simultaneously using a number line,
Solving simultaneously with
In Case 2:
Split the second inequality into two parts:
i.
ii.
In Case i,
Solving simultaneously using a number line, we get
Solving simultaneously with
In Case ii,
Solving simultaneously using a number line, we have no solutions.
Hmmm...there should be one other solutions...
Post by: the.watchman on January 14, 2010, 08:00:30 pm
This thread has grown exponentially while I was gone!!! :D
Post by: xD_aQt on January 14, 2010, 08:03:59 pm
xD_aQt, you have many questions, young Padawan.Haha, welcome back! :) missed ya ;)
This thread has grown exponentially while I was gone!!! :D
Post by: brightsky on January 14, 2010, 08:09:13 pm
The domain of the function f(x) =is
y is existent when
After much inequality calculations, remembering that
They are your domains. :)
Post by: brightsky on January 14, 2010, 08:11:00 pm
xD_aQt, you have many questions, young Padawan.
This thread has grown exponentially while I was gone!!! :D
Help, the.watchman. For the inequality in th previous question, out of just mere observation, there should be one more solution:
Post by: xD_aQt on January 14, 2010, 08:11:19 pm
Thanks I understand everything up to the domain. I don't understand how you got your domain :-\The domain of the function f(x) =
y is existent when
After much inequality calculations, remembering thatand
this yields
and
.
They are your domains. :)
Post by: brightsky on January 14, 2010, 08:13:55 pm
Post by: xD_aQt on January 14, 2010, 08:17:07 pm
Thanks I understand everything up to the domain. I don't understand how you got your domain :-\The domain of the function f(x) =
y is existent when
After much inequality calculations, remembering that
They are your domains. :)
Post by: brightsky on January 14, 2010, 08:17:12 pm
Split into cases
Solve
:)
TT, shouldn't it be:
Post by: brightsky on January 14, 2010, 08:17:59 pm
Thanks I understand everything up to the domain. I don't understand how you got your domain :-\The domain of the function f(x) =
y is existent when
After much inequality calculations, remembering that
They are your domains. :)and
Oh yes, my bad, arithmetic error! :p
Post by: xD_aQt on January 14, 2010, 08:21:24 pm
I thought it was meant to be?Split into cases
Solve
:)
TT, shouldn't it be:
Mabye I'm wrong :P
Post by: TrueTears on January 14, 2010, 08:21:57 pm
SKETCH IT FTW!!!!111ONE
DO NOT UNDERESTIMATE SKETCHING
Post by: brightsky on January 14, 2010, 08:24:45 pm
This is all so trivial lulz
SKETCH IT FTW!!!!111ONE
DO NOT UNDERESTIMATE SKETCHING
lolz....
Post by: xD_aQt on January 14, 2010, 08:25:19 pm
Post by: brightsky on January 14, 2010, 08:28:24 pm
Post by: superflya on January 14, 2010, 08:29:42 pm
Post by: xD_aQt on January 14, 2010, 08:30:30 pm
^^ asking like every second question in the book :PHaha, am I? What book would this be? :)
Post by: brightsky on January 14, 2010, 08:30:55 pm
Post by: xD_aQt on January 14, 2010, 08:31:34 pm
|c|>a and c <0?Could you explain it please? I'm not sure what the question is asking.
Post by: superflya on January 14, 2010, 08:32:45 pm
Good practice for everyone, superflya! :)
true. :)
oh n btw u didnt answer my question, are u in yr 12?
Post by: brightsky on January 14, 2010, 08:35:06 pm
Good practice for everyone, superflya! :)
true. :)
oh n btw u didnt answer my question, are u in yr 12?
Nope.
Post by: xD_aQt on January 14, 2010, 08:36:08 pm
WOW :DGood practice for everyone, superflya! :)
true. :)
oh n btw u didnt answer my question, are u in yr 12?
Nope.
Post by: superflya on January 14, 2010, 08:36:24 pm
Post by: brightsky on January 14, 2010, 08:37:28 pm
|c|>a and c <0?Could you explain it please? I'm not sure what the question is asking.
Ok,
Post by: xD_aQt on January 14, 2010, 08:42:42 pm
Post by: TrueTears on January 14, 2010, 08:49:03 pm
Post by: superflya on January 14, 2010, 08:49:30 pm
Post by: xD_aQt on January 14, 2010, 08:51:28 pm
Long divide.You mean long division?! :o
Post by: xD_aQt on January 14, 2010, 08:52:15 pm
y=1 and x=-3 ?Haha, did you use what TT suggested? Lol, how'd you get there? :)
Post by: superflya on January 14, 2010, 08:53:11 pm
damn u latex -.-
rest is trivial :P yes, wat TT suggested
Post by: xD_aQt on January 14, 2010, 08:54:59 pm
Your expression didn't come out properly :)
rest is trivial :P
EDIT: never mind :P LOL
^Thanks
Post by: brightsky on January 14, 2010, 08:58:46 pm
Post by: superflya on January 14, 2010, 09:00:09 pm
Your expression didn't come out properly :)
rest is trivial :P
EDIT: never mind :P LOL=
^Thanks
Post by: /0 on January 14, 2010, 09:01:18 pm
Post by: superflya on January 14, 2010, 09:01:56 pm
:p
ceebs learning how to use latex :P
Post by: xD_aQt on January 14, 2010, 09:05:24 pm
I didn't know how to use it till this morning and I've been a part of VNOTES since last year :P
ceebs learning how to use latex :P
Post by: xD_aQt on January 14, 2010, 09:06:48 pm
wtf this thread grew 9 PAGES while I was sleeping? That's great dedication xDHaha, more to come my friend! :)
Post by: superflya on January 14, 2010, 09:08:10 pm
Post by: xD_aQt on January 14, 2010, 09:10:42 pm
this is only 1 exercise ey? lolHaha, its just some questions from the topic Functions
Post by: xD_aQt on January 14, 2010, 09:13:53 pm
Post by: brightsky on January 14, 2010, 09:15:42 pm
Post by: xD_aQt on January 14, 2010, 09:16:28 pm
The maximal domain D of the function f: D->R with rule f(x) = loge(2x +1) is
Post by: /0 on January 14, 2010, 09:18:35 pm
So
So the domain is
Post by: xD_aQt on January 14, 2010, 09:20:47 pm
Post by: superflya on January 14, 2010, 09:21:09 pm
Yes, I DESPISE probability!!!! >:(
probability is one of my fave topics :P
Post by: xD_aQt on January 14, 2010, 09:22:37 pm
LOL :DYes, I DESPISE probability!!!! >:(
probability is one of my fave topics :P
Post by: brightsky on January 14, 2010, 09:23:12 pm
This translates to:
Now x has to be greater than 0 for the x-value to exist.
This is because
Post by: Over9000 on January 14, 2010, 09:28:37 pm
Take the general log equation:I disagree, i^2=-1
This translates to:
Now x has to be greater than 0 for the x-value to exist.
This is becausecannot be negative, and
is undefined.
Post by: brightsky on January 14, 2010, 09:29:07 pm
Take the general log equation:I disagree, i^2=-1
This translates to:
Now x has to be greater than 0 for the x-value to exist.
This is because
Maths Methods don't deal with complex numbers....:p
Post by: Over9000 on January 14, 2010, 09:33:12 pm
Post by: xD_aQt on January 14, 2010, 09:34:38 pm
Post by: TrueTears on January 14, 2010, 09:35:22 pm
Let f: R->R, f(x) = ex + e-x . For all u (element of) R , f(2u) is equal toHint
Post by: xD_aQt on January 14, 2010, 09:36:10 pm
Post by: TrueTears on January 14, 2010, 09:40:12 pm
Post by: brightsky on January 14, 2010, 09:42:52 pm
Post by: xD_aQt on January 14, 2010, 09:57:04 pm
A hyperbolic function, namelyOh, I've never heard it before.
Post by: xD_aQt on January 14, 2010, 10:16:47 pm
Won't it be merely,I guessing that is still right but??
The answer is f(2u) = (f(u))2 - 2
Post by: brightsky on January 14, 2010, 10:30:20 pm
Post by: the.watchman on January 14, 2010, 10:34:35 pm
(oh and i'm back :P)
13 pages!!! Wow!
Post by: xD_aQt on January 15, 2010, 09:58:17 am
Post by: the.watchman on January 15, 2010, 10:01:44 am
That is, never.
So the domain is
Post by: xD_aQt on January 15, 2010, 10:02:33 am
Post by: the.watchman on January 15, 2010, 10:04:03 am
so then would the domain be R ?
Yup. If you plug it into your calc, you can see that it is R
Post by: xD_aQt on January 15, 2010, 10:05:48 am
Post by: the.watchman on January 15, 2010, 10:12:20 am
For the first root:
That is, when
So its domain is
For the second root:
That is, when
So its domain is
If you think about when
The domain is
Post by: xD_aQt on January 15, 2010, 10:13:46 am
Post by: the.watchman on January 15, 2010, 10:15:08 am
I'm just wondering but shouldn't it include, not exclude?
Oops, my bad! :P
Post by: xD_aQt on January 15, 2010, 10:25:19 am
Post by: the.watchman on January 15, 2010, 10:32:51 am
1. Denominator = 0
2.
