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

.

We'd have to do a little more work to find the exact value of c. The Mean Value Theorem just tells us that there's a value of c that will make this happen.

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