Author Topic: VCE Methods Question Thread! (Read 5763020 times) Tweet Share

IndefatigableLover · « **Reply #4710 on:** May 10, 2014, 05:14:52 pm »

Quote from: Marco Reus on May 10, 2014, 05:07:24 pm

Could anyone please do this for me? I ended up with
y=√x-9 + 6
D: x≥9
R:R

Not too sure with it though, so if anyone could tell me if it's correct? As my teacher didn't provide answers with the sheet haha
Thanks in advance.

For $f^-1(x)$ to exist it needs to be a one-to-one function. In this case you should just look at the turning point of the graph and restrict it from there meaning that for an inverse function to exist for this, the domain could either be $[-\infty, 6] \text{or} [6, \infty]$

Champ101 · « **Reply #4711 on:** May 10, 2014, 05:22:05 pm »

Quote from: Marco Reus on May 10, 2014, 05:07:24 pm

Could anyone please do this for me? I ended up with
y=√x-9 + 6
D: x≥9
R:R

Not too sure with it though, so if anyone could tell me if it's correct? As my teacher didn't provide answers with the sheet haha
Thanks in advance.

The domain that you chose (x≥9) is assuming you chose to restrict the domain of the original from (6,infinity). If however you had y=-√x-9 + 6 than it would mean that you restricted the original domain from (-infinity,6).

Reus · « **Reply #4712 on:** May 10, 2014, 05:39:26 pm »

Ahhhh fml really struggling atm haha, i have a = 3, n = ?, b = -1
Anyone?
Thanks in advance.

soNasty · « **Reply #4713 on:** May 10, 2014, 05:42:48 pm »

since the period is 4pi

use your knowledge of how to determine the period of a cos/sin graph using 2pi/n

so you since you know the period is 4pi

you can solve it as:
4pi=2pi/n
n=2pi/4pi
n=1/2

Reus · « **Reply #4714 on:** May 10, 2014, 05:46:48 pm »

Quote from: andrew2910 on May 10, 2014, 05:42:48 pm

since the period is 4pi

use your knowledge of how to determine the period of a cos/sin graph using 2pi/n

so you since you know the period is 4pi

you can solve it as:
4pi=2pi/n
n=2pi/4pi
n=1/2

Was just about to delete this post haha! Yes i got it, i knew it was a logical question, but i was using n/2pi instead of 2pi/n to find n.
Hence why i was confused, thanks though

alchemy · « **Reply #4715 on:** May 10, 2014, 05:49:26 pm »

Quote from: Marco Reus on May 10, 2014, 05:46:48 pm

Was just about to delete this post haha! Yes i got it, i knew it was a logical question, but i was using n/2pi instead of 2pi/n to find n.
Hence why i was confused, thanks though

You shouldn't delete your posts though; other people may benefit from it

Reus · « **Reply #4716 on:** May 10, 2014, 06:09:47 pm »

Big ask I know! I've already done it, but we were never given answer sheet and so i have no idea if i am right and if I'm wasting my time, so if anyone could do 8 a-e, I'd honestly really appreciate it!

Thanks in advance.

swagsxcboi · « **Reply #4717 on:** May 10, 2014, 06:37:27 pm »

Quote from: Marco Reus on May 10, 2014, 06:09:47 pm

Big ask I know! I've already done it, but we were never given answer sheet and so i have no idea if i am right and if I'm wasting my time, so if anyone could do 8 a-e, I'd honestly really appreciate it!
Thanks in advance.

a
i) x=-20 and x=0
ii) -28<x<-20 0<x<12
iii) -20<x<0

b) should be a quadratic with intersection points at -20 and 0, and turning point at x=-10 and y=-1/2 (dont think you need to label that)

c) look at the highest point on the gradient graph, the x value will be the answer

d) dy/dx = 3/200x^2 + 3/10x

e) just sub in each x value into dy/dx to get the answer

Reus · « **Reply #4718 on:** May 10, 2014, 09:51:51 pm »

Quote from: swagsxcboi on May 10, 2014, 06:37:27 pm

b) should be a quadratic with intersection points at -20 and 0, and turning point at x=-10 and y=-1/2 (dont think you need to label that)

Thanks for the answers! Had everything right

But could you draw the gradient function?
Thanks.

Reus · « **Reply #4719 on:** May 10, 2014, 10:22:06 pm »

If anyone has the time to do these questions, I'd really appreciate it!

Zealous · « **Reply #4720 on:** May 10, 2014, 11:10:53 pm »

Quote from: Marco Reus on May 10, 2014, 10:22:06 pm

If anyone has the time to do these questions, I'd really appreciate it!

So I did the working out for most of the question but then realised I misinterpreted part C haha. I don't have time right now to fix it up but here's what I've got anyway if you're interested. For parts c-f I've assumed that "a" is still an unknown value.

Spoiler

a.
$\\\begin{eqnarray*}\\f(x) & = & e^{x}(x^{2}-a)\\let\ f(x) & = & 0\\e^{x} & = & 0\ \rightarrow no\ solution\\or\ x^{2}-a & = & 0\ \rightarrow x=\pm\sqrt{a}\\\therefore x-intercepts & are & (\sqrt{a},0)\ or\ (-\sqrt{a},0)\\\end{eqnarray*}\\\\$

b.
$\begin{eqnarray*}\\f(x) & = & e^{x}(x^{2}-a)\\f'(x) & = & e^{x}(2x)+e^{x}(x^{2}-a)\ \leftarrow chain\ rule\\f'(x) & = & e^{x}(x^{2}+2x-a)\ \leftarrow take\ out\ e^{x}\\\\\end{eqnarray*}\\$

c. I've completed the square because I don't really like their method.
$\begin{eqnarray*}\\let\ f'(x) & = & 0\\x^{2}+2x-a & = & 0\\x^{2}+2x+1-1-a & = & 0\ \\(x+1)^{2}-1-a & = & 0\\(x+1)^{2} & = & 1+a\\x+1 & = & \pm\sqrt{1+a}\\x & = & \pm\sqrt{a+1}-1\\Sub\ this\ into\ f(x):\\f(\sqrt{a+1}-1) & = & -2(\sqrt{a+1}-1)e^{\sqrt{a+1}-1}\\f(-\sqrt{a+1}-1) & = & 2(\sqrt{a+1}-1)e^{-\sqrt{a+1}-1}\\\end{eqnarray*}\\$

