Integration

Is this your question $\displaystyle \left(x -\frac1{x}\right)^2$?

For your second question, $\displaystyle f(x) = 4x^3-6x^2+2x$
Differentiate f(x), then make the dy/dx =0, because dy/dx is zero at turning points. So by solving it equal to zero you will get the x-coordinates. Now substitute the values you will find for 'x' in the f(x) expression to find the y-coordinates.

Yes, that is the equation. I got x^2 + 2 + 1/x^2 when I opened the brackets, is that right? What is meant by second derivative?

4x^3-6x^2+2x is the dy/dx, what would I sub in, I know when there is a number at the end without x, we just sub all the multiples of that number, but what do we use when it has a x?

$\displaystyle \left(x-\frac1{x}\right)^2 = x^2 -2+\frac1{x^2}$

You have made a mistake in you expansion.

Now differentiate the expand expression twice.

e.g. $\displaystyle f(x) = (x+1)^2 = x^2 +2x +1$
Differentiate it twice,
$\displaystyle f'(x) = 2x + 2$
$\displaystyle f''(x)=2$

$\displaystyle f''(x)$ is the 2nd derivative.

Now for the next equation,

e.g. $\displaystyle f(x)= x^2 + 2x$

We have to find the turning points of f(x).
First differentiate f(x),
$\displaystyle f'(x)=2x+2$

Make f'(x) equal to zero,
$\displaystyle f'(x)=2x+2 =0 \implies x=-1$

So the turning point is at x=-1, now find the y-coordinate of it,
$\displaystyle f(-1)=(-1)^2+2(-1)=1-2=-1$

So the turning point is at, $\displaystyle (-1, -1)$

Now try your questions yourself.

My bad, its (x+1/x)

Also, when its x at the end, what do I sub in, in respect to the second equation.

What do you mean by the text in red? I don't get you.

Original equation is y=x ^4-2x^3+x^2

dy/dx= 4x^3-6x^2+2x

what do I sub in to find the turning points?

To find the turning points, make the dy/dx equal to zero.
So you get, 4x^3-6x^2+2x=0
Factorise the LHS, and find the values of 'x'.

Post the question in the maths forum, you can get more help there.

What do you mean? I know 1 is a turning point as well as zero and .5 (1/2), is that right?.

Yes, it is right, you have figured out the correct turning points.

Now substitute them into, y=x^4-2x^3+x^2, to find their y-coordinate.

Yes, it is right, you have figured out the correct turning points.

Now substitute them into, y=x^4-2x^3+x^2, to find their y-coordinate.

Thanks, I appreciate it...

For Q)17, what do I do?

Sorry, i am busy, so can't help you more.

I hope someone else helps you out.

No problem, thanks anyway.

Sorry, i am busy, so can't help you more.

I hope someone else helps you out.

Are you free atm?

Well by taking log of both sides:

logy= log4x^3

Can you think of how to split this up further:

Note: log(ab) = log(a) + log(b)

Thanks, its becomes

log y=3 log x + log 4

Which is the equation of the line but what do I plot and why?

That's right.

Now compare this to a standard straight line equation:

Y = MX + C

Here Y= log y
M= 3

X= log x

C= log 4

You took logs to reduce this to a straight line graph (see above).

So I would plot log x against log y????


Also, I'm stuck on Q)10

The other way round: log y against log x.

Have you expanded (x+1/x)^2?

It might be less confusing if you let u= 1/x and expand (x+u)^2 then sub 1/x back in.

Then you just need to differentiate twice.