So find when
So the domain is
For the range:
As
As
As
As
So the range is
Post by: xD_aQt on January 15, 2010, 10:43:33 am
Post by: the.watchman on January 15, 2010, 10:50:13 am
Compare this to the endpoints:
When
As
Because
The range is
NOTE: You can do this question this way because the domain is less than a period.
Post by: xD_aQt on January 15, 2010, 10:53:42 am
Post by: xD_aQt on January 15, 2010, 11:04:09 am
Determine the domain and range of the following function y = |x|
Post by: the.watchman on January 15, 2010, 11:07:07 am
Would the domain be R ?
Of which function???
Post by: xD_aQt on January 15, 2010, 11:07:42 am
Post by: the.watchman on January 15, 2010, 11:09:53 am
So yes the domain is
Post by: GerrySly on January 15, 2010, 11:10:19 am
Would the domain be R ?Yes, only thing you need to remember about that function is the range is
Determine the domain and range of the following function y = |x|
Post by: xD_aQt on January 15, 2010, 11:11:43 am
What do you mean by R \ {0} ? The range should be R+ right ?Would the domain be R ?Yes, only thing you need to remember about that function is the range is
Determine the domain and range of the following function y = |x|and the domain of the derivative function is
Post by: xD_aQt on January 15, 2010, 11:19:39 am
The linear function f: D-> R, f(x) = 6 -2x has range [-4,12] The domain D is
Post by: the.watchman on January 15, 2010, 11:21:07 am
What do you mean by R \ {0} ? The range should be R+ right ?Would the domain be R ?Yes, only thing you need to remember about that function is the range is
Determine the domain and range of the following function y = |x|
He means that the derivative function has the implied domain R \ {0}.
Oh, and the range isn't
It's
Post by: xD_aQt on January 15, 2010, 11:23:25 am
Oh, how come its R+ U {0} ?What do you mean by R \ {0} ? The range should be R+ right ?Would the domain be R ?Yes, only thing you need to remember about that function is the range is
Determine the domain and range of the following function y = |x|
He means that the derivative function has the implied domain R \ {0}.
Oh, and the range isn't.
It's{0}
Post by: the.watchman on January 15, 2010, 11:23:43 am
So the domain is
Post by: the.watchman on January 15, 2010, 11:25:18 am
Oh, how come its R+ U {0} ?What do you mean by R \ {0} ? The range should be R+ right ?Would the domain be R ?Yes, only thing you need to remember about that function is the range is
Determine the domain and range of the following function y = |x|
He means that the derivative function has the implied domain R \ {0}.
Oh, and the range isn't
It's
Post by: xD_aQt on January 15, 2010, 11:28:55 am
Post by: xD_aQt on January 15, 2010, 11:34:57 am
y = ||x+1|-1|-1
Post by: the.watchman on January 15, 2010, 11:42:15 am
Range is
This is because the min. value of the 'big' modulus is 0, thus the min. value of the function is
Maybe this is a better way to think of it:
Take
Then think of
Then by moving everything down one unit, the range becomes
Post by: xD_aQt on January 15, 2010, 11:55:06 am
Post by: xD_aQt on January 15, 2010, 11:59:51 am
4 f (
Post by: the.watchman on January 15, 2010, 12:06:40 pm
Solve the following functional equations for f:
4 f () - f (x) =
Do you mean solve for f(x)?
Functional equations are probably my weakness, I'd better go do an exercise or two :P
What chapter are they in???
Post by: xD_aQt on January 15, 2010, 12:08:34 pm
What book are you using?
Post by: the.watchman on January 15, 2010, 12:11:01 pm
Yeah solve for f (x) :) Haha
What book are you using?
Essential, thanks!
Post by: xD_aQt on January 15, 2010, 12:12:23 pm
Post by: Ahmad on January 15, 2010, 12:16:26 pm
Post by: xD_aQt on January 15, 2010, 12:19:34 pm
Replace x by 1/x in the functional equation and solve the resulting simultaneous equation in f(x) and f(1/x).So I'll have two separate equations to solve from ?
Post by: Ahmad on January 15, 2010, 12:22:15 pm
Post by: xD_aQt on January 15, 2010, 02:20:25 pm
f(x+2)+f(x-1)=11-6x
Post by: brightsky on January 15, 2010, 03:23:35 pm
Ok, my screwed up answer, but it at least gives a solution:
Let
Hence
Hence,
Hence
Post by: the.watchman on January 15, 2010, 03:25:23 pm
Hang on, is the function necesarily linear??? It just says polynomial...
Post by: brightsky on January 15, 2010, 03:26:27 pm
Post by: brightsky on January 15, 2010, 03:27:23 pm
Post by: the.watchman on January 15, 2010, 03:28:49 pm
I've got a question completely irrelevant to this. Why doeshave real solutions?
Well, say when
Post by: cipherpol on January 15, 2010, 03:29:38 pm
I've got a question completely irrelevant to this. Why does
The
Post by: brightsky on January 15, 2010, 03:30:20 pm
I've got a question completely irrelevant to this. Why does
Well, say when, y is defined
Oh I see. Back to functional equations..:p
Post by: xD_aQt on January 15, 2010, 03:32:51 pm
Post by: xD_aQt on January 15, 2010, 03:34:12 pm
f(x-1) x f(x+2) = 4x2+16x+7
Post by: /0 on January 15, 2010, 03:41:42 pm
The degree of f must be 1, since if you multiply together 2 linear functions you will get a quadratic (
Let
Then do what brightsky did in his/her previous solution.
These problems are quite non-standard, and a bit beyond what's normally required in methods
Also, for
Giving that f is a polynomial in x, solve
f(x+2)+f(x-1)=11-6x
f must be a linear function.
If f is a quadratic, then
And the same for all higher powers.
Post by: brightsky on January 15, 2010, 03:44:22 pm
Turns out the same doesn't it?? :p
Post by: the.watchman on January 15, 2010, 03:45:39 pm
The degree of f must be 1, since if you multiply together 2 linear functions you will get a quadratic ()
Let
Then do what brightsky did in his previous solution.
By the way, where do you get these problems? I don't remember ever doing this in Methods
I understand where you're coming from, but what about the previous question?
It could have been two quadratics that cancel their
Post by: the.watchman on January 15, 2010, 03:46:49 pm
Let's try making
Turns out the same doesn't it?? :p
LOL, good luck doing that in an exam for each question :P
Post by: brightsky on January 15, 2010, 03:48:59 pm
Post by: the.watchman on January 15, 2010, 03:51:18 pm
Functional equations make my head feel dizzy :p
Same, that's why I'm not answering anymore, I'm editing my MM project instead :P
Post by: xD_aQt on January 15, 2010, 03:53:24 pm
Functional equations make my head feel dizzy :p
Lol, sorry for all the questions! :PFunctional equations make my head feel dizzy :p
Same, that's why I'm not answering anymore, I'm editing my MM project instead :P
Post by: /0 on January 15, 2010, 03:53:43 pm
The degree of f must be 1, since if you multiply together 2 linear functions you will get a quadratic (
Let
Then do what brightsky did in his previous solution.
By the way, where do you get these problems? I don't remember ever doing this in Methods
I understand where you're coming from, but what about the previous question?
It could have been two quadratics that cancel their, couldn't it?
If we pick for f to be of degree 'n', then when we multiply
Post by: xD_aQt on January 15, 2010, 03:54:44 pm
So, we're unlikely to find questions like these in the exam ?
The degree of f must be 1, since if you multiply together 2 linear functions you will get a quadratic (
Let
Then do what brightsky did in his/her previous solution.
These problems are quite non-standard, and a bit beyond what's normally required in methods
Also, forGiving that f is a polynomial in x, solve
f(x+2)+f(x-1)=11-6x
f must be a linear function.
If f is a quadratic, then
And the same for all higher powers.
Post by: /0 on January 15, 2010, 03:56:25 pm
I'm not saying it won't happen though. I'm just saying it hasn't happened... yet.
Post by: the.watchman on January 15, 2010, 03:57:03 pm
Lol, sorry for all the questions! :PFunctional equations make my head feel dizzy :p
Same, that's why I'm not answering anymore, I'm editing my MM project instead :P
I'm joking, it's more that I struggle with functionals than anything else.
I'm still around, to see if I can help with something I'm more capable of doing :P
Post by: xD_aQt on January 15, 2010, 03:57:16 pm
In all my ~50 methods practice exams (including past vcaa) I have never seen a question like this.Lol, gotcha! Thanks :)
I'm not saying it won't happen though. I'm just saying it hasn't happened... yet.
Post by: the.watchman on January 15, 2010, 03:58:33 pm
In all my ~50 methods practice exams (including past vcaa) I have never seen a question like this.
Nor have I (not that I've done many :P), the only functionals that I've seen are much simpler, like:
Prove that if
And the like.
Post by: xD_aQt on January 16, 2010, 09:58:22 am
f(x) = sin2(x) - 6sin(x) + 10
Post by: brightsky on January 16, 2010, 10:33:36 am
Since the range of
Hence the maximum and minumum of the function is 5 and 17.
