ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Mr. Study on February 05, 2012, 05:41:21 pm

Title: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 05, 2012, 05:41:21 pm: Hey,

Thought I would start this topic, so I can tackle my problems with Methods!

Essential Math Methods

Question 1: f(x)=3x^2+x-2. Find {x:f(x)>0}
Could someone explain what this means: {x:f(x)>0} and then show me how to solve it?

Question 2: (http://i1054.photobucket.com/albums/s481/BondJames5/MM-Q1.png)
I can transpose other log equations to find x but this one has me stumped, as it's not the usual equation for me. (If that makes sense. :-[)

Question 3: (http://i1054.photobucket.com/albums/s481/BondJames5/MM-q2.png)

Question 4: (http://i1054.photobucket.com/albums/s481/BondJames5/MM-q3.png)
Its suppose to be f circle g and not the word fog. I have never seen the circle symbol used before in math. xD

Question 5: (http://i1054.photobucket.com/albums/s481/BondJames5/MM-q4.png)

If someone explains Question 1 dont worry about Question 5.

Thanks!!
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Greatness on February 05, 2012, 05:52:35 pm: 1) It means find the values of x for which the function is greater than 0. There are probably a few ways to do this, you can tell its a positive parabola so the values of x which are positive are the left side of the left x intercept and the right side of the right intercept - if that makes sense to you. lol
2) x= 10^2 because log b (a) = p so a=b^p
3) skip lol
4) i think it's a composite function... lol really cant remember
5) its asking for when the derivative is greater than 0
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 05, 2012, 05:59:02 pm: 1. Ahh!! I get it! I was thinking of some other things and NOT of it being positive.
2. Ah, I was thinking of doing it like that but I wasn't too sure it was the correct way of showing working out.

3. xD

4. Out of curiousity, What is a composite function?
5. So would I do f'(x)=6(x-1).
Then let f'(x) = 0
So it looks like 0 < 6(x-1)
0 < x-1
1 < x
Am I on the right track?? I didn't do enough of these types of problems last year. :(

Just incase. No need for Questions 1 and 2 now. Yay!

Whoops. Thank you for the help swarley. :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Greatness on February 05, 2012, 06:18:49 pm: 4. http://en.wikipedia.org/wiki/Function_composition you'll learn it some stage of the year, i hated them.
5. looks good :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 05, 2012, 06:23:39 pm: 5. It does??? Wow.

Thank you so much for your help swarley. :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Greatness on February 05, 2012, 06:32:42 pm: Quote from: Mr. Study on February 05, 2012, 06:23:39 pm
5. It does??? Wow.

Thank you so much for your help swarley. :)
Yeah i think so.... Altho i havent touched any math since exams... Hopefully most of what i've said is right :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 06, 2012, 07:09:10 pm: Heres my question:

I cant really draw the cubic graph on paint/microsoft but heres all the information.

Maximum turning point: (-1,6)
Minimum T.P: (1,4)
Y-Intercept: (0,-1)

Find {x:f'(x)>0}. We are also not given the cubics equation. :(

I checked the answers but it is {x:x<-1} u {x:x>1}. I am not too familiar with that way of writing so could someone also explain what it essentially means?

Thanks! :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: pi on February 06, 2012, 07:40:22 pm: Removed last post, just realised my mistake (a big one!), will repost soon

edit: I give up, someone else find an equation to the cubic. For me, the y-int does not add up :( I think I've forgotten all of methods lol! Ignoring the y-int, the equation can be given by: $y=\frac{1}{2}x^3 - \frac{3}{2}x +5$ ... I think. But that doesn't add up to the info, sorry Mr. Study :(
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Phy124 on February 06, 2012, 08:50:01 pm: Quote from: Mr. Study on February 06, 2012, 07:09:10 pm
Heres my question:

I cant really draw the cubic graph on paint/microsoft but heres all the information.

Maximum turning point: (-1,6)
Minimum T.P: (1,4)
Y-Intercept: (0,-1)

Find {x:f'(x)>0}. We are also not given the cubics equation. :(

I checked the answers but it is {x:x<-1} u {x:x>1}. I am not too familiar with that way of writing so could someone also explain what it essentially means?

