Author Topic: Questions thread (Read 19067 times) Tweet Share

kyzoo · « **Reply #15 on:** December 02, 2009, 11:30:50 am »

Wow I only started the Methods book at the beginning of the year, and finished maybe a month and a half before the exams. I don't really see the point in doing the Methods course so early since you'll just be really bored in class.

Hielly · « **Reply #16 on:** December 02, 2009, 09:04:59 pm »

How would you do this on the CAS calc?
I found that x=-(my-5)/3 and y=m(my-2) + 2(my-5)/15

I dont know how to find m. The answer says -5, i understand this by subbing -5 to the 2 equations and they are the same equation and therefore will have infinity number of solutions. But how would you find out m ?

Thanks!

kenhung123 · « **Reply #17 on:** December 02, 2009, 09:52:00 pm »

What is the domain and range of $y\geq 3x+2$
I thought it would be domain and range $[-1,\infty ]$

Why is a circle with the centre (0,0), radius of 4 units and domain of [-4,0] and $\pm \sqrt{-x}$ not an one to one function?

kenhung123 · « **Reply #18 on:** December 03, 2009, 11:52:04 am »

What is the implied domain of:
$f(x)=\sqrt{x^2-7x+12}$
$f(x)=\sqrt{x^2+4}$
$\frac{(x^2-1)}{x+1}$

How do you draw these graphs?

/0 · « **Reply #19 on:** December 03, 2009, 04:01:25 pm »

I think the domain and range of $y\geq 3x+2$ would be $(-\infty, \infty)$

If you draw the graph, it's the whole region 'above' the line $y=3x+2$ . This region includes all values of $y$ and all values of $x$ , so the domain and range is $(-\infty,\infty)$ .

A one-to-one function has 1 y-value for each x-value, and 1 x-value for each y-value.
A graph is one-to-one if it obeys BOTH the horizontal and vertical line tests.
i.e, If you draw a horizontal line anywhere on the page, it does not intersect the graph at more than 1 point, and if you draw a vertical line anywhere on the page, it does not intersect the graph at more than 1 point.

$x^2+y^2 = 16$ , $x \in [-4, 0]$ does not obey the vertical line test, so it is not one-to-one. However, it does obey the horizontal line test, so it is one-to-many (each x-value has more than 1 y-value)

$\pm \sqrt{-x}$ also does not obey the vertical line test, but it obeys the horizontal line test, so it is one-to-many.

For the domain questions, there are certain things to remember:
- the expression under a square root cannot be negative
- you can't /0

So for $f(x) = \sqrt{x^2-7x+12}$ you require $x^2-7x+12 \geq 0$ . Solving this for x (by graphing or otherwise) will get you the domain.

You can graph this by first graphing $y = x^2-7x+12$ , then taking the square root of the whole graph. Any y-value between 0 and 1 will INCREASE when you take the square root, and any y-value above 1 will DECREASE when you take the square root. The x-intercepts will stay where they are.

For $\sqrt{x^2+4}$ , the expression under the square is always positive, so the domain is R.

Once again, graph $y= x^2+4$ and then square root every y-value.

For $\frac{x^2-1}{x+1}$ , the domain is x $\in$ R \ {-1}, since if x = -1 you divide by 0.

To graph this, first simplify: $f(x)=\frac{x^2-1}{x+1} = \frac{(x+1)(x-1)}{x+1} = x-1$ , $x \neq -1$

So the graph will be the straight line $y=x-1$ with an open circle at $x = -1$ .

kenhung123 · « **Reply #20 on:** December 03, 2009, 04:27:01 pm »

Quote from: /0 on December 03, 2009, 04:01:25 pm

I think the domain and range of $y\geq 3x+2$ would be $(-\infty, \infty)$

If you draw the graph, it's the whole region 'above' the line $y=3x+2$ . This region includes all values of $y$ and all values of $x$ , so the domain and range is $(-\infty,\infty)$ .

A one-to-one function has 1 y-value for each x-value, and 1 x-value for each y-value.
A graph is one-to-one if it obeys BOTH the horizontal and vertical line tests.
i.e, If you draw a horizontal line anywhere on the page, it does not intersect the graph at more than 1 point, and if you draw a vertical line anywhere on the page, it does not intersect the graph at more than 1 point.

$x^2+y^2 = 16$ , $x \in [-4, 0]$ does not obey the vertical line test, so it is not one-to-one. However, it does obey the horizontal line test, so it is one-to-many (each x-value has more than 1 y-value)

$\pm \sqrt{-x}$ also does not obey the vertical line test, but it obeys the horizontal line test, so it is one-to-many.

For the domain questions, there are certain things to remember:
- the expression under a square root cannot be negative
- you can't /0

So for $f(x) = \sqrt{x^2-7x+12}$ you require $x^2-7x+12 \geq 0$ . Solving this for x (by graphing or otherwise) will get you the domain.