Post by: the.watchman on January 16, 2010, 10:35:08 am
Let
When
When
So max & min are 17 & 5
(I know it's longer, but I prefer algebraic methods :P)
Post by: xD_aQt on January 16, 2010, 10:43:47 am
Mmm .. how do you go from the first one to the second one? :-\,
has no solutions
Post by: the.watchman on January 16, 2010, 10:46:07 am
Mmm .. how do you go from the first one to the second one? :-\
Post by: xD_aQt on January 16, 2010, 10:47:11 am
Oh okay :)Mmm .. how do you go from the first one to the second one? :-\is the general solution for
Post by: xD_aQt on January 16, 2010, 10:53:20 am
f(1-2x) = 8x2+2x
Post by: brightsky on January 16, 2010, 11:09:26 am
So
So
Post by: brightsky on January 16, 2010, 11:11:07 am
Let,
When,
,
When,
,
So max & min are 17 & 5
(I know it's longer, but I prefer algebraic methods :P)
Why do you sub
Post by: xD_aQt on January 16, 2010, 11:15:22 am
LetThanks! :)
Soand
So = 2x^2 -5x + 3 )
Post by: the.watchman on January 16, 2010, 11:29:11 am
Let
When
When
So max & min are 17 & 5
(I know it's longer, but I prefer algebraic methods :P)
Why do you suband
?
Consecutive integer values of n will alternate between maximum points and minimum points, so any values would have worked (0,1 are easier to work with)
Post by: xD_aQt on January 16, 2010, 11:31:30 am
Post by: the.watchman on January 16, 2010, 11:35:16 am
now you've sort of lost me .. could you explain why again ? :-\
The points where the gradient is equal to 0 are alternating between maxima and minima, so any consecutive integer values of n would yield two stationary points (one maximum and one minimum).
Post by: xD_aQt on January 16, 2010, 11:38:30 am
Oh right! Gotchanow you've sort of lost me .. could you explain why again ? :-\
The points where the gradient is equal to 0 are alternating between maxima and minima, so any consecutive integer values of n would yield two stationary points (one maximum and one minimum).
Post by: the.watchman on January 16, 2010, 11:39:19 am
LetThanks! :)
So
So
Yeh, nice! Now I know how to do these sorts of questions! :D
Post by: xD_aQt on January 17, 2010, 10:14:04 am
If y =
Ta,b
Post by: the.watchman on January 17, 2010, 10:25:39 am
So the translation is 1 unit up and 2 units right
Sorry, I don't know how to Latex a matrix :P
Post by: xD_aQt on January 17, 2010, 10:27:07 am
Post by: the.watchman on January 17, 2010, 10:28:00 am
Haha, Thanks! :)
No prob :)
Post by: xD_aQt on January 17, 2010, 10:29:17 am
f(x) =
Post by: the.watchman on January 17, 2010, 10:36:00 am
By grouping method,
It is a negative cubic, so it is less than 0 when
So the domain is (-
Post by: xD_aQt on January 17, 2010, 10:38:46 am
f (x) =
Post by: the.watchman on January 17, 2010, 10:43:49 am
So Domain is [1,6]
Post by: xD_aQt on January 17, 2010, 10:45:55 am
Post by: the.watchman on January 17, 2010, 10:46:55 am
Are the domains included or excluded?
For most root domains, the endpoint(s) are included as the expression inside the root can equal 0.
Post by: xD_aQt on January 17, 2010, 10:50:01 am
Post by: xD_aQt on January 17, 2010, 10:56:55 am
f: [
where a b c are postive constants.
Post by: the.watchman on January 17, 2010, 10:58:13 am
The maximum is
NOTE: If it didn't say positive constants, then put a modulus around
Post by: xD_aQt on January 17, 2010, 11:05:18 am
f (x) = 6 + 4 cos (x) - sin2 (x)
Post by: the.watchman on January 17, 2010, 11:12:00 am
Let
When
When
So max is 10, min is 2
Post by: xD_aQt on January 17, 2010, 11:15:12 am
Post by: the.watchman on January 17, 2010, 11:17:03 am
Is there another way without having to differentiate?
With something like that (more than one type of trig function), it is best to diff and work it out that way, rather than try to recognise values of x that give max/min.
Post by: xD_aQt on January 17, 2010, 11:24:57 am
f (x) = 5 - 2cosx, -
Post by: the.watchman on January 17, 2010, 11:29:36 am
Then differentiate to ensure no stationary points in the domain:
None of these occur during the domain (except the endpoint, but that doesn't matter)
So the range is [3,4)
Post by: xD_aQt on January 17, 2010, 11:33:33 am
f (x) =
Post by: the.watchman on January 17, 2010, 11:41:33 am
When
SO max is
Post by: xD_aQt on January 17, 2010, 11:44:24 am
f (x) = |x+3| - |x-4|
Post by: the.watchman on January 17, 2010, 11:47:23 am
Find the range
f (x) = |x+3| - |x-4|
I graphed it, it's R
Post by: xD_aQt on January 17, 2010, 11:48:24 am
Post by: the.watchman on January 17, 2010, 11:49:41 am
Oh, would graphing it be the only way to find the range?
Probs not, but I cbf doing it any other way :P
Post by: xD_aQt on January 17, 2010, 11:51:41 am
Thanks anyway :)Oh, would graphing it be the only way to find the range?
Probs not, but I cbf doing it any other way :P
Post by: xD_aQt on January 17, 2010, 12:09:30 pm
f (x) = 1 - 3 sin (2x) , 0 < x <
Post by: the.watchman on January 17, 2010, 12:14:07 pm
Compare this to the endpoints:
So the range is [-2,1)
Post by: the.watchman on January 17, 2010, 12:17:10 pm
Post by: xD_aQt on January 17, 2010, 12:18:12 pm
Post by: the.watchman on January 17, 2010, 12:26:00 pm
Haha .. sorry :) hope its not too much trouble - I struggle with Functions the most! Can't wait till its all over :P
No not at all, I basically skipped that chapter coz I don't like it, so it's good for me to do some :D
Post by: brightsky on January 17, 2010, 12:27:32 pm
Post by: the.watchman on January 17, 2010, 12:29:42 pm
Me <3 Functions. Me >:( Probability.
I hate both. I loooooove calculus!!! :P
YES! I now make up just over 1% of the MM board :P
Post by: brightsky on January 17, 2010, 12:35:05 pm
Post by: xD_aQt on January 17, 2010, 01:02:29 pm
Post by: xD_aQt on January 17, 2010, 01:09:42 pm
f (x) =
I got [-1,3) is that right?
f (x) =
Is it (-
Post by: brightsky on January 17, 2010, 01:17:36 pm
Split into cases:
1.
2.
Case 1:
Solving simultaneously:
Case 2:
No solutions.
So you are correct. :)
Post by: the.watchman on January 17, 2010, 01:18:29 pm
I'm not so sure about the second one:
So domain is [10-e^2,
Post by: xD_aQt on January 17, 2010, 01:22:21 pm
The first one is rightthe second one was 2 - log (small 10) (10-x) i forgot to put the 10 its not e
I'm not so sure about the second one:
So domain is [10-e^2,
Post by: brightsky on January 17, 2010, 01:24:36 pm
2.
But
So
Post by: xD_aQt on January 17, 2010, 01:33:15 pm
Post by: brightsky on January 17, 2010, 01:35:31 pm
so would the domain be (-90,10] ? .. or am I wrong?
Other way around. [-90, 10).
This is because x can equal to -90 as
But x cannot equal to 10 as
Post by: xD_aQt on January 17, 2010, 01:37:00 pm
Post by: xD_aQt on January 17, 2010, 01:44:43 pm
Find the domain
f (x) =
what do you do again when its a fraction?
Post by: the.watchman on January 17, 2010, 01:46:39 pm
So the domain is (-
Post by: xD_aQt on January 17, 2010, 01:52:14 pm
Post by: the.watchman on January 17, 2010, 01:59:30 pm
would (-
Yes, it is incorrect, because
Post by: xD_aQt on January 20, 2010, 03:04:45 pm
f (x) 5 - 2
Post by: xD_aQt on January 20, 2010, 03:05:17 pm
Post by: the.watchman on January 20, 2010, 03:12:33 pm
Min of
The min occurs when
So the min is
Post by: cipherpol on January 20, 2010, 03:13:52 pm
Well the max is when the root = 0, so 3
The min occurs whenis at its maximum (
So the min is
Shouldn't max be -3?
Post by: the.watchman on January 20, 2010, 03:17:27 pm
Post by: xD_aQt on January 20, 2010, 03:25:02 pm
Post by: the.watchman on January 20, 2010, 03:30:09 pm
If the function has a positive root (eg.
Post by: xD_aQt on January 20, 2010, 03:49:51 pm
y = -
Post by: the.watchman on January 20, 2010, 03:59:24 pm
The max is reached when
So the range is (-
Post by: xD_aQt on January 20, 2010, 04:41:56 pm
Post by: simonhu81292 on January 20, 2010, 04:42:47 pm
Post by: the.watchman on January 20, 2010, 04:44:38 pm
Post by: xD_aQt on January 20, 2010, 04:47:10 pm
Post by: simonhu81292 on January 20, 2010, 04:47:48 pm
Grrr...NOT AGAIN!!! >:(:D
Post by: the.watchman on January 20, 2010, 04:48:52 pm
how would you solve the equation on the ti nspire cas ?
