Question

How to find out when the derived graph>0?

Can someone please explain to me how to do question 10 b? Thanks!

Answer 1

I believe, `{x : g'(x)>0} = (-oo,0) uu (0,15/4)`.

Explanation:

This is tricky.

Of course the actual differentiation is no problem.

`g(x) = (2x^2 - 15x)^3`, then by the chain rule:
`g'(x) = 3(2x^2-15x)^2(4x-15)`

Solving part b is made simpler by simplifying `g'(x)`.

`g'(x) = 3(2x^2-15x)^2(4x-15)`
`= 3(2x(x-15/4))^2(4(x-15/4))`
`= 3*4x^2(x-15/2)^2(4(x-15/4))`
`= 48x^2(x-15/2)^2(x-15/4)`

Now you know that the graph of `y=g'(x)` has `x`-intercepts at `x = 0`, `15/4` and `15/2`, the first and last of which are also stationary points.

The shape of a standard negative quintic (=degree 5 polynomial? I don't know) comes down from a greatly positive `y` value, turns four times, going back down to a very negative `y` value.

If you have access to a calculator, you can solve for `x` when `g'(x) = 0` or differentiate again to find `g''(x)`. This tells you that you have two more stationary points, though these do not really matter, as one you know will be between `0` and `15/4` and the other between `15/4` and `15/2`.

Since we know the shape of the curve, we know that `g'(x)` will be positive before turning at `0`, then again between `0` and `15/4`. The rest of the time it will be negative or zero.

Hence, `{x : g'(x)>0} = (-oo,0) uu (0,15/4)`.

Hope this helps!