You can graph this by first graphing $y = x^2-7x+12$ , then taking the square root of the whole graph. Any y-value between 0 and 1 will INCREASE when you take the square root, and any y-value above 1 will DECREASE when you take the square root. The x-intercepts will stay where they are.

For $\sqrt{x^2+4}$ , the expression under the square is always positive, so the domain is R.

Once again, graph $y= x^2+4$ and then square root every y-value.

For $\frac{x^2-1}{x+1}$ , the domain is x $\in$ R \ {-1}, since if x = -1 you divide by 0.

To graph this, first simplify: $f(x)=\frac{x^2-1}{x+1} = \frac{(x+1)(x-1)}{x+1} = x-1$ , $x \neq -1$

So the graph will be the straight line $y=x-1$ with an open circle at $x = -1$ .

Wow, great explanation and lots of detailed knowledge /0.
However I thought essentials only said a one to one function is one that passes the horizontal line test? Unless they mean a one to one is what passes the horizontal line test and a function is one which passes the vertical line test, so a one to one function is what passes both? Lol. Thats confusing!
Btw, are we expected to graph $\sqrt{x^2+4}$ and $f(x) = \sqrt{x^2-7x+12}$ ?
How did you solve this one sorry $y\geq 3x+2$ ?
Also, for the graph f(x)=x-1 when simplified, why is there still domain restriction?

m@tty · « **Reply #21 on:** December 03, 2009, 04:36:19 pm »

Think of it as x-to-y, how many x-values correspond how many y-values.

A graph which passes the vertical line test can only have one y-value to each x-value and thus becomes either a many-to-one or a one-to-one function.

The horizontal line test determines how many x-values correspond to the same y-value, if the relation passes the horizontal line test then it is either a one-to-one[function] or a many-to-one[relation]

If a graph passes both the vertical and the horizontal line tests then it is a one-to-one function. As there is only one y value to one x value and only one x value to each y value.

Quote from: kenhung123 on December 03, 2009, 04:27:01 pm

Btw, are we expected to graph $\sqrt{x^2+4}$ and $f(x) = \sqrt{x^2-7x+12}$ ?

Yes

Quote from: kenhung123 on December 03, 2009, 04:27:01 pm

How did you solve this one sorry $y\geq 3x+2$ ?

Solve for what? If you mean sketch you would draw the straight line $y=3x+2$ a full line as it is included, if it wasn't included you would draw a broken line, and shade the area above the line as y is equal to all values greater than or equal to $3x+2$ .

Quote from: kenhung123 on December 03, 2009, 04:27:01 pm

Also, for the graph f(x)=x-1 when simplified, why is there still domain restriction?

When an equation is simplified any initial restrictions remain, in this case as the denominator in the original equation would equal zero when $x=-1$ in the simplified version x cannot equal 1 either.

A common problem with this arises with logarithms.

$y=2\log_{e}{x}$ When you sketch this you find a basic logarithm dilated by a factor of 2 from the x-axis.

Using logarithm laws it follows that $y=2\log_{e}{x}=\log_e{x^2}$

Yet if you sketch both of these graphs you will notice they are different. This affects the validity of any solutions found. It is the domain of the altered function which causes the problem, you need to keep any domain restrictions from the initial equation and apply it in any changed forms of the equation. In this case $x > 0$ . With this restriction they are identically equal.
This caused trouble for some in Methods exam 1 this year.

/0 · « **Reply #22 on:** December 03, 2009, 04:56:35 pm »

Yeah, if it passes
Only horizontal line test: one-to-many (one x value for many y-values)
Only vertical line test: many-to-one (many x values for one y-value)
Both lines tests: one-to-one
Neither line test: many-to-many

You might be expected to graph $y = \sqrt{f(x)}$ , I think it's part of the 'transformations' section of the course. Square rooting a function is mapping every $y \mapsto \sqrt{y}$ , so it counts as a transformation.

I attached a picture of $y \geq 3x+2$ (orange region). ANY point in that region will satisfy the relation, and since $y=3x+2$ goes from $-\infty$ to $\infty$ in both the x and y directions, the region $y \geq 3x+2$ also extends to $-\infty$ and $\infty$ in the x and y directions. (imagine zoooming out, and you will see the region goes to infinities)
So the domain is R and the range is R.

Also, if $f(x) = \frac{x^2-1}{x+1}$ is undefined for $x = -1$ , then if you want to simplify it to $f(x)=x-1$ , you must keep that restriction.

Otherwise, you would have changed the function, by allowing an extra value of x!

The statements
$f(x) = \frac{x^2-1}{x+1}$
and
$f(x) = x-1, \ \ x \neq -1$
are equivalent.

kenhung123 · « **Reply #23 on:** December 03, 2009, 05:30:45 pm »

How do you do in equations such as
{x:y>x^2+4}
Do you think of it like: when is the graph above y=0 or something?

/0 · « **Reply #24 on:** December 03, 2009, 05:37:00 pm »

Oh do you mean how to graph: $\lbrace (x,y): y > x^2+4 \rbrace$

Normally I just draw $y=x^2+4$ (and make it dotted :/ I nearly forgot to say that!!) then pick a random point... say (0,0)

$0 < 0^2+4$ , so (0,0) does not lie in the required region.