Just sketch, or use fmin/fmax function, really handy!
Post by: xD_aQt on January 20, 2010, 04:50:57 pm
Post by: the.watchman on January 20, 2010, 04:53:56 pm
okay ive sketched the graph then where do i go from there ?
Go back to calc mode and type fmin (or fmax). The menu shortcut is menu,4,6 (or 7 for max)
Then inside the brackets, type in the function (or function name eg. f1(x)) then comma x)
Use the x -value given to find the min/max values
Post by: xD_aQt on January 20, 2010, 04:55:51 pm
Post by: the.watchman on January 20, 2010, 05:23:39 pm
Post by: xD_aQt on January 20, 2010, 05:26:54 pm
Post by: the.watchman on January 20, 2010, 05:27:43 pm
the essentials one ?
Yup, pg. 703 onwards is pretty handy
Post by: superflya on January 20, 2010, 05:37:19 pm
Post by: xD_aQt on January 20, 2010, 05:39:01 pm
Post by: the.watchman on January 20, 2010, 05:41:06 pm
Post by: xD_aQt on January 20, 2010, 05:47:30 pm
what does the second part mean ?
Post by: the.watchman on January 20, 2010, 05:50:31 pm
solve inequality . graph the solution on the number line< 1
what does the second part mean ?
Well the answer is
This can be drawn on a number line like so:
<------------o o----------------------------------o
______________________________________________________
-3 -2 -1 0 1 2 3
Post by: superflya on January 20, 2010, 05:51:15 pm
Post by: superflya on January 20, 2010, 05:51:49 pm
Post by: the.watchman on January 20, 2010, 05:53:48 pm
damn his soo fast, solved it and all :P
Lol, like the diagram? :P
Post by: superflya on January 20, 2010, 05:57:54 pm
Post by: xD_aQt on January 20, 2010, 05:59:43 pm
Post by: brightsky on January 20, 2010, 09:30:56 pm
The min is reached whenis at its minimum, which is
The max is reached when(max is therefore
)
So the range is (-
Hmmm...
Solutions:
Isn't this the range? *bangs confused head against desk*
Post by: superflya on January 20, 2010, 09:40:01 pm
Post by: brightsky on January 20, 2010, 10:01:22 pm
Post by: superflya on January 20, 2010, 10:15:08 pm
Post by: xD_aQt on January 23, 2010, 10:17:04 am
f (x) = 4 -
I got (-
Post by: the.watchman on January 23, 2010, 10:19:40 am
Post by: xD_aQt on January 23, 2010, 10:21:20 am
Post by: the.watchman on January 23, 2010, 11:14:08 am
what do you mean transformations occur after the dilations ?
DRT (dilation/reflection/translation)
You reflect in the x-axis first (asymp. y=0), then translate (asymp. y=4), so the range is (-
Sorry about the poor explanation :P
Post by: superflya on January 23, 2010, 11:16:57 am
horizontal asymptote at y=4.
Post by: xD_aQt on January 23, 2010, 11:32:53 am
f(x) = 2x3sin(x)
i'm not sure what the question is asking ... :-\
Post by: /0 on January 23, 2010, 11:51:47 am
So it's even
Post by: brightsky on January 23, 2010, 11:56:18 am
Odd functions, on the other hand, means that the graph remains unchanged after a rotation of 180 degrees about the point of origin. For instance,
Now, to your question.
Algebraically, an even function is given by
Since
Hence,
Post by: Studyinghard on January 23, 2010, 12:43:35 pm
find the range
f (x) = 4 -
I got (-
From the little knowledge that I have I think it would be (-
f (x) = -
Post by: superflya on January 23, 2010, 02:22:17 pm
Post by: the.watchman on January 23, 2010, 02:23:51 pm
^^ yea that's correct.
+1, remember DRT, very helpful
Post by: xD_aQt on January 23, 2010, 04:50:15 pm
y= 2x2 and y =
Post by: the.watchman on January 23, 2010, 04:52:00 pm
What dilation transforms the curve with equation y = x2 to that with equation
y= 2x2
It is a dilation factor of 2 from the x-axis
OR
REASON:
Let
ALSO
NEXT QUESTION:
It is dilation factor of
OR
Post by: Studyinghard on January 24, 2010, 09:12:05 pm
If
Thanks
PS: I bet the.watchman will be here first even though he is offline :P
Post by: xD_aQt on January 24, 2010, 09:23:19 pm
Post by: superflya on January 24, 2010, 09:24:11 pm
Post by: xD_aQt on January 24, 2010, 09:26:09 pm
Post by: brightsky on January 24, 2010, 09:32:45 pm
I'm assuming the question is as above.
From (2),
Sub into (1):
Substitute that into equation 1 to find A.
Post by: superflya on January 24, 2010, 09:33:54 pm
Post by: Studyinghard on January 24, 2010, 09:40:56 pm
Ifand y = 19.3 when t = 2, and y = 19.02 when t = 5, find the value of the constants A and k. Give answers correct to 2 decimal places.
aha yeah i fail at latex.
With that step you did
Shouldnt you add them there instead of subtracting?
and look like this:
Post by: brightsky on January 24, 2010, 09:44:15 pm
If
aha yeah i fail at latex.
With that step you did
Shouldnt you add them there instead of subtracting?
and look like this:
Nope as you're dividing.
It would make it easier if you let
Then
Post by: Studyinghard on January 24, 2010, 10:05:50 pm
If
aha yeah i fail at latex.
With that step you did
Shouldnt you add them there instead of subtracting?
and look like this:
Nope as you're dividing.
It would make it easier if you letand
Then
but aren't you multiplying, as in :
like when you do:
you times the a by all of them to get :
Post by: brightsky on January 24, 2010, 10:08:42 pm
And nope, I'm not multiplying through in that step. I'm using the laws of indices in that
Post by: superflya on January 24, 2010, 11:42:33 pm
Post by: m@tty on January 24, 2010, 11:46:12 pm
Post by: Studyinghard on January 25, 2010, 12:14:20 am
Post by: xD_aQt on January 25, 2010, 10:26:16 am
Find the rules and domains for
- f x g
Post by: fady_22 on January 25, 2010, 11:22:50 am
domain: intersection of both domains:
For f/g, the domain is the intersection of both domains, except where g(x)=0:
domain:
Post by: Studyinghard on January 25, 2010, 12:10:06 pm
Given f (x) =and g (x) =
Find the rules and domains for
- f x g
What chapter was that ? because i kid you not my tutor who just came to assess me on that same thing and see if im "good enough", came and gave me that question to do. like how the f*** am I supposed to already know that from yr 11?
Post by: superflya on January 25, 2010, 01:43:05 pm
for the above question
before u attempt to do anything with composite functions, u MUST check if there defined.
Post by: Blakhitman on January 25, 2010, 01:57:32 pm
Post by: m@tty on January 25, 2010, 02:20:53 pm
I thought composite was things like f(g(x))?That's right. This is not a composite function, it is a product function. A product function is only defined where BOTH or all of the constituent functions are, i.e. the intersection of the domains.
So
Post by: xD_aQt on January 25, 2010, 07:26:26 pm
Post by: TrueTears on January 25, 2010, 07:30:32 pm
Post by: kyzoo on January 25, 2010, 07:31:49 pm
You will see that the TP is at (0, 1)
Then sub x = -2 and x = 1 into the equation
The lowest point in the domain [-2, 1] is the TP (0, 1)
The highest point is (-2, 5)
Therefore the range is [1, 5]
Post by: brightsky on January 25, 2010, 07:35:32 pm
Find the range of the function represented by {(x,y): y = x2+1, x [-2,1]}
The function
The restricted domain is:
Hence, the range is
Post by: brightsky on January 25, 2010, 07:37:07 pm
Find the range of the function represented by {(x,y): y = x2+1, x [-2,1]}
This type of notation still confuses me. Can someone explain it to me? (Sorry for hijacking your question thread, xD_aQt.)
Post by: TrueTears on January 25, 2010, 07:39:23 pm
Post by: Studyinghard on January 25, 2010, 07:41:49 pm
Find the range of the function represented by {(x,y): y = x2+1, x [-2,1]}
This type of notation still confuses me. Can someone explain it to me? (Sorry for hijacking your question thread, xD_aQt.)
I think it is just. First you plot the graph of x2 + 1 and then sub in the domain for the x values.
So when x = - 2 what does y equal? and when x = 1 what does y equal? and stop the graph there and those will be inclusive.
Post by: brightsky on January 25, 2010, 07:52:04 pm
Find the range of the function represented by {(x,y): y = x2+1, x [-2,1]}
This type of notation still confuses me. Can someone explain it to me? (Sorry for hijacking your question thread, xD_aQt.)
I think it is just. First you plot the graph of x2 + 1 and then sub in the domain for the x values.
So when x = - 2 what does y equal? and when x = 1 what does y equal? and stop the graph there and those will be inclusive.
