Author Topic: VCE Methods Question Thread! (Read 5644084 times) Tweet Share

vashappenin · « **Reply #1725 on:** March 25, 2013, 12:16:55 pm »

A few more questions that I need clarification with:
1. Let f:R→R,where $f(x)=3+6x^2-2x^3$ . Determine the values of x for which the graph of y = f (x) has a positive gradient.
2. Find the derivative of $(4x + 9/x)^2$ and values of x at which the derivative is zero.
3. For $y=(2x-3)/(x^2+4)$ , Show that $dy/dx= (8+6x-2x^2)/(x^2+4)^2$ . Also, find values of x for which y and dy/dx are both positive.
4. $x(1+x^2)^1/2$ (dunno how to do the square root function but it's supposed to be the square root of 1+x^2)

clıppy · « **Reply #1726 on:** March 25, 2013, 03:44:19 pm »

Hopefully someone can help me with these two questions. I'm not exactly sure what I'm looking for or how to solve it so any help would be appreciated

b^3 · « **Reply #1727 on:** March 25, 2013, 05:57:55 pm »

1. Solving for the solutions in terms of $b$ using the quadratic formula gives
$\begin{alignedat}{1}x^{2}+bx+15 & =0 \\ x & =\frac{-b\pm\sqrt{b^{2}-4(1)(15)}}{2(1)} \\ & =\frac{-b\pm\sqrt{b^{2}-60}}{2} \end{alignedat} $
Now for solutions to exist we need $b^{2}-60>0$ which implies $b^{2}>60$ . Out of the values given the only answer that accomplishes this is option E. Now we should just check that our values of $b$ fit the information given, by substituting them back into the equation and solving (since its MC its quicker to do it on calc). So for $b=8$ we get $=-5,\:-3$ and for $b=-8$ we get $x=3,\:5$ . Which fits the question, thus the answer is option E.

2)
For $y=x$ to be tangent to the second curve, we need it to just 'touch' the curve, that is only have one intersection point. i.e. we would have
$\begin{alignedat}{1}x & =\frac{k}{x-1} \\ x^{2}-x-k & =0 \end{alignedat}$
For there to be only one solution, the discriminate, $b^{2}-4ac=0$ .
$\begin{alignedat}{1}b^{2}-4ac & =0 \\ \left(-1\right)^{2}-4\left(1\right)\left(-k\right) & =0 \\ 1+4k & =0 \\ k & =-\frac{1}{4} \end{alignedat}$
That is option D.

(if you've done the calculus part of methods you could do this a different way, you could even check the answer using calculus).

EDIT: 2500^th post!

e^1 · « **Reply #1728 on:** March 25, 2013, 06:25:49 pm »

I post this only to realise it has been answered. Anyway here's the Calculus way of doing it.

Calculus way

$y = \frac{k}{x-1}$

Okay, so we must find the value of k so that the tangent $y = x$ touches a point of the curve. To find this, I will use two clues.

The first one will be finding the value k for which the tangent $y=x$ intersects the equation.

$\begin{align*} x &= \frac{k}{x-1} \\ k &= x^2 - x \end{align*}$

This will be useful in finding the value of x where the tangent's gradient is equal to 1.

$\begin{align*} \frac{dy}{dx} = -\frac{k}{(x-1)^2} \end{align*}$

The gradient of the tangent is 1, and using the equation of k above:

$\begin{align*} 1 = -\frac{x^2 - x}{(x-1)^2} \\ x^2 - 2x + 1 = x - x^2 \\ 2x^2 - 3x + 1 = 0 \\\\ \therefore x=\frac{1}{2} \text{ or } x=1 \\\\ \text{But since equation is undefined at } x = 1 \\ \therefore x = \frac{1}{2} \text{ only. }\end{align*}$

Now use the solution above which will give the result.

$\begin{align*} 1 = -\frac{k}{(\frac{1}{2}-1)^2} \\\\ \therefore k = -\frac{1}{4} \end{align*}$

So the answer is D.
Good questions btw

507 · « **Reply #1729 on:** March 25, 2013, 09:26:48 pm »

Quote from: vashappenin on March 25, 2013, 12:16:55 pm

A few more questions that I need clarification with:
1. Let f:R→R,where $f(x)=3+6x^2-2x^3$ . Determine the values of x for which the graph of y = f (x) has a positive gradient.
2. Find the derivative of $(4x + 9/x)^2$ and values of x at which the derivative is zero.
3. For $y=(2x-3)/(x^2+4)$ , Show that $dy/dx= (8+6x-2x^2)/(x^2+4)^2$ . Also, find values of x for which y and dy/dx are both positive.
4. $x(1+x^2)^1/2$ (dunno how to do the square root function but it's supposed to be the square root of 1+x^2)

1. The derivative of a function is essentially its gradient graph. So when $f'(x)>0$ the gradient is positive.
$f(x)=-2x^3+6x^2+3$
$f'(x)=-6x^2+12x$
$0<-6x^2+12x$
$0<x<2$ (graphing may help in visualising this.)

