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Let $f$ be a countinously differentiable function on $[a,b]$. Suppose that there exists $c \in(a,b)$ such that $f'(c)=0$. Prove that there exist $\xi\in (a,b)$ such that $$f(\xi)=f(a)+f'(\xi)(b-a) $$

I tried looking at function $H(x)=f(x)-f(a)-f'(x)(b-a)$ no good came from it. How to solve this ?

As suggesed in an answer lets solve for $f'(\xi)$: $$f'(\xi)=\frac{f(\xi)-f(a)}{b-a} $$ This is not the statement of mean value theorem, if it were $f(b)$ it would have been

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  • $\begingroup$ I missed that it was an $f(\xi)$ on the left instead of an $f(b)$. If we had an $f(b)$, this would be the mean value theorem. $\endgroup$ Jan 16 '20 at 21:37
  • $\begingroup$ You seem to have copied something incorrectly. Note that for the function $f(x) = x$ over the interval $[0,1]$, there is no such $\xi$. $\endgroup$ Jan 16 '20 at 22:34
  • $\begingroup$ @Omnomnomnom $f(x)=x$ doesn't satisfy the condition that there exist a $c\in(a,b)$ such that $f'(c)=0$ $\endgroup$
    – Milan
    Jan 16 '20 at 22:38
  • $\begingroup$ @Omnomnomnom maybe i didn't phrase that part of the question well. I am not familiar with how it is usually said in English $\endgroup$
    – Milan
    Jan 16 '20 at 22:41
  • $\begingroup$ I think I misread that part as well. Anyway, now that I understand the question I'm not sure where to start either, good luck $\endgroup$ Jan 16 '20 at 23:02
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For simplicity, one may prove a special case (see Proposition below), and then use it to derive OP's result (see Corollary below).

Proposition. Let $g(x)$ be a continuously differentiable function on $[0,1]$ such that $g(0)=0$ and there exists $c\in (0,1)$ with $g'(c)=0$. Then there exists $\xi\in (0,1)$ such that $g(\xi)=g'(\xi).$

Proof. One needs to prove the existence of $\xi\in (0,1)$ such the $g(\xi)=g'(\xi)$ or equivalently $H(\xi)=0$ for $H(x)=g(x)-g'(x)$. Up to multiplication by $-1$, there are two cases to consider:

Case 1: $g'(0)>0$.

Let $c_1=\inf\{c|g'(c)=0\}$. By continuity of $g'$, one has $g'(c_1)=0$. Furthermore $g'(x)\geq 0$ on $[0,c_1]$ (otherwise by intermediate value theorem for $g'$, there would exist $c_2\in (0,c_1)$ with $g'(c_2)=0$ contradicting the assumption of $c_1$), so $g$ is increasing on $[0,c_1]$ and $g(c_1)>0$. It follows that $$H(0)=g(0)-g'(0)=-g'(0)<0,~H(c_1)=g(c_1)-g'(c_1)=g(c_1)>0.$$ So by intermediate value theorem, there exists $\xi\in (0,c_1)\subseteq (0,1)$ such that $H(\xi)=0$.

Case 2: $g(0)=g'(0)=0.$

One has $H(0)=0$. If $H(x)=0$ or changes sign in $(0,1)$, then the result is either trivial or follows from the intermediate value theorem. It remains to rule out the case when $H(x)\neq 0$ in $(0,1).$ Up to multiplication by $-1$, one proves that the following case is impossible: $$g(0)=g'(0)=0,g(x)-g'(x)>0,\forall x\in(0,1),g'(c)=0~ ({\rm so~}g(c)>0){\rm~for~some~}c\in (0,1)$$

By mean value theorem, there exists $\eta\in(0,c)$ such that $$g'(\eta)=\frac{g(c)-g(0)}{c-0}=\frac{g(c)}c>g(c)~(\because 0<c<1)$$ $$\Rightarrow g(\eta)>g'(\eta)>g(c),$$ which shows that the absolute maximum of $g$ on $[0,c]$ occurs at some $\eta_1\in (0,c),$ so $g'(\eta_1)=0$. By mean value theorem as above, there exists $\eta_2\in (0,\eta_1)$ such that $$g'(\eta_2)=\frac{g(\eta_1)-g(0)}{\eta_1-0}=\frac{g(\eta_1)}{\eta_1}>g(\eta_1)$$ $$\Rightarrow g(\eta_2)>g'(\eta_2)>g(\eta_1),$$ contradicting the assumption that $g(\eta_1)$ is the absolute maximum of $g$ on $[0,c]$. This completes the proof of Case 2.

Combining Case 1 and Case 2, the Proposition is proven.

Corollary. Let $f$ be continuously differentiable on $[a,b]$ such that there exists $c\in (a,b)$ with $f'(c)=0$. Then there exists $\xi\in (a,b)$ such that $$f(\xi)=f(a)+f'(\xi)(b-a).$$

Proof. Let $f$ be as given. Let $c_1$ be such that $c=a+(b-a)c_1$ (so $0<c_1<1$). Define $g$ on $[0,1]$ by $$g(x)=f(a+(b-a)x)-f(a).$$ Then one has $g(0)=0$ and $g'(c_1)=f'(a+(b-a)c_1)(b-a)=f'(c)(b-a)=0.$ So $g(x)$ satisfies the assumptions of the Proposition. It follows that there exists $\xi_1\in (0,1)$ such that $$g(\xi_1)=g'(\xi_1)$$ $$\Leftrightarrow f(a+(b-a)\xi_1)-f(a)=f'(a+(b-a)\xi_1)(b-a)$$ $$\Leftrightarrow f(\xi)=f(a)+f'(\xi)(b-a),$$ where $\xi:=a+(b-a)\xi_1\in (a,b)$, as required. QED

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  • $\begingroup$ What function do you mean when writing $f$ $\endgroup$
    – Milan
    Jan 17 '20 at 22:13
  • $\begingroup$ What is the relationship of your f to the original f. Use of g confuses me $\endgroup$
    – Milan
    Jan 17 '20 at 22:52
  • $\begingroup$ @Milan I have rewritten the proof. Hope this is no longer confusing. $\endgroup$
    – Pythagoras
    Jan 18 '20 at 2:20
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Taking $\phi(x) = f(x) - f(a) - f'(x)(b-a)$ we have $\phi(a) = -f'(a)(b-a)$ and $\phi(c) = f(c) - f(a)$ (where $f'(c) = 0$). Also since $f$ is continuously differentiable it follows that $\phi$ is continuous.

If $f'(a) > 0$ and $f(c) > f(a)$, then $\phi(a) < 0$ and $\phi(c) > 0$ and there exists a point $\xi \in (a,c)$ such that $\phi(\xi) = 0$ by the intermediate value theorem. On the other hand if $f(c) \leqslant f(a)$ then by continuity with $f'(a) > 0$ there exists a relative maximum at some point $c_1 < c$ where $f(c_1) > f(a)$ and $f'(c_1) = 0$. Again we find a point $\xi$ such that $\phi(\xi) = 0$.

Note that $\phi(\xi) = 0$ implies

$$f(\xi) = f(a) + f'(\xi)(b-a)$$

A similar argument applies if $f'(a) < 0$. See if you can take care of the case where $f'(a) = 0$.

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