Yeah, I know that, but what does the notation mean part by part. I mean, I've seen stuff like:
Post by: Studyinghard on January 25, 2010, 07:55:46 pm
Find the range of the function represented by {(x,y): y = x2+1, x [-2,1]}
This type of notation still confuses me. Can someone explain it to me? (Sorry for hijacking your question thread, xD_aQt.)
I think it is just. First you plot the graph of x2 + 1 and then sub in the domain for the x values.
So when x = - 2 what does y equal? and when x = 1 what does y equal? and stop the graph there and those will be inclusive.
Yeah, I know that, but what does the notation mean part by part. I mean, I've seen stuff like:, but I've never seen notation like that...
Find the range of the function represented by {(x,y): y = x2+1, x [-2,1]}
Which part? because the (x,y) part just means function. Then there is the equation and then x [-2,1] is just the domain
Post by: brightsky on January 25, 2010, 08:00:24 pm
Shouldn't it be, either:
or
Post by: kyzoo on January 25, 2010, 08:00:57 pm
It means find the relation (set of ordered pairs - (x,y), or points on the Cartesian plane) such that
And like TT said, there is a mistake in the notation, there's meant to be an "is an element of" sign between the "x" and the "[-2, 1]"
Edit: I put the {...} brackets in the latex but it didn't show up, and I dunno the latex coding for the "is an element of" sign that should go between "x" and "[-2, 1]"
Post by: Studyinghard on January 25, 2010, 08:04:19 pm
Lol I use that notation all the time, especially in working in.: y = x^2+1, x [-2,1] } )
It means find the relation (set of ordered pairs - (x,y), or points on the Cartesian plane) such that, when the domain (or x-coordinates) are restricted to the interval [-2, 1].
And like TT said, there is a mistake in the notation, there's meant to be an "is an element of" sign between the "x" and the "[-2, 1]"
Edit: I put the {...} brackets in the latex but it didn't show up, and I dunno the "is an element of" latex that should go between "x" and "[-2, 1]"
probably a better explanation compared to mine :P
Post by: brightsky on January 25, 2010, 08:04:41 pm
Lol I use that notation all the time, especially in working in.
It means find the relation (set of ordered pairs - (x,y), or points on the Cartesian plane) such that
And like TT said, there is a mistake in the notation, there's meant to be an "is an element of" sign between the "x" and the "[-2, 1]"
Edit: I put the {...} brackets in the latex but it didn't show up.
Ah, yeah that was what I was looking for.
Two questions,
Is there any reason it's ordered like that?
Why do you need to put a curled bracket around it?
Post by: kyzoo on January 25, 2010, 08:06:32 pm
And I dunno why it's ordered like that =/
Post by: brightsky on January 25, 2010, 08:10:03 pm
The {...} bracket is to indicate "the set of"
And I dunno why it's ordered like that =/
I see, I see. Is it essential to know the correct ordering of things?
Post by: xD_aQt on January 25, 2010, 08:11:24 pm
Post by: m@tty on January 25, 2010, 08:12:54 pm
Edit: I put the {...} brackets in the latex but it didn't show up, and I dunno the latex coding for the "is an element of" sign that should go between "x" and "[-2, 1]"
Curly brackets are: \{ and \}
Element of is: \in
Post by: kyzoo on January 25, 2010, 08:15:48 pm
Post by: kyzoo on January 25, 2010, 08:16:42 pm
The {...} bracket is to indicate "the set of"
And I dunno why it's ordered like that =/
I see, I see. Is it essential to know the correct ordering of things?
Yea it would, I have never seen this notation written in another order.
Post by: xD_aQt on January 25, 2010, 08:17:40 pm
Post by: brightsky on January 25, 2010, 08:18:15 pm
The {...} bracket is to indicate "the set of"
And I dunno why it's ordered like that =/
I see, I see. Is it essential to know the correct ordering of things?
Yea it would, I have never seen this notation written in another order.
Cool, thanks. *Goes and tries to familiarise with it*. :p
Post by: brightsky on January 25, 2010, 08:18:38 pm
Is the range [1,5] or is that still wrong ?
Yeah that's right. :)
Post by: xD_aQt on January 25, 2010, 08:19:30 pm
Post by: xD_aQt on January 25, 2010, 08:31:51 pm
Express f (x) in the form f (x) =
Post by: brightsky on January 25, 2010, 08:41:22 pm
Let f : D --> R, f (x) =where D is the largest subset of R for which f is defined
Express f (x) in the form f (x) =+ B
Do you use Essentials? What chapter is it in?
Post by: m@tty on January 25, 2010, 08:45:36 pm
Let f : D --> R, f (x) =Easiest way to do these is 'synthetic division' though you can long divide.
Express f (x) in the form f (x) =
From there it is a familiar hyperbola.
Post by: Blakhitman on January 25, 2010, 08:48:05 pm
Post by: kyzoo on January 25, 2010, 08:48:48 pm
Post by: xD_aQt on January 25, 2010, 08:49:05 pm
>.< Never use long division.I agree ;D
Post by: m@tty on January 25, 2010, 08:51:52 pm
>.< Never use long division.I only properly figured how to do it ~2 weeks before the exams. ::)
Post by: xD_aQt on January 25, 2010, 08:52:40 pm
State D .. what would that be ? :-\Let f : D --> R, f (x) =Easiest way to do these is 'synthetic division' though you can long divide.
Express f (x) in the form f (x) =
From there it is a familiar hyperbola.
Post by: brightsky on January 25, 2010, 08:53:47 pm
>.< Never use long division.I only properly figured how to do it ~2 weeks before the exams. ::)
Can you even do it for this question? It turns out as
Post by: m@tty on January 25, 2010, 08:54:28 pm
Solve
Post by: brightsky on January 25, 2010, 08:54:42 pm
State D .. what would that be ? :-\Let f : D --> R, f (x) =Easiest way to do these is 'synthetic division' though you can long divide.
Express f (x) in the form f (x) =
From there it is a familiar hyperbola.
I think you just find the domain for the function. Which, as matty said would be: R\{-1}
Post by: xD_aQt on January 25, 2010, 08:57:03 pm
Post by: Blakhitman on January 25, 2010, 08:59:01 pm
>.< Never use long division.I only properly figured how to do it ~2 weeks before the exams. ::)
Can you even do it for this question? It turns out as, which is quite useless...
Long division? yea it works.
>.< Never use long division.
Why is that?
Post by: Studyinghard on January 25, 2010, 09:00:10 pm
>.< Never use long division.I only properly figured how to do it ~2 weeks before the exams. ::)
Can you even do it for this question? It turns out as
Long division? yea it works.>.< Never use long division.
Why is that?
cause its too long :P.
drumset - /budum-sh
Post by: m@tty on January 25, 2010, 09:00:33 pm
What do you mean?>.< Never use long division.I only properly figured how to do it ~2 weeks before the exams. ::)
Can you even do it for this question? It turns out as
Post by: brightsky on January 25, 2010, 09:11:01 pm
What do you mean?>.< Never use long division.I only properly figured how to do it ~2 weeks before the exams. ::)
Can you even do it for this question? It turns out as
Maybe I'm just doing it incorrectly. :p
This is how I did it:
1
------------------
x + 1 | x + 0
x + 1
_______________
0 - 1
EDIT:
Post by: m@tty on January 25, 2010, 09:15:54 pm
For long division, you should have:What do you mean?>.< Never use long division.I only properly figured how to do it ~2 weeks before the exams. ::)
Can you even do it for this question? It turns out as
Maybe I'm just doing it incorrectly. :p
Long division is very hard to communicate online.
EDIT: You did pretty well showing it :P Though it is wrong.
Post by: brightsky on January 25, 2010, 09:20:41 pm
Post by: m@tty on January 25, 2010, 09:23:06 pm
Should be:
1
-----------
x+1| x+0
| x-1
| -----
| 0-1
|_________
-1
Top number is how many full times the numerator goes into the denominator, the bottom number is the remainder.
Post by: xD_aQt on January 25, 2010, 09:41:48 pm
Which functions satisfies the above functional equation ?
a) f (x) = x4
b) f (x) = 2x
c) f (x) =
d) f (x) = 5
Post by: kyzoo on January 25, 2010, 09:45:15 pm
cause its too long :P.
Not just that, but because there's a much better, faster, "space-efficient" method.
Post by: superflya on January 25, 2010, 09:46:46 pm
Post by: Blakhitman on January 25, 2010, 09:50:14 pm
Post by: m@tty on January 25, 2010, 09:52:21 pm
Consider f (xy) = f(x) f(y)I don't know how you do it in CAS, but, what I would do is just see if they're equivalent.
Which functions satisfies the above functional equation ?
a) f (x) = x4
b) f (x) = 2x
c) f (x) =
d) f (x) = 5
Long division is, as kyzoo said, too time consuming, annoying, and takes up too much space. Though there are times where it is necessary.
Post by: kyzoo on January 25, 2010, 09:53:39 pm
Post by: brightsky on January 25, 2010, 09:53:58 pm
Consider f (xy) = f(x) f(y)
Which functions satisfies the above functional equation ?
a) f (x) = x4
b) f (x) = 2x
c) f (x) =
d) f (x) = 5
a)
Keep on doing that for the rest.