2. $f(x)=(4x + \frac{9}{x})^2$
$f'(x)=2(4x+\frac{9}{x})(\frac{-9}{x^2}+4)$
$f'(x)=32x-\frac{162}{x^3}$
$0=32x-\frac{162}{x^3}$
$x=\frac{3}{2}, x=\frac{-3}{2}$

3. $y=\frac{2x-3}{x^2+4}$
Using the quotient rule $\frac{dy}{dx}=\frac{2(x^2+4)-(2x)(2x-3)}{(x^2+4)^2}$
Expanding the numerator $\frac{dy}{dx}=\frac{-2x^2+6x+8}{(x^2+4)^2}$
Use the same method used in #1 for the next bit.

4. Not sure what you want to do with this, I'm assuming differentiate.
$f(x)=x(1+x^2)^{1/2}$
Using the product and chain rule $f'(x)=1*\sqrt{x^2+1}+x(\frac{1}{2})(x^2+1)^\frac{-1}{2}(2x)$
$=\sqrt{x^2+1}+\frac{x^2}{\sqrt{x^2+1}}$
$=\frac{2x^2+1}{\sqrt{x^2+1}}$

jono88 · « **Reply #1730 on:** March 28, 2013, 04:56:22 pm »

I have a SAC 2nd day back from term 2 covering the first 3 chapters (Polynomial and modulus functions, exponential and logs, and circular functions. Our sac will have 2 parts (non-calc and calc). With calc section, we get to bring in a bound reference or our textbook. I want to bring in my textbook with extra notes, what should i do?

*Or go YOLO without any notes just my brain and CAS xD*

Daenerys Targaryen · « **Reply #1731 on:** March 28, 2013, 05:54:01 pm »

Quote from: jono88 on March 28, 2013, 04:56:22 pm

I have a SAC 2nd day back from term 2 covering the first 3 chapters (Polynomial and modulus functions, exponential and logs, and circular functions. Our sac will have 2 parts (non-calc and calc). With calc section, we get to bring in a bound reference or our textbook. I want to bring in my textbook with extra notes, what should i do?

*Or go YOLO without any notes just my brain and CAS xD*

If you're confident enough you wont even refer to the book. It would only be there for insurance. So if you now your stuff just bring either; but if you're someone who needs all the extra help write your own, or the notes your teacher gives you because you wrote them yourself so it should make more sense

Sanguinne · « **Reply #1732 on:** April 02, 2013, 02:10:25 pm »

does anyone know how to differentiate the following

y=cos³(x)

e^1 · « **Reply #1733 on:** April 02, 2013, 02:18:06 pm »

Quote from: Sanguinne on April 02, 2013, 02:10:25 pm

does anyone know how to differentiate the following

y=cos³(x)

Hint: $\cos^3(x) = (\cos(x))^3$

Sanguinne · « **Reply #1734 on:** April 02, 2013, 06:23:10 pm »

thanks for the hint

made it much more simpler

ashoni · « **Reply #1735 on:** April 02, 2013, 07:39:01 pm »

could someone please explain how

e raised to the natural log of x is equal to x?

i've completely forgot about logarithm rules and stuff

Daenerys Targaryen · « **Reply #1736 on:** April 02, 2013, 07:55:05 pm »

Quote from: ashoni on April 02, 2013, 07:39:01 pm

could someone please explain how

e raised to the natural log of x is equal to x?

i've completely forgot about logarithm rules and stuff

The process 'un-do' each other. Like squaring undoes a square root.

thus $e^{log_{e}(x)}$ becomes just $x$

b^3 · « **Reply #1737 on:** April 02, 2013, 08:24:17 pm »

While I think HatersGonnaHate's way of looking at it is the best way of looking at it, there is a longer way to see it too (kinda tedious though), making use of log rules.
$\begin{alignedat}{1}\text{Let }y & =e^{\log_{e}x} \\ \text{Take the log of both sides} \\ \log_{e}y & =\log_{e}\left(e^{\log_{e}x}\right) \\ \log_{e}y & =\log_{e}x\times\log_{e}e \\ \log_{e}y & =\log_{e}x\times1 \\ \log_{e}y & =\log_{e}x \\ \text{Equate the inside of the logs} \\ y & =x \\ \implies e^{\log_{e}x} & =x \end{alignedat} $

Planck's constant · « **Reply #1738 on:** April 02, 2013, 08:39:40 pm »

Quote from: HatersGonnaHate on April 02, 2013, 07:55:05 pm

The process 'un-do' each other. Like squaring undoes a square root.

thus $e^{log_{e}(x)}$ becomes just $x$

The functions loge(x) and e^x are the inverse of each other.
We also know that the composition of a function with its inverse returns x, i.e. f o f-1 (x) = x

$e^{log_{e}(x)}$

Is one such composition. Therefore it must return x

loge(e^x) is another such composition. It must also return x

polar · « **Reply #1739 on:** April 02, 2013, 09:03:08 pm »

also, $\log_a(b)$ is the power we need to be raise $a$ by to get $b$ (for example, $\log_2(8)=3$ means we need to raise 2 to the power to 3 to get 8 ). therefore, $a^{\log_a(b)}$ is $a$ raised to the power of a needed to get $b$ . hence, we get $b$ . $2^{\log_2(8)}=8$ is 2 raised to the power need to get from 2 to 8 (which is 3). similarly, $e^{\log_e(x)} = x$

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vashappenin

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