Question

Suppose f is an odd function and is 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

To prove the given statement, we can use the Mean Value Theorem and define a new function g(x) to show that there exists a number c in (-b,b) such that f'(c) = f(b)/b.

Step-by-step explanation:

To prove that for every positive number b, there exists a number c in (-b,b) such that f'(c)= f(b)/b, we can use the Mean Value Theorem. Since f is an odd function, we know that f(-b) = -f(b). Let's define a new function g(x) = f(x) - (f(b)/b)x. Note that g(x) is also an odd function and is differentiable everywhere. By the Mean Value Theorem, there exists a number c in (-b,b) such that g'(c) = (g(b) - g(-b))/(b - (-b)). Since g(b) = f(b) - (f(b)/b)b = 0 and g(-b) = f(-b) + (f(b)/b)b = 0, we have g'(c) = 0. Therefore, f'(c) - (f(b)/b) = 0, which implies f'(c) = f(b)/b.