Post by: xD_aQt on January 25, 2010, 09:54:58 pm
OMGOSH .. Thanks heaps! I tripped out for a sec --> made it look harder than it is :)Consider f (xy) = f(x) f(y)I don't know how you do it in CAS, but, what I would do is just see if they're equivalent.
Which functions satisfies the above functional equation ?
a) f (x) = x4
b) f (x) = 2x
c) f (x) =
d) f (x) = 5so a) satisfies the functional equation.
so b) does not satisfy the requirement.
so c) does satisfy the functional equation.
therefore d) does not satisfy the requirement.
Long division is, as kyzoo said, too time consuming, too annonying and takes up too much space. Though there are times where it is necessary.
Post by: m@tty on January 25, 2010, 09:55:21 pm
0.o I have never found a situation where it is necessary.I was thinking big equations, like cubics and quartics, in both the numerator and denominator.
Post by: xD_aQt on January 25, 2010, 09:55:30 pm
Cheers :)Consider f (xy) = f(x) f(y)
Which functions satisfies the above functional equation ?
a) f (x) = x4
b) f (x) = 2x
c) f (x) =
d) f (x) = 5
a)Hence a) satisfies the functional equation.
Keep on doing that for the rest.
Post by: Blakhitman on January 25, 2010, 09:55:39 pm
0.o I have never found a situation where it is necessary.
What about Factorising a cubic equation? Don't tell me there's an easier way!
Post by: superflya on January 25, 2010, 09:55:47 pm
0.o I have never found a situation where it is necessary.
probably on gma exams/tests in yr 11 where they specifically tell u to use it :P
Post by: superflya on January 25, 2010, 09:56:46 pm
0.o I have never found a situation where it is necessary.
What about Factorising a cubic equation? Don't tell me there's an easier way!
haha, synthetic? first u gotta find a linear factor though.
Post by: xD_aQt on January 25, 2010, 09:58:38 pm
Post by: Blakhitman on January 25, 2010, 10:00:45 pm
0.o I have never found a situation where it is necessary.
What about Factorising a cubic equation? Don't tell me there's an easier way!
haha, synthetic? first u gotta find a linear factor though.
I'm gonna have to look up this synthetic stuff.
Post by: brightsky on January 25, 2010, 10:02:16 pm
Hate trial and error!
Yeah, you can use the rational roots test, but that requires work as well! :p
Post by: kyzoo on January 25, 2010, 10:04:52 pm
Can't be bothered to write it out in Latex, just attached image
Post by: superflya on January 25, 2010, 10:05:10 pm
explains it pretty well.
Post by: m@tty on January 25, 2010, 10:07:35 pm
There is a root where
You could go long dividing now, or, you could divide synthetically.
Synthetic division just makes use of the above fact and now you find which numbers work for
Logically, a must be 1, b must be 3 and c must be 1.
There, you have just divided
After doing this all you have to do is factorise the quadratic, which should be easy work.
Post by: Blakhitman on January 25, 2010, 10:12:44 pm
Divideby
Can't be bothered to write it out in Latex, just attached image
Wow, long dividing that would take aaaaages, thanks man, real useful!
Post by: xD_aQt on January 26, 2010, 02:14:23 pm
I sketched the graph onto my calculator and I found that the only coordinates of the points of intersection was at (0,0) .. is that right ?Let f : D --> R, f (x) =Easiest way to do these is 'synthetic division' though you can long divide.
Express f (x) in the form f (x) =
From there it is a familiar hyperbola.
And I forgot but what is an asymptotes ? Is there an asymptote in that equation ?
Post by: brightsky on January 26, 2010, 02:30:59 pm
An asymptote is where a graph y-value (horizontal asymptote) or x-value (vertical asymptote) doesn't exist in the graph.
For this, it is clear that
Now, there is also a y-value that cannot exist, and hence there is also a horizontal asymptote. It is clear that because 1 is a numerator for the fraction in
Sketching that would give you the conventional hyperbola.
Post by: xD_aQt on January 26, 2010, 02:32:59 pm
Umm .. the question said
clearly mark the coordinates of the points of intersection with the axes
and there was only one point which is at (0,0)
Post by: Blakhitman on January 26, 2010, 02:35:19 pm
Post by: brightsky on January 26, 2010, 02:37:45 pm
Oh okay!
Umm .. the question said
clearly mark the coordinates of the points of intersection with the axes
and there was only one point which is at (0,0)
Oh right, with the axis.
So to find the y-intercepts (or the point of intersection with
So there's an intercept at
To find x-intercepts, let y = 0.
Hence, intercept at
So the only point of intersection with the axes is (0,0).
Post by: xD_aQt on January 26, 2010, 02:38:48 pm
Post by: xD_aQt on January 26, 2010, 02:40:10 pm
f (x) - 3 f (
Post by: brightsky on January 26, 2010, 02:58:14 pm
Let
So
Post by: /0 on January 26, 2010, 03:09:11 pm
Let
Now, 'a' is really just a pronumeral. You could replace it with any other letter and the above equation would still be true. So let's replace 'a' with 'x'
So
And you can solve (1) and (2) for 'f(x)' and 'f(1/x)'
Post by: xD_aQt on January 26, 2010, 03:59:34 pm
The numerator is meant to be -1 not xOh okay!
Umm .. the question said
clearly mark the coordinates of the points of intersection with the axes
and there was only one point which is at (0,0)
Oh right, with the axis.
So to find the y-intercepts (or the point of intersection with(i.e. the y axis)) let
.
So there's an intercept at
To find x-intercepts, let y = 0.
Hence, intercept at.
So the only point of intersection with the axes is (0,0).
Post by: Blakhitman on January 26, 2010, 04:03:22 pm
Post by: Blakhitman on January 26, 2010, 04:11:26 pm
y-int:
x-int:
Post by: brightsky on January 26, 2010, 04:32:11 pm
Yea must've been typo, but he also left out the +1 so it'd still give you x=0.
No, I worked with
Post by: xD_aQt on January 26, 2010, 04:36:13 pm
Post by: Blakhitman on January 26, 2010, 04:41:39 pm
Yea must've been typo, but he also left out the +1 so it'd still give you x=0.
No, I worked withwhich is the same as
but much easier to work with.
ah fair enough, explaining that would've helped :p
Post by: xD_aQt on January 26, 2010, 05:03:05 pm
f (x+1) + f (x-1) = - 8x2 + 4x - 18
Post by: brightsky on January 26, 2010, 05:13:48 pm
Let
Hence,
P.S. Tell me if I've interpreted the question wrongly. :)
Post by: xD_aQt on January 26, 2010, 08:34:22 pm
a) f (x) = cos2x + 8cosx + 19
b) f (x) = cos2x + 8sinx + 19
Post by: /0 on January 26, 2010, 08:36:19 pm
b) Convert to 'sin's and repeat
Post by: brightsky on January 26, 2010, 08:43:10 pm
Find the range
a) f (x) = cos2x + 8cosx + 19
b) f (x) = cos2x + 8sinx + 19
a)
Hence,
b) Same range as
Post by: xD_aQt on January 27, 2010, 10:53:34 am
f (x) =
f (x) =
Post by: m@tty on January 27, 2010, 11:05:39 am
determine the domain and range
f (x) =
f (x) =
The range for both will be
But the domain is restricted so you have to work it out after with the actual domain.
The domain for both will be where the denominator is defined and is not equal to zero.
The square root portion is defined where:
Therefore the domain is:
Post by: xD_aQt on January 27, 2010, 11:20:26 am
Post by: m@tty on January 27, 2010, 11:21:55 am
Post by: xD_aQt on January 27, 2010, 11:26:34 am
so is the domain (-
Post by: superflya on January 27, 2010, 11:41:42 am
Post by: xD_aQt on January 27, 2010, 11:42:33 am
Post by: m@tty on January 27, 2010, 11:43:08 am
From there you can see it is a squareroot function trailing to the left from the point where
Therefore
Post by: superflya on January 27, 2010, 11:45:52 am
That
Post by: xD_aQt on January 27, 2010, 11:52:10 am
x <
Post by: superflya on January 27, 2010, 11:54:40 am
Post by: m@tty on January 27, 2010, 11:56:55 am
For the square root to be defined:
Where is this quadratic greater than zero? It is a basic parabola moved up four units... ;)
Now make sure not to include x-intercepts(if you find any). And there you have you domain.
Post by: xD_aQt on January 27, 2010, 11:58:44 am
Post by: superflya on January 27, 2010, 11:58:51 am
Post by: m@tty on January 27, 2010, 11:59:07 am
^ so then the domain ends up being R like superflya said ?Yes.
Post by: m@tty on January 27, 2010, 12:00:38 pm
there wont be any x-ints as its moved up 4 units :PNot in
Post by: xD_aQt on January 27, 2010, 12:02:15 pm
Post by: xD_aQt on January 27, 2010, 12:07:30 pm
find f(a - 1) in terms of a
I ended up with f (a - 1) = 1 +
is that wrong ? is the question asking that ?
Post by: m@tty on January 27, 2010, 12:15:19 pm
f (x) = 1 +, x is equal/greater than 3
find f(a - 1) in terms of a
I ended up with f (a - 1) = 1 +
is that wrong ? is the question asking that ?
