Let f(x) = x^{2}. Use the Mean Value Theorem to show that there's some value of c in (0,2) with f'(c) = 2.

f(x) is continuous on [0,2] and differentiable on (0,2). The Mean Value Theorem says there is some c in (0,2) for which f'(c) is equal to the slope of the secant line between (0,f(0) and (2,f(2)), which is

$$\frac{f(2)-f(0)}{2-0} = \frac{4-0}2 = 2.$$

Here's the picture:

Example 2

Let f(x) = x^{2}. What does the Mean Value Theorem guarantee on the interval (-2,0)?

Since f is continuous on [-2,0] and differentiable on (-2,0), the Mean Value Theorem guarantees that there is some c in (-2,0) with