How to find out when the derived graph>0?

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

1 Answer
Oct 13, 2017

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!