How to prove $f'(\xi)=a$ ,if $ f(0)=0,f(a)=a,f'(x_{0})=0$?

Question

let $a\in (0,1)$,and $f(x)$ is continuous on $[0,a]$, and is differentiable on $(0,a)$, and $x_{0}\in (0,a)$,such $f'(x_{0})=0$,and such $$f(0)=0, f(a)=a$$ show that there exsit $\xi\in(0,a)$,such $$f'(\xi)=a$$

My try: I want to put $$F(x)=f(x)-ax$$ since $$F(0)=f(0)-0=0$$ so if we find another point $c\in(0,a)$ such that $F(c)=f(c)-ac$

then we can use Rolle theorem to prove it.

Now How to prove there exist $c\in(0,a)$,such that $f(c)=ac$

maybe this method is not useful,Thank you.

By MVT there exists $x_1\in (0,a)$ such that $f'(x_1)=\frac{f(a)-f(0)}{a-0}=1$. So $f'$ takes the values $0$ and $1$. By Darboux, it also takes the value $a\in (0,1)$. Now surely this is overkill... — Julien, Dec 24 '13 at 14:30
@CWeid Did you mean: "It's time to make efforts to learn English"? Or can we really say "to take efforts in doing something"? — Julien, Dec 24 '13 at 15:19

score 1 · Answer 1 · answered Dec 24 '13 at 14:37

1

using Rolle theorem you obtain

$\exists c \in (0,1)\qquad F'(c)=0$

$F'(x)=f'(x)-a$

$F'(c)=f'(c)-a=0$ Finally

$f'(c)=a$

answered Dec 24 '13 at 14:37

Laki888

133

why $F'(c)=0?$,I think we can't have this – math110 Dec 24 '13 at 14:57
My mistake. I thought that $F(a)=0$, but it isn't. – Laki888 Dec 24 '13 at 15:12

How to prove $f'(\xi)=a$ ,if $ f(0)=0,f(a)=a,f'(x_{0})=0$?

1 Answers1