Answer to Question #320678 in Calculus for Aune Ndeutenge

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

$Since\,\,f\,\,is\,\,odd, f\left( -b \right) =-f\left( b \right) \\By\,\,Lagrange‘s\,\,theorem\\\frac{f\left( b \right) -f\left( -b \right)}{b-\left( -b \right)}=f'\left( c \right) ,c\in \left( -b,b \right) \\\frac{2f\left( b \right)}{2b}=f'\left( c \right) \Rightarrow f'\left( c \right) =\frac{f\left( b \right)}{b}$