So you are right, just remember the restriction, though I have no idea what the question is asking.
Post by: xD_aQt on January 27, 2010, 12:16:28 pm
Post by: superflya on January 27, 2010, 12:45:45 pm
Post by: S2paramore on January 27, 2010, 01:23:03 pm
An awesome thread btw. Thanks guys!
Post by: superflya on January 27, 2010, 02:10:33 pm
Post by: xD_aQt on January 27, 2010, 02:18:57 pm
This one should be quite straight-forward, given that I've interpreted the question correctly:Your right! +1
Let
Hence,
P.S. Tell me if I've interpreted the question wrongly. :)
Post by: S2paramore on January 27, 2010, 03:26:30 pm
I have seen some very similar ones in the essentials book. Which text are u using?I'm using Essentials too. Haha maybe I just haven't done enough questions. :P
Post by: superflya on January 27, 2010, 03:43:00 pm
I have seen some very similar ones in the essentials book. Which text are u using?I'm using Essentials too. Haha maybe I just haven't done enough questions. :P
this stuff shood be covered in the first chapter if i recall correctly :P
Post by: xD_aQt on January 27, 2010, 04:03:20 pm
Dh,k
Post by: Studyinghard on January 27, 2010, 04:07:50 pm
Post by: xD_aQt on January 27, 2010, 04:09:26 pm
h = 2 and k = 3 :-\
Post by: m@tty on January 27, 2010, 04:47:26 pm
Or dilation of factor
This can be seen by substituting
So the answer you gave would be right, assuming I am interpreting the notation correctly.
Post by: superflya on January 27, 2010, 04:51:42 pm
Dilation of factor of: 2 in x-direction, 3 in y-direction.
Or dilation of factorin y-direction.
This can be seen by substitutingand
So the answer you gave would be right, assuming I am interpreting the notation correctly.
yea that does work
Post by: xD_aQt on January 27, 2010, 04:53:06 pm
Dilation of factor of: 2 in x-direction, 3 in y-direction.Too good! :) +1
Or dilation of factor
This can be seen by substituting
So the answer you gave would be right, assuming I am interpreting the notation correctly.
Post by: xD_aQt on January 27, 2010, 05:36:03 pm
Dh,k
Post by: m@tty on January 27, 2010, 05:40:37 pm
Second one:
Post by: kyzoo on January 27, 2010, 05:42:33 pm
What is this
Post by: m@tty on January 27, 2010, 05:54:36 pm
I assume, though, that it is:
Post by: QuantumJG on January 27, 2010, 05:59:55 pm
What is thisnotation?
No idea. Is it some CAS notation?
Post by: superflya on January 27, 2010, 06:10:26 pm
What is this
No idea. Is it some CAS notation?
havnt seen it in the essentials book :s
Post by: xD_aQt on January 27, 2010, 08:12:24 pm
Post by: brightsky on January 27, 2010, 08:37:48 pm
So the fourth term is:
Tell me if this is wrong. :)
Post by: superflya on January 27, 2010, 08:40:58 pm
The "4th term" is pretty ambiguous...
So the fourth term is:
Tell me if this is wrong. :)
tis correct :)
Post by: xD_aQt on January 27, 2010, 08:41:42 pm
Post by: brightsky on January 27, 2010, 08:49:42 pm
The answer says 540x3y3 but I'll have a look at your working out first!
Edited. Looks like I can't tell the difference between
Post by: superflya on January 27, 2010, 08:55:55 pm
The answer says 540x3y3 but I'll have a look at your working out first!
Edited. Looks like I can't tell the difference betweenand
. :p
ur not alone brightsky :buck2:
Post by: xD_aQt on February 14, 2010, 05:59:23 pm
What would the list of transformation be .. I get a little confused when it comes to the order and where the dilations take place :-\
Post by: kyzoo on February 14, 2010, 06:03:25 pm
Dilation by factor 0.5 fromt the y-axis
Translation 2 units to the right
Post by: superflya on February 14, 2010, 06:05:11 pm
Post by: kyzoo on February 14, 2010, 06:06:47 pm
To find transformation parallel to the x-axis
Let 2x - 4 = 0
2x = 4
x = 2
Thus translation of 2 units in the positive x-direction
Translation of 2 units to the right.
Post by: m@tty on February 14, 2010, 06:07:48 pm
Therefore, dilation by factor of 8 from the x-axis and translation of 2 units right.
Post by: xD_aQt on February 14, 2010, 06:21:53 pm
dilations of factor _ parallel to the y-axis/x-axis
dilations from x-axis/y-axis by a factor of _
and so on .. but my point is that, how do I know what dilates from which axis ?
Post by: kyzoo on February 14, 2010, 06:28:18 pm
From the y-axis/parallel to the x-axis = horizontal dilation
y = a(bx - c)2 + d
This is the graph of y = x2
Dilated by factor "a" from the x-axis (vertical)
Dilated by factor "1/b" from the y-axis (horizontal)
Translated by "d" units up
Translated by {x:bx-c=0} => "c/b" units to the right.
Post by: xD_aQt on February 14, 2010, 06:31:24 pm
Post by: xD_aQt on February 14, 2010, 07:00:11 pm
i know that there will be two dilations but which comes first .. from the x or the y axis ?
Post by: m@tty on February 14, 2010, 07:04:13 pm
So you can do either order.
Post by: xD_aQt on February 14, 2010, 07:07:17 pm
a dilation from the x axis by a factor of 3 and
a dilation from the y axis by a factor of 2 ?
Post by: m@tty on February 14, 2010, 07:12:46 pm
Post by: xD_aQt on February 14, 2010, 07:16:03 pm
Post by: the.watchman on February 14, 2010, 11:12:31 pm
Each different transformation WITHIN THE SAME GROUP can be done in any order
eg. translations can be done in any order, but must collectively be done last
Post by: xD_aQt on February 20, 2010, 08:38:42 pm
f : [ -2, 2 ] --> R, f (x) = 2x + 1
g: (
Find the domain of f(g(x)) and g(f(x))
Post by: TrueTears on February 20, 2010, 08:40:03 pm
make sure ran g subset dom f
same thing for g o f
Post by: superflya on February 20, 2010, 08:41:27 pm
domain f(g(x)) = domain g
edit: beat :P
Post by: TrueTears on February 20, 2010, 08:42:05 pm
Post by: the.watchman on February 21, 2010, 08:11:38 am
First check if the composites are defined,
Because Ran g = [0,\infty) is not a subset of Dom f
Therefore f[g(x)] is not defined
Because Ran f = [-3,5] is not a subset of Dom g
Therefore g[f(x)] is not defined
Post by: xD_aQt on February 24, 2010, 07:40:17 pm
g (x) = cos (x)
state the transformations required to change f (x) to g (x)
i've thought about the question and i wonder is it even possible :o
Post by: TrueTears on February 24, 2010, 07:42:11 pm
Post by: xD_aQt on February 24, 2010, 07:45:18 pm
Post by: TrueTears on February 24, 2010, 07:47:43 pm
Post by: superflya on February 24, 2010, 11:17:27 pm
the transformations should be easier to work with.
Post by: xD_aQt on February 26, 2010, 04:12:23 pm
y =
Ta,b
Post by: the.watchman on February 26, 2010, 04:27:23 pm
Therefore transformations occurred are:
1) Translation 2 units right
2) Translation 1 unit up
Hope this is what you want!
Post by: xD_aQt on February 26, 2010, 04:33:01 pm
Post by: xD_aQt on February 26, 2010, 04:38:01 pm
Ta,b
Post by: the.watchman on February 26, 2010, 04:41:22 pm
Therefore the transformations are:
1) Translation 4 units right
2) Translation -3 units down
Post by: xD_aQt on February 26, 2010, 04:43:52 pm
y =
Ta,b
Post by: the.watchman on February 26, 2010, 04:45:20 pm
Post by: xD_aQt on February 26, 2010, 04:46:03 pm
Post by: the.watchman on February 26, 2010, 04:49:05 pm
Therefore transformations are:
1) translation 3 units right
2) translation 4 units down
EDIT: No more from me, I'm going out soon! :)
Post by: xD_aQt on February 26, 2010, 04:51:18 pm
Post by: the.watchman on February 26, 2010, 10:50:19 pm
Cheers buddy :)
No prob, I like these ones ;)
Post by: TrueTears on February 27, 2010, 09:49:43 am
cubic formula0.o I have never found a situation where it is necessary.
What about Factorising a cubic equation? Don't tell me there's an easier way!
lol jk
Post by: xD_aQt on February 28, 2010, 03:42:46 pm
- dilations by a factor of 2 from y-axis
- dilations by a factor of 3 from x-axis
- translation of 4 units parallel to the x-axis (positive direction)
- reflection in the x-axis
I got y = -3e2(x-2) is that right ?
Post by: the.watchman on February 28, 2010, 03:48:13 pm
For dilations from the y-axis, the image is the original multiplied by the reciprocal of the factor:
Post by: TrueTears on February 28, 2010, 03:50:59 pm
y = exhi there,
- dilations by a factor of 2 from y-axis
- dilations by a factor of 3 from x-axis
- translation of 4 units parallel to the x-axis (positive direction)
- reflection in the x-axis
I got y = -3e2(x-2) is that right ?
