How to use the mean value theorem

Question

I have the following question:

I answered the first part of the question (i) by saying

I am unsure if i have answered correctly and I dont know how to show that for g''(b)

Answer 1

For the first part, yes you've done a correct computation of the average rate of change. You should note that the mean value theorem (MVT) says that there is some $a$ such that $g'(a)=1/2$, which is exactly what the question asks (I am specifying the some a to make it clear where the a is coming from).

For the second part, notice how in the question statement they also gave you that $g(1)=0$. You didn't need that for part a, so you should be suspicious that it will be used in part b. Also, since the MVT turns a rate of change of a function into a statement about the derivative, to find out something about $g''$, you'll need to know two different values of $g'(x)$.

So how can we get values of $g'(x)$? We can do that with MVT! If you apply MVT to $g(x)$ on the interval $0\leq x\leq 1$, then you get some $a_1$ with $g'(a_1) = 0$, and if you apply MVT to $g(x)$ on $1\leq x \leq 2$, then you get some $a_2$ with $g'(a_2)=1$. Now we can apply MVT to $g'(x)$ on the interval $a_1\leq x \leq a_2$ and get that there is some $b$ with $$g''(b) = (g')'(b) = \frac{g'(a_2)-g'(a_1)}{a_2-a_1} = \frac{1-0}{a_2-a_1} = \frac{1}{a_2-a_1}.$$ While you don't actually know what $a_1$ and $a_2$ are, you do know that $0<a_1<1<a_2<2$, and so $a_2-a_1 < 2$. So $g''(b)>1/2$, which is what you wanted.