Example 2: Rolle's Theorem

f(x) = (x - 1)²(x - 2)³, a = 1, b = 2.

Figure 3.8.12

The function f is continuous and differentiable everywhere (Figure 3.8.12). Moreover, f(1) = f(2) = 0. Therefore by Rolle's Theorem there is a point c in (1, 2) with f'(c) = 0.

Let us find such a point c. We have

f'(x) = 3(x - 1)²(x - 2)² + 2(x - 1)(x - 2)³ = (x - 1)(x - 2)²(5x - 7).

Notice that f'(1) = 0 and f'(2) = 0. But Rolle's Theorem says that there is another point c which is in the open interval (1, 2) where f'(c) = 0. The required value for c is c = || because f'() = 0 and 1 < < 2.