Mean Value Theorem

Question

(a): Determine whether each of the conditions of the Mean Value Theorem is met for the function f(x) = x+ 1/x on the interval [1, 3]; in each case, explain or show why the condition is (or is not) met.

(b): If both conditions are met, then the Mean Value Theorem guarantees there is a value of x = c, 1 < c < 3, such that f '(c) = (f(3) − f(1)) / (3 − 1)

If both conditions are not met, then there is no guarantee: but for this particular case, there is such a c. Find a c between 1 and 3 that makes the equation true. Give clear steps and explanations.

Answer 1

The conditions are just that the function is continuous on [a, b] and differentiable on a[ b]

The derivative of x + 1/x, 1 - 1/x^2, exists whenever x # 0. x is continuous on this same interval. Thus, as [a, b] in this problem is [1, 3], the conditions are met.

(f(3) - f(1))/(3 - 1) = (3 + 1/3 -(1 + 1/1))/(3 - 1) = 4/3/2 = 2/3

1 - 1/x^2 = 2/3

1/3 = 1/x^2

x = √3 (we are looking on the interval [1, 3]