So shade in the region that doesn't have (0,0) in it

OR

You could have picked (0,5)

$5 > 0^2+4$ , so (0,5) does lie in the required region.

So shade in the region that has (0,5) in it

(Just pick a test point that isn't on the curve $y=x^2+4$ )

m@tty · « **Reply #25 on:** December 03, 2009, 08:11:30 pm »

Top left is a one-to-one function as it passes both the vertical and horizontal line test.

Top right is a many-to-one function as it passes the vertical line test [hence is a function] but does not pass the horizontal line test.

Bottom left is a one-to-many relation as it does not pass the vertical line test but does pass the horizontal line test.

Bottom right is a many-to-many relation as it passes neither the vertical or horizontal line test.

"Passing" a test means that the line intersects the graph only once.
You may think that bottom left is a one-to-one function by placing the lines in a position similar to that of the pink and green lines, but you would be wrong. When conducting one of these line tests you must consider the whole domain/range and find if there exists any position where the line cuts the graph multiple times.

Souljette_93 · « **Reply #26 on:** December 03, 2009, 09:49:36 pm »

Heys, i need help as well. i don't know if this is the right place to post it, but i will.

1- Find the coeffient of p^4 in the expansion of (p+3)^5*(2p-5)

2- in expansion of (2a-1)^n the coeffient of the second term is -192. find the value of n

Thanks in Advance.
Souljette<3

Cataclysmic · « **Reply #27 on:** December 03, 2009, 10:42:04 pm »

1st one looks like a boring and very long expansion and simplification -_-

I'm not sure about this for 1 but the 1st part,(p+3)^5 will have a part that is to the power of
5,4,3,2,1 and a constant. If you were to times that to the other part ( 2p-5)
the ^5 would go to ^6 and ^5. So the ^4 part would have ^5 and ^4 ( 1 part here) and ^3 would have ^4 and ^3.

so maybe it could just be:
$(_1^5) (p)^{4}(3)^{1} * -5$ = 1 part that will have p^4
and $(_2^5) (p)^{3}(3)^{2} * 2p$ = 2nd part
so I'd say -75p^4 + 180p^4 = 105 as the coefficient.

2nd one: All you want is the 2nd expansion, so using the binomial theorem, you would have $(_1^n)(2a)^{n-1}(-1)^{1}$ but you're only after the co-efficient so you don't need the a.
This leads to $192 = n(2^{n-1})$ $( _1^n) = \frac{n!}{(n-1)!1!} = \frac{(n(n-1)(n-2)(n-3))....}{(n-1)(n-2)(n-3)..}$ leaving n on top.
I end up with that thing above or simplifying it to $384 = n(2)^{n}$ ( by breaking $2^{n-1}$ into $2^{n}*2^{-1}$ The answer is n = 6 but I don't see how to solve it from there ( maybe you need a calc). Maybe someone else can continue from here.

kenhung123 · « **Reply #28 on:** December 04, 2009, 12:25:30 pm »

Someone help with modulus functions?

/0 · « **Reply #29 on:** December 04, 2009, 02:28:15 pm »

The basic definition of the modulus means that |x| can never be negative.

So if you're asked to graph $y=|f(x)|$ , your graph can never be below the x-axis. If $f(x)$ goes below the x-axis, you must reflect those sections about the x-axis so they become positive in $|f(x)|$ .

If you're asked to graph $y=f(|x|)$ , the graph will be symmetrical about the y-axis. In other words it will be an 'even' graph. This is because if you plug in a negative value of $x$ , the absolute value signs will make it positive, so it will have the same y-value as if you plugged in a positive value of $x$ . So it's the graph to the left of the y-axis that changes to match the graph to the right of the y-axis.

If you get more complicated functions, such as $f(x) = x+|x|+|x+1|$ , you will need to use the definition of the modulus to break it up into a hybrid function.

$|x| = \begin{cases} x & \mbox{if} \ x\geq 0 \\ -x & \mbox{if} \ x < 0 \end{cases}$

$|x+1| = \begin{cases} x+1 & \mbox{if} \ x+1 \geq 0 \\ -(x+1) & \mbox{if} \ x+1 < 0 \end{cases} \ \ = \begin{cases} x+1 & \mbox{if} \ x \geq -1 \\ -(x+1) & \mbox{if} \ x < -1 \end{cases}$

So your hybrid function would look like this:

$f(x) = \begin{cases} x-x-(x+1) & \mbox{if}\ x<-1 \\ x-x+(x+1) & \mbox{if}\ -1 \leq x < 0 \\ x+x+(x+1) & \mbox{if} \ x \geq 0\end{cases}$

Also, other things to watch out for is $x^2 =|x|^2$ (when x is real)

And also $\sqrt{x^2} = |x|$ (since the square root function always give a positive answer)
(This caught out a lot of people in the Methods CAS 2009 Exam apparently)

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