Mean Value Theorem

Question

Let f(x) = x - x^1/5 on [0,1]

Find c such as f'(c) = 0, or if Rolle's Theorem does not apply, enter DNE

c =

Answer 1

f is continuous on [0,1] and is differentiable on (0,1) and f(0) = f(1). So by Rolle's Theorem, there is at least one number c in the interval (0,1) for which f'(c) = 0.

f'(x) = 1 - (1/5)x^-4/5 = (5x^4/5 - 1) / (5x^4/5) = 0

5x^4/5 = 1 So, x^4/5 = 1/5. Therefore, x = c = (1/5)^5/4.