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I have a question that involves the application of the Mean Value Theorem:

Suppose $f$ is a function differentiable on $[a, b]$ with $a<b$ and $f'(a)=f'(b)=0$. Show that there exists a point $c\in(a,b)$ such that $\frac{f(c)-f(a)}{c-a}=f'(c)$.

I understand that the idea is to apply Mean Value Theorem. In particular, I considered the function $h(x)=\frac{f(x)-f(a)}{x-a}$ if $x\in(a,b]$, and $h(a)=f'(a)=0$, and tried to show that $h$ cannot be strictly increasing or strictly decreasing (and hence there must be two points $x_1, x_2\in(a, b)$ such that $h(x_1)=h(x_2)=0$, and the result follows). However, I can't seem to be able to use the condition that $f'(b)=0$.

Is there any hints that I could use for this problem?

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1 Answer

Define, for $a<x \le b$, the function $$h(x)=\frac{f(x)-f(a)}{x-a}$$ and note that we have $$h'(x)=\frac{(x-a)f'(x)-(f(x)-f(a))}{(x-a)^2}.$$ Since $f'(a)=0$ we see that $\lim_{x \to a}h(x)=0$, so that on defining $h(a)=0$ we have $h(x)$ continuous on $[a,b]$, and at a max or min of $h(x)$ we have $h'(x)=0$. Suppose for now there is a point $x=c$ in the open interval $(a,b)$ at which $h(x)$ is max or min; then $h'(c)=0$, which means $$(c-a)f'(c)-(f(c)-f(a)=0,$$ i.e. $$f'(c)=\frac{f(c)-f(a)}{c-a}.$$

EDIT: Part of the following approach (to ensure some $c \in (a,b)$) was suggested by user1296727 in a comment.

Now $h$ has both a max and a min in $[a,b]$. If both of these were at the same point, then $h$ would be constant, so it's derivative would vanish everywhere. So the max and min are at two different points, and the only bad case would be if the min were at $a$ and the max were at $b$, or vice-versa.

Now if $f(a)=f(b)$ then from $h(a)=0=h(b)$ and Rolle's theorem there is $c \in (a,b)$ with $h'(c)=0$ as desired. Note now that from the assumption $f'(b)=0$ we have $$h'(b)=-\frac{f(b)-f(a)}{(x-a)^2}.$$ Since we've taken care of $f(a)=f(b)$ there are two cases remaining.

If $f(a)<f(b)$ then $h'(b)<0$ while $h(b)>0$ so that the max of $h$ is not at $a$, and also cannot be at $b$ because $h'(b)<0$. Hence in this case the max of $h$ occurs at some $c \in (a,b).$

If on the other hand $f(a)>f(b)$ then we have $h'(b)>0$ while $h(b)<0$ so that the min of $h$ is not at $a$, and also not at $b$ because $h'(b)>0$. So in this case the min of $h$ occurs at some $c \in (a,b).$

In all three cases of possible ordering between $f(a)$ and $f(b)$ we have the desired $c$ in $(a,b)$ at which $h'(c)=0$, to finish the argument.

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I just fixed an erroneous answer in which I didn't think about the possibility of a max at $b$ rather than interior to $(a,b)$. That's where the $f'(b)=0$ assumption came in to save it. See rewrite. –  coffeemath Jan 3 '13 at 6:04
    
It's probably easier to just cite the fact that both a max and a min must occur; if both occur at $b$, the function is plainly constant. Otherwise, one of 'em happens in the open interval....Anyway, good show. –  user1296727 Jan 3 '13 at 6:06
    
Yes, that is simpler than the detour through Rolle's I took! –  coffeemath Jan 3 '13 at 6:10
    
@coffeemath Might be worth it to rollback and implement user's fix instead. –  Calvin Lin Jan 3 '13 at 6:16
    
To anyone who read this answer above: I have fixed it up as to showing the max or min must occur at an interior point, without the previous "intuitive/handwaving" approach. –  coffeemath Jan 3 '13 at 12:11
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