I am stuck somewhere in this analysis problem

Question

Consider a real valued continuous function f on [0,1] such that f is differentiable on (0,1) and f(0)=f(1)=0. Does there exist some c in (0,1) where f(c)=f'(c)?

It seems like the answer is yes. I am trying to show that the graph of f and its derivative cut each other somewhere in (0,1). I tried using Rolle's theorem and the continuity of the derivative, but its not helping. Any constructive suggestions would be appreciated.

Is it possible for a derivative to exist everywhere on an open interval but be discontinuous? — Robert Shore, Feb 17 '19 at 07:44
@angryavian Oops... I meant the intermediate value property of the derivative. It comes under the heading "the continuity of the derivatives" in Rudin. — krianaaa, Feb 17 '19 at 07:45
@robert shore yes, please look at example 5.6(b) of Rudin (Third Edition) on page 106. — krianaaa, Feb 17 '19 at 07:47

score 7 · Accepted Answer · answered Feb 17 '19 at 08:14

7

Define $g:[0,1]\rightarrow \mathbb R$ such that $x \mapsto e^{-x}f(x)$ This is continuous on $[0,1]$ and differentiable on $(0,1)$. Also the values at the end points are equal. So use Rolle's theorem to conclude that $g'(c)=0$ for some $c\in (0,1)$ which gives you $f(c)=f'(c)$

answered Feb 17 '19 at 08:14

user6

4,092

1

Nice....thanks...onek sundor! – krianaaa Feb 17 '19 at 09:22

I am stuck somewhere in this analysis problem

1 Answers1