Question

Suppose 𝑓 is odd and differentiable everywhere. Prove that for every positive

number 𝑏, there exists a number 𝑐 in (−𝑏, 𝑏) such that 𝑓′(𝑐) = 𝑓(𝑏)/𝑏.

Answer 1

ANSWER.

The function $f$ is differentiable everywhere, hence $f$ is continuous everywhere.Let $b>0$ , so $f$ is differentiable on a closed interval $[-b,b]$ ( hence $f$ is continuous on $[-b,b]$ ). Then , by Lagrange's Mean Value Theorem , there exists a point $c\in(-b,b)$ , such that

$f(b)-f(-b)=f'(c)\cdot[b-(-b)]=2b\cdot f'(c)$ . (1)

Since $f$ is odd, so $f(-b)=-f(b)$ or $f(b)-f(-b)=2f(b).$ From (1) we get

$2f(b)=2b\cdot f'(c).$

Therefore, for every $b>0$ there exists $c\in(-b,b)$ , such that

$f'(c)=\frac{f(b)}{b}$ .