# Prove there exists $\xi$ such that $f(\xi)=f(a)+f'(\xi)(b-a)$

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

• 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. Jan 16 '20 at 21:37
• You seem to have copied something incorrectly. Note that for the function $f(x) = x$ over the interval $[0,1]$, there is no such $\xi$. Jan 16 '20 at 22:34
• @Omnomnomnom $f(x)=x$ doesn't satisfy the condition that there exist a $c\in(a,b)$ such that $f'(c)=0$ Jan 16 '20 at 22:38
• @Omnomnomnom maybe i didn't phrase that part of the question well. I am not familiar with how it is usually said in English Jan 16 '20 at 22:41
• 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 Jan 16 '20 at 23:02

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 $$g'(\eta)=\frac{g(c)-g(0)}{c-0}=\frac{g(c)}c>g(c)~(\because 0 $$\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 $$0). Define $$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

• What function do you mean when writing $f$ Jan 17 '20 at 22:13
• What is the relationship of your f to the original f. Use of g confuses me Jan 17 '20 at 22:52
• @Milan I have rewritten the proof. Hope this is no longer confusing. Jan 18 '20 at 2:20

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 $$f'(a) = 0$$.