Thanks! :)
Given those turning points, I don't think the y-intercept can be -1? (I'm with you Rohit ;))

Let's pretend it says the y-intercept = (0,5)

We want to find when f'(x) >0 (When the graph is increasing)

Given the stationary points and our intercept, our graph will look like this https://www.desmos.com/calculator/kb0hymstlt (The equation Rohitpi worked out)

Therefore, it is increasing from (-oo,-1) then decreasing (-1,1) and increasing again from (1,oo)

We only want when it is increasing so

(-oo,-1) and (1,oo)

Which can also be written as

{x:x<-1} and {x:x>1}

If anyone else can find a logical solution that doesn't involve changing what is given then by all means call me out :P
Title: Re: Mr. Study's 3 and 4 Questions
Post by: b^3 on February 06, 2012, 09:13:47 pm: Agreed that the y-intercept is pretty screwed up. Also to add to Phy124's graph
https://www.desmos.com/calculator/nsivedqhna
f'(x)>0 when that tangent has a positive gradient, i.e. when it is not between those two green lines, try moving the slider for the value of a and having a look at what happens
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 06, 2012, 09:15:26 pm: Heres the image, I probably should've uploaded it but I couldn't find mine CD.
(http://i1054.photobucket.com/albums/s481/BondJames5/mmessentialmaths9bq7.png).

Thank you all for the help, I actually get it now!! :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: b^3 on February 06, 2012, 09:17:18 pm: Quote from: Mr. Study on February 06, 2012, 09:15:26 pm
Heres the image, I probably should've uploaded it but I couldn't find mine CD.
(http://i1054.photobucket.com/albums/s481/BondJames5/mmessentialmaths9bq7.png).

Thank you all for the help, I actually get it now!! :)
Makes sense then, (1,-4) not (1,4). But yeh in this case, you can see by the graph when the tangents will be positive, without having to find an equation.
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 06, 2012, 09:21:44 pm: Ah crap, Sorry about that!!!
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 17, 2012, 06:52:20 pm: Todays theme is Log questions. (Woo... :D).

(http://i1054.photobucket.com/albums/s481/BondJames5/logqs.png).

I'll try Latex next time. (Hopefully. :) ).

Thanks
Title: Re: Mr. Study's 3 and 4 Questions
Post by: pi on February 17, 2012, 07:00:20 pm: Probably made some stupid mistakes, but this is the way to go about them :)

1.
$\log_3{(x^5)} + \log_3{(x^2)} - \log_3{(x^7)}$
$\log_3{(\frac{(x^5)(x^3)}{x^7})$
$\log_3{(x^0)}$
$0$

2.
Let $\log_{10}{(x)} = a$
$a^2 + a -2 = 0$
$(a+2)(a-1)=0$
$a = -2$ or $a=1$
$\log_{10}{(x)}=-2$ or $\log_{10}{(x)}=1$
$x= \frac{1}{100}$ or $x=10$

3.
Multiply both sides by $e^x$
$e^{2x} - 4=0$
$(e^x -2)(e^x +2)=0$
$e^x = 2$ or $e^x = -2$
$x = \ln(2)$ as $e^x>0$
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 17, 2012, 07:06:10 pm: oO, You obliterated those questions easily... Thanks for the help Rohit $3.14159265....$
~~No one solve 3. I THINK I know what to do now.~~ xD, NVM.
Title: Re: Mr. Study's 3 and 4 Questions
Post by: brightsky on February 17, 2012, 07:06:27 pm: e^x - 4e^(-x) = 0
e^x - 4/e^(x) = 0
e^(2x) - 4 = 0 (note that e^x cannot equal 0)
e^(2x) = 4
x = 1/2 * ln(4)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on February 18, 2012, 07:36:50 pm: Quartic's Question

1. $y=x^2(x-2)(x-3)$ . Is there any way to find the turning points WITHOUT a CAS calculator?
_
2. $f(x)=|x^2-3x|+2$ Rewrite as hybrid. I get this far: $f(x)=| x^2-3x+2, ......$
|_ $-(x^2-3x)+2, .......$ .
I'm not too sure how to write the less/greater than for x.

Thanks. :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: monkeywantsabanana on February 18, 2012, 07:42:20 pm: Quote from: Mr. Study on February 18, 2012, 07:36:50 pm
Quartic's Question

1. $y=x^2(x-2)(x-3)$ . Is there any way to find the turning points WITHOUT a CAS calculator?
you derive it then equate it to 0, solve for x, sub it back into original equation to find the y.
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Panicmode on February 18, 2012, 07:46:04 pm: Quote from: Mr. Study on February 18, 2012, 07:36:50 pm
Quartic's Question

1. $y=x^2(x-2)(x-3)$ . Is there any way to find the turning points WITHOUT a CAS calculator?
_
2. $f(x)=|x^2-3x|+2$ Rewrite as hybrid. I get this far: $f(x)=| x^2-3x+2, ......$
|_ $-(x^2-3x)+2, .......$ .
I'm not too sure how to write the less/greater than for x.

Thanks. :)

2. Think about it in terms of where it "cuts" the negative region. Since $x^2-3x$ is a positive quadratic, it will be positive for x < x-intercept 1 and x> x intercept 2. It will be negative between those values.

