Calculus Problems

Question

By applying the Mean Value Theorem to the function f(x)=\sqrt[5]{x}$ on [32,33]

show that 2<\sqrt[5]{33}<2.015

Answer 1

If f(x) = x^1/5 , then f'(x) = 1/5 x^-4/5 and f is continuously differentiable on [32, 33]

f'(32) = 1/5 * 1/16 = 1/80, and 0 < f'(x) < 1/80 on [32,33], as f'(x) is monotone decreasing on the interval.

f(32) = 2.

Thus, if f(33) <= 2, there exist a c in (32,33) such that f'(c) = (f(33) - f(32))/(33-32) = (2-2)/1 = 0

Yet, the derivative is greater than 0 throughout the interval. Therefore f(33) > 2

If f(33) >= 2.015, then there exist a c in [32,33] such that f'(c) >= (2.015 - 2)/(33-32) = .015

Yet, 1/80 < .015, and f'(c) < 1/80 for all c in (32,33).

This is a contradiction, so f(33) < 2.015

Thus, 2 < f(33) < 2.015