Prove that there exists $c$ such that $f'(\xi)=\frac{f(c)-f(a)}{c-a}$ or $f'(\xi)=\frac{f(c)-f(b)}{c-b}$

Question

Let $f(x) \in C^2 [a,b]$ and $f''(x) \neq 0$. Prove that for any $\xi \in (a.b)$ , there exists $c \in [a,b]$ such that $$f'(\xi)=\frac{f(c)-f(a)}{c-a}\qquad \text{or} \qquad~f'(\xi)=\frac{f(c)-f(b)}{c-b}.$$

Someone asked me the problem above, but I'm not sure whether it holds or not. Can anyone help to verify that? Thanks.

Answer 1

Since $f''$ is continuous and never $0$ it is always positive or always negative. I will give the proof in the case $f''(t)>0$ for all $t$. (The other case is similar). In this case $f$ is strictly convex adn $f'$ is strictly increasing.

Consider two cases:

a) $ (f'(a)<)f'(\xi) \leq \frac {f(b)-f(a)} {b-a}$. In this case the continuous function $\frac {f(x)-f(a)} {x-a}$ must attain the value $f'(\xi)$ (by IVP) because this function tends to $f'(a)$ as $x$ decreases to $a$ and tends to $\frac {f(b)-f(a)} {b-a}$ as $c \to b$.

b) $(f'(b)>) f'(\xi) >\frac {f(b)-f(a)} {b-a}$. In this case $\frac {f(x)-f(b)} {x-b}$ tends to $f'(b)$ as $x \to b$ and it tends to $\frac {f(b)-f(a)} {b-a}$ as $x \to a$ so once again it must attain the value $f'(\xi)$.

Answer 2

Since $f\in C’’[a,b])$, $f’\in C’[a,b]$. Now that $f’’(\xi)\ne 0$, we can choose a small open interval with endpoint $p$, $q$ and $f’(p)<f(\xi)<f’(q)$. Now let $c$ move in the interval, $\frac{f(p)-f(c)}{p-c}$ tend to $f’(p)$ when $c\to p$, and is $\frac{f(p)-f(q)}{p-q}$ as $c\to q$. Consider another ratio: $\frac{f(c)-f(q)}{c-q}$, it tends to $f’(q)$ and $\frac{f(p)-f(q)}{p-q}$ on each side respectively. Since the two ratios is continuous with respect to $c$, we know that there exists a $c$ in that interval such that one of the ratio is $f’(\xi)$.

Answer 3

I'm lucky to figure out a proof as follows. Please check it in case of any flaw.

Proof

Since $\forall x \in (a,b):f''(x)\neq 0$, as per Darboux's theorem，$f''(x)$ can not change its sign over $(a,b)$, namely, either $f''(x)>0$ or $f''(x)<0$ holds. Suppose $f''(x)>0$ now. If $f''(x)<0$，the proof is similar.

Denote $$g(x):=\begin{cases}\dfrac{f(x)-f(a)}{x-a},&x \in (a,b],\\f'(a+),&x=a,\end{cases}~~~\text{and}~~~h(x):=\begin{cases}\dfrac{f(x)-f(b)}{x-b},&x \in [a,b),\\f'(b-),&x=b.\end{cases}$$ It's simple to verify that $g(x),h(x)$ are both continuous over $[a,b]$，and $g(b)=h(a)$. Since $f''(x)>0$, $f'(x)$ is increasing over $[a,b]$，thus $g(a)<f'(\xi)<h(b).$ If $g(a)<f'(\xi)<g(b)$，as per the continuity of $g(x)$，$\exists c\in (a,b):f'(\xi)=g(c)$；otherwise，it must hold that $g(b)=h(a)\leq f'(\xi)\leq h(b)$. Therefore，by the continuity of $h(x)$，$\exists c \in [a,b]:f'(\xi)=h(c)$. So far, the proof is completed.