$x (x-3)$

$x = 0, x = 3$

so it will be $x^2-3x+2$ for $x \ge 3$ and $x \le 0$

and
$-(x^2-3x)+2$ for $0<x<3$
Title: Re: Mr. Study's 3 and 4 Questions
Post by: b^3 on February 18, 2012, 07:50:58 pm: Quote from: Mr. Study on February 18, 2012, 07:36:50 pm
2. $f(x)=|x^2-3x|+2$ Rewrite as hybrid. I get this far: $f(x)=| x^2-3x+2, ......$
|_ $-(x^2-3x)+2, .......$ .
I'm not too sure how to write the less/greater than for x.

Thanks. :)
Now remember a mod makes whatever is inside it positive, so its is negative (y<0) then it will be made positive.

So when $x^{2}-3x\geq0$ then you are just left with the positive of it, i.e. $x^{2}-3x+2$ . When $x^{2}-3x<0$ , it i s negative and being made positive hence we have $-(x^{2}-3x)+2$ .

So you have to work out when $x^{2}-3x\geq0$
Because it is not linear, we need to draw it out and visually work out when it is above 0. So intercepts are needed x(x-3)=0, x=0,3
https://www.desmos.com/calculator/bntqyrvf32

So it is above or equal to 0 when x<=0 and x=>3
So that is your domain for the $x^{2}-3x+2$
For $-(x^{2}-3x)+2=-x^{2}+3x+2$ it will be when its less then 0, so 0<x<3

Hope that helps :)

EDIT: Beaten.
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on March 07, 2012, 04:46:28 pm: $y=ax^2+b/x.$ . Gradient is 5 at (2,-2). Find a and b

This is what I have done:
$y=ax^2+bx^-^1$
$dy/dx =2ax-bx^-^2$

Let $dy/dx =5$
$5=2ax-bx^-^2$

Let y=-2 and x=2 and sub into first equation.
$-2=4a+b/2$

I am not too sure what to do next. Also, If My working out is either wrong or I'm reasoning the wrong way, please say so. :)

Thank you.

EDIT: Just need my working out to be checked.

Find acute angle lying between $3x+4y-2=0$ and $2x-3y-4=0$

$3x+4y-2=0$
$3x+4y=2$
$4y=-3x+2$
$y=-3/4x+1/2$

$2x-3y-4=0$
$2x-3y=4$
$-3y=4-2x$
$y=2/3x-4/3y$

I let m₁=-3/4 and m₂=2/3.
$tan\Theta=-3/4$
$\Theta=tan^-1(-3/4)$
But I know we can't get tan^-1(negative number). So would someone be able to look through what I've done so far?

Thanks. :)
Title: Re: Mr. Study's 3 and 4 Questions
Post by: illuminati on March 07, 2012, 05:31:56 pm: Quote from: Mr. Study on March 07, 2012, 04:46:28 pm
$y=ax^2+b/x.$ . Gradient is 5 at (2,-2). Find a and b

This is what I have done:
$y=ax^2+bx^-^1$
$dy/dx =2ax-bx^-^2$

Let $dy/dx =5$
$5=2ax-bx^-^2$

Let y=-2 and x=2 and sub into first equation.
$-2=4a+b/2$

I am not too sure what to do next. Also, If My working out is either wrong or I'm reasoning the wrong way, please say so. :)

Thank you.

EDIT: Just need my working out to be checked.

Okay so you have
dy/dx = 5
and then you go on to
5 = 2ax - bx^(-2)
you can sub in x = 2 here, because the gradient at x = 2 is 5
so you'll get
5 = 4a - b/4 ----> 20 = 16a - b
and then you have two simultaneous equations with a and b.
Title: Re: Mr. Study's 3 and 4 Questions
Post by: illuminati on March 07, 2012, 05:40:42 pm: Quote from: Mr. Study on March 07, 2012, 04:46:28 pm
Find acute angle lying between $3x+4y-2=0$ and $2x-3y-4=0$

$3x+4y-2=0$
$3x+4y=2$
$4y=-3x+2$
$y=-3/4x+1/2$

$2x-3y-4=0$
$2x-3y=4$
$-3y=4-2x$
$y=2/3x-4/3y$

I let m₁=-3/4 and m₂=2/3.
$tan\Theta=-3/4$
$\Theta=tan^-1(-3/4)$
But I know we can't get tan^-1(negative number). So would someone be able to look through what I've done so far?

Thanks. :)

You can tan inverse a negative.
You'll get ~-0.64 radians
Your angle is tan^-1(2/3) - tan^-1(-3/4), because if you recall, the angle between a line and the x axis is tan(theta) = m
So you can firstly find the angle your positive gradient line makes with the x axis, and then subtract the negative angle your negative gradient line makes
and you'll get the answer
Hope this helps :D
Title: Re: Mr. Study's 3 and 4 Questions
Post by: Mr. Study on March 07, 2012, 05:50:34 pm: Hey,

Thank you so much for that. :)

I had to change my calculator settings to do the tan inverse, otherwise it was a syntax error. :O