(x,y) -> (2x,y) -> (2x,3y) -> (2x+4,3y) -> (2x+4, -3y)
Post by: xD_aQt on February 28, 2010, 03:51:53 pm
Post by: the.watchman on February 28, 2010, 03:54:30 pm
so y = -3e0.5x+4 ?
No, I think it is
Post by: TrueTears on February 28, 2010, 03:59:17 pm
solve for x and y, then sub back in ;)
Post by: the.watchman on February 28, 2010, 04:00:29 pm
well id let y' = -3y and x' = 2x+4
solve for x and y, then sub back in ;)
Absolutely (although I would just sub
Good on you TT!
Post by: TrueTears on February 28, 2010, 04:03:40 pm
Post by: xD_aQt on February 28, 2010, 04:06:18 pm
Post by: xD_aQt on February 28, 2010, 04:08:05 pm
Post by: the.watchman on February 28, 2010, 04:09:34 pm
The new graph has a vertical asymptote at y=0, but is reflected in the x-axis, so the new range is
Post by: TrueTears on February 28, 2010, 04:09:40 pm
Would the range just be R ?for which
Post by: xD_aQt on February 28, 2010, 04:11:26 pm
Would the range just be R ?for which
y = ex
- dilations by a factor of 2 from y-axis
- dilations by a factor of 3 from x-axis
- translation of 4 units parallel to the x-axis (positive direction)
- reflection in the x-axis
Post by: TrueTears on February 28, 2010, 04:12:29 pm
Post by: the.watchman on February 28, 2010, 04:13:53 pm
yea watchmen is probs right
Thanks lol, it's singular (not related to the movie ;))
Post by: TrueTears on February 28, 2010, 04:15:20 pm
Post by: xD_aQt on March 26, 2010, 05:07:59 pm
do you just swap the x and y around to find the inverse?
Post by: Blakhitman on March 26, 2010, 05:09:49 pm
Post by: xD_aQt on March 26, 2010, 05:11:29 pm
Post by: m@tty on March 26, 2010, 05:15:07 pm
yea
Post by: Blakhitman on March 26, 2010, 05:15:38 pm
Post by: m@tty on March 26, 2010, 05:16:53 pm
Post by: Blakhitman on March 26, 2010, 05:18:34 pm
Swapping x and y is reflecting in the line.
lol, I meant graphically sketch it and reflect it,
Post by: m@tty on March 26, 2010, 05:19:15 pm
Post by: xD_aQt on March 26, 2010, 05:20:50 pm
is the inverse of the domain the range of the function and vice versa?
Post by: m@tty on March 26, 2010, 05:21:44 pm
yea
:P
You can see that as
and that has a domain of
Post by: Blakhitman on March 26, 2010, 05:22:33 pm
Post by: Blakhitman on March 26, 2010, 05:24:32 pm
Post by: xD_aQt on March 26, 2010, 05:31:01 pm
Post by: m@tty on March 26, 2010, 05:36:13 pm
Line of symmetry is at
Post by: xD_aQt on March 26, 2010, 05:50:44 pm
Post by: xD_aQt on April 02, 2010, 03:41:06 pm
Find the inverse function
f(x) =
Given that -8
Post by: /0 on April 02, 2010, 04:01:52 pm
Set
Post by: xD_aQt on April 02, 2010, 04:26:52 pm
Post by: /0 on April 02, 2010, 04:42:48 pm
Essentially what we want to do is factor
We can see that
So we know
We can set
Post by: xD_aQt on April 02, 2010, 04:45:05 pm
Post by: xD_aQt on April 02, 2010, 04:48:53 pm
Find the equation of the tangent to the curve y = f -1 (x) at the point (4,1)
Post by: stonecold on April 02, 2010, 04:54:23 pm
Post by: stonecold on April 02, 2010, 04:58:26 pm
thats not the answer by any chance?
Post by: superflya on April 02, 2010, 05:02:37 pm
Post by: /0 on April 02, 2010, 05:03:11 pm
EDIT: gj stonecold, looks right to me
Post by: qshyrn on April 02, 2010, 05:03:35 pm
Given f (x) = x3 + x2 + x + 1 and f -1 (x) is the inverse function of y = f (x).from point (4,1) on inverse we know that the original eq passes (1,4) find the equation of the tangent line to f(x) at (1,4) then you reflect that in line y=x (finding the inverse of that)
Find the equation of the tangent to the curve y = f -1 (x) at the point (4,1)
edit:beaten by /0
Post by: stonecold on April 02, 2010, 05:04:27 pm
Post by: xD_aQt on April 02, 2010, 05:13:34 pm
Post by: xD_aQt on July 01, 2010, 04:02:47 pm
Post by: Whatlol on July 01, 2010, 04:18:59 pm
For the graph whose equation is y = - 4 cos () + 10, find the equation of the tangent to the graph at the point where x = 4.
Find the derivative function and find the gradient when x =4 by subbing x=4 into gradient function.
Then use y=mx+c , since you know x=4 and you have found gradient all you need to do is find the y value(for x=4) by subbing x=4 into original function and then simply solving for c.
then just write out the equation.
Post by: brightsky on July 01, 2010, 04:22:28 pm
When
So the equation of the tangent is:
When
Substituting that back into (1):
Hence the equation of the tangent is:
EDIT: Thanks Blakhitman! :p
Post by: Whatlol on July 01, 2010, 04:27:30 pm
The gradient of the graph is given by:
When, the gradient is:
So the equation of the tangent is:where C is a constant....(1)
When
Substituting that back into (1):
Hence the equation of the tangent is:
Thats what i said :p but i dont know how to enter fractions etc.. )=
Post by: xD_aQt on July 01, 2010, 04:28:13 pm
Post by: brightsky on July 01, 2010, 04:30:12 pm
For fractions, type in the code:
Code: [Select]
[tex] \frac{m}{n} [/tex]
It would look like this:
Post by: Blakhitman on July 01, 2010, 04:32:11 pm
The gradient of the graph is given by:
You forgot the
Post by: Whatlol on July 01, 2010, 04:33:09 pm
Whatlol, search up a Latex Guide on Google (should come up with heaps). There's one on VN but I forgot where it is...
For fractions, type in the code:Code: [Select][tex] \frac{m}{n} [/tex]
It would look like this:.
Hmm ok ill have to check it out. ill get it eventually. thanks
Post by: brightsky on July 01, 2010, 04:36:51 pm
The gradient of the graph is given by:
You forgot thein
in your post. :P
Woops! :p
Post by: 99.95 on July 05, 2010, 12:06:43 pm
loge X+1 / X-1
Post by: /0 on July 05, 2010, 12:17:13 pm
Post by: Stroodle on July 05, 2010, 04:37:56 pm
Quote
Thats what i said :p but i dont know how to enter fractions etc.. )=
You can also use this site, then copy and paste the code over:
http://www.codecogs.com/latex/eqneditor.php
Post by: Whatlol on July 05, 2010, 04:57:02 pm
thats awesome, thanks.
Post by: xD_aQt on July 21, 2010, 05:00:01 pm
Post by: 98.40_for_sure on July 21, 2010, 05:25:03 pm
At x=2, y=2(4)+1=9
When x=2, dy/dx=8
9=16+c
c=-7
Now find the normal to the equation
At x=3, y=1
When x=3, dy/dx=1/2
1=-6+c
c=7
y=-2x+7
Let
-2x+7=8x-7
x=
Sub into any y equation
When x=
y=
Sorry if it's bad, first time using LaTeX :)
Post by: xD_aQt on July 22, 2010, 05:59:31 pm
a) find average rate of change of volume over the first 10 minutes
b) find the instantaneous rate of change of volume when t = 10
I'm not too sure how to start, do I differentiation?
Post by: Blakhitman on July 22, 2010, 06:07:51 pm
b) Differentiate and sub in t=10 into the derivative.
Post by: xD_aQt on July 22, 2010, 07:12:19 pm
Post by: 98.40_for_sure on July 22, 2010, 07:18:12 pm
for a) I got -25.1 and b) I got -25.2 ... is that right :S
Yep, exactly what i got
Post by: xD_aQt on July 22, 2010, 07:19:09 pm
Post by: xD_aQt on July 25, 2010, 10:28:01 pm
safe to swim after 6 am only if speed of current less than 3 km/hr ... find the times during which swimming is safe
Post by: Martoman on July 26, 2010, 12:38:33 am
Find when the v = 3 over [6,24]
By using the graph interpret. We want when the cos graph is lower than the 3. There will be multiple times.
Post by: xD_aQt on July 29, 2010, 04:53:19 pm
Q) According to this model, the platform is exactly 6 metres above the ground for the first time about 58 seconds into the ride. Find this time correct to two decimal places of a second.
The answer is 58.03 seconds but I'm not sure how you get to it.
Post by: Yitzi_K on July 29, 2010, 05:55:48 pm
Post by: xD_aQt on July 29, 2010, 06:05:01 pm
Post by: Yitzi_K on July 29, 2010, 06:18:15 pm
Post by: xD_aQt on July 29, 2010, 06:43:12 pm