Question

Suppose \(f\) is an odd function and is differentiable everywhere. Prove that for every positive number \(b\), there exists a number \(c\) in \(\left( { - b,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right)}}{b}\).

Short answer

It is proved that for every positive number \(b\), there exists a number \(c\) in \(\left( { - b,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right)}}{b}\) .

Step-by-step solution

The mean value theorem

Let \(f\) be a function that is continuous on the closed interval \(\left( {a,b} \right)\) and differentiable on the open interval \(\left( {a,b} \right)\). Then there is a number \(c \in \left( {a,b} \right)\) such as that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( b \right)}}{{b - a}}\).

The odd function

A function \(f\) is said to be an odd function if \(f\left( { - a} \right) = - f\left( a \right)\).

Proof the statement

Since the given function is continuous and differentiable everywhere, therefore, it satisfies the conditions of the Mean Value Theorem.

The given interval is \(\left( { - b,b} \right)\). So, by Mean Value Theorem, there exists a number \(c \in \left( { - b,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( { - b} \right)}}{{b - \left( { - b} \right)}}\).

Solve the above equation as follows:

\(\begin{aligned}{c}f'\left( c \right) &= \frac{{f\left( b \right) - \left( { - f\left( b \right)} \right)}}{{b + b}}\\ &= \frac{{f\left( b \right) + f\left( b \right)}}{{2b}}\\ &= \frac{{2f\left( b \right)}}{{2b}}\\ &= \frac{{f\left( b \right)}}{b}\end{aligned}\)

Hence, it is proved that for every positive number \(b\), there exists a number \(c\) in \(\left( { - b,b} \right)\) such that \(f'\left( c \right) = \frac{{f\left( b \right)}}{b}\).