$\therefore stationary \ points \ are \ (\sqrt{a+1}-1,-2(\sqrt{a+1}-1)e^{\sqrt{a+1}-1}) \$ $and \ (-\sqrt{a+1}-1, 2(\sqrt{a+1}-1)e^{-\sqrt{a+1}-1})$

d. I'm assuming this is a CAS question, so the simplest thing to do is to sketch the graph.
https://www.desmos.com/calculator/4eiibseipw
You can see that for any value of a, we will have a local minimum (first of the above stationary points) and a local maximum (second of the above stationary points).

e.
$\begin{eqnarray*}\\f(0) & = & e^{0}(0^{2}-a)\\f(0) & = & -a\\\therefore y-intercept & = & (0,-a) \\\end{eqnarray*}\\$

f.
From the graph link I posted previously, you can see that as "x" approaches negative infinity, "y" approaches 0.
Think about:
$y=e^x$
$as \ x\rightarrow \-infty, \ \ e^x \rightarrow 0, \ \ hence f(x) \rightarrow 0$

g.
Again, refer to part d to see a sketch of the graph.

Reus · « **Reply #4721 on:** May 10, 2014, 11:24:51 pm »

Quote from: Zealous on May 10, 2014, 11:10:53 pm

So I did the working out for most of the question but then realised I misinterpreted part C haha. I don't have time right now to fix it up but here's what I've got anyway if you're interested. For parts c-f I've assumed that "a" is still an unknown value.

Spoiler
a.
$\\\begin{eqnarray*}\\f(x) & = & e^{x}(x^{2}-a)\\let\ f(x) & = & 0\\e^{x} & = & 0\ \rightarrow no\ solution\\or\ x^{2}-a & = & 0\ \rightarrow x=\pm\sqrt{a}\\\therefore x-intercepts & are & (\sqrt{a},0)\ or\ (-\sqrt{a},0)\\\end{eqnarray*}\\\\$

b.
$\begin{eqnarray*}\\f(x) & = & e^{x}(x^{2}-a)\\f'(x) & = & e^{x}(2x)+e^{x}(x^{2}-a)\ \leftarrow chain\ rule\\f'(x) & = & e^{x}(x^{2}+2x-a)\ \leftarrow take\ out\ e^{x}\\\\\end{eqnarray*}\\$

c. I've completed the square because I don't really like their method.
$\begin{eqnarray*}\\let\ f'(x) & = & 0\\x^{2}+2x-a & = & 0\\x^{2}+2x+1-1-a & = & 0\ \\(x+1)^{2}-1-a & = & 0\\(x+1)^{2} & = & 1+a\\x+1 & = & \pm\sqrt{1+a}\\x & = & \pm\sqrt{a+1}-1\\Sub\ this\ into\ f(x):\\f(\sqrt{a+1}-1) & = & -2(\sqrt{a+1}-1)e^{\sqrt{a+1}-1}\\f(-\sqrt{a+1}-1) & = & 2(\sqrt{a+1}-1)e^{-\sqrt{a+1}-1}\\\end{eqnarray*}\\$

$\therefore stationary \ points \ are \ (\sqrt{a+1}-1,-2(\sqrt{a+1}-1)e^{\sqrt{a+1}-1}) \$ $and \ (-\sqrt{a+1}-1, 2(\sqrt{a+1}-1)e^{-\sqrt{a+1}-1})$

d. I'm assuming this is a CAS question, so the simplest thing to do is to sketch the graph.
https://www.desmos.com/calculator/4eiibseipw
You can see that for any value of a, we will have a local minimum (first of the above stationary points) and a local maximum (second of the above stationary points).

e.
$\begin{eqnarray*}\\f(0) & = & e^{0}(0^{2}-a)\\f(0) & = & -a\\\therefore y-intercept & = & (0,-a) \\\end{eqnarray*}\\$

f.
From the graph link I posted previously, you can see that as "x" approaches negative infinity, "y" approaches 0.
Think about:
$y=e^x$
$as \ x\rightarrow \-infty, \ \ e^x \rightarrow 0, \ \ hence f(x) \rightarrow 0$

g.
Again, refer to part d to see a sketch of the graph.

You sir are a legend. Thank you so much

Reus · « **Reply #4722 on:** May 11, 2014, 12:36:32 am »

Genuine help needed for e, f and g! Honestly have no idea and do have answer to them this time :p

alchemy · « **Reply #4723 on:** May 11, 2014, 10:25:48 am »

It is known that f(1)=2 and f'(1)=4. Find the derivative of f(x)log_ex when x=1.

alchemy · « **Reply #4724 on:** May 11, 2014, 10:38:37 am »

Quote from: alchemy on May 11, 2014, 10:25:48 am

It is known that f(1)=2 and f'(1)=4. Find the derivative of f(x)log_ex when x=1.

Never mind. I figured it out.
For anyone who's interested in solving this:
First, let y=f(x)log_ex
We can calculate dy/dx to be f'(x)log_ex + f(x)*1/x. When x=1: dy/dx=2.

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