Let $f: [a, b]\rightarrow R$ be differentiable at each point of $[a, b ]$ and $f'(a)=f'(b)$, prove that there's a line passing to $a$ tangent to $f$

Question

Let $f: [a, b]\rightarrow R$ be differentiable at each point of $[a, b ]$, and suppose that $f'(a) = f'(b)$. Prove that there is at least one point $c$ in $(a,b)$ such that

$$ f'(c) = \dfrac{f(c)-f(a)}{c-a} $$

My attempt:

define $h(x) = \dfrac{f(x)-f(a)}{x-a}$ on $(a,b]$ and $h(a) = f'(a)$. Notice that $h$ is continuous on $[a,b]$.

Now $$h'(x) = \dfrac{f'(x)}{x-a}-\dfrac{f(x)-f(a)}{(x-a)^2}$$

Note that we define $h'$ on $(a,b]$

Our goal is to show that an extremum point of $h(x)$ lies in $(a,b)$ so we can claim $h'(c)=0$ for some $c\in (a,b)$.

Moving things around we see that $f'(x) = h'(x)(x-a)+h(x)$ on $(a,b]$. We observe that if $h(x)$ is strictly increasing (or strictly decreasing), then $f'(x)$ is also strictly increasing (or strictly decreasing). Hence a contradiction to $f'(a)=f'(b)$ and so there's an extremum $c$ for $h(x)$. Here, we obtain a contradiction because if we were to avoid a contradiction then $f'(a)>d>f'(d+\epsilon)$ (assuming $f'$ is increasing) for any $\epsilon>0$. Applying an intermidate-value-theorem type lemma to $f'$ we contradict monotonicity. Hence, $f(a)<f(a+\epsilon)$ for any $\epsilon>0$.

Therefore, $h'(c)=0$ implies $$\dfrac{f(c)-f(a)}{c-a}=f'(c)$$

I am only looking for proof verifications. If my proof is wrong, please $\textbf{only respond with hints}$.

Answer 1

A few small critiques:

Note that we define ℎ′ on (,]

should instead be "for $a < x \le b$". You're not defining $h'$; you're evaluating it (using the quotient rule, etc.) and then noting the domain in which this evaluation is valid.

Our goal is to show that an extremum point of $ℎ()$ lies in $(,)$ so we can claim $ℎ′()=0$ for some $ \in (,)$.

I'd rewrite this as

Our goal is to show that for some $c$ strictly between $a$ and $b$, $c$ is an extreme point for $h$. We can then conclude that $h'(c) = 0$.

Given how sensitive the argument is, it might be worth clarifying what you mean by "extreme point." If $f$ is constant, for instance, then $h$ is also constant, and many folks would find it odd to say that a constant function has extreme points (although many others would be fine with it).

....and that's where I stopped reading, because I need to go do other things.