Decide whether Rolle's theorem can be applied to f(x) = x2 + 2x − 1 on the interval [–3, 1].

Question

 If Rolle's theorem can be applied, find all value(s) of c in the interval such that f '(c) = 0.

Answer 1

The function f(x) = x² + 2x - 1

By the Rolle's theorem

If a real valued function f is continuous on a closed interval [a, b],differentiable on the open interval (a, b), and f(a) = f(b), then there exist c in the open interval (a, b) such that f'(c) = 0.

f(x) = x² + 2x - 1

The closed interval [-3, 1]

a = - 3, b = 1

f(a) = (- 3)² + 2(- 3) - 1 = 9 - 6 - 1 = 2

f(a) = 2

f(b) = (1)² + 2(1) - 1 = 1 + 2 - 1 = 2

f(b) = 2

f(a) = f(b)

f(x) = x² + 2x - 1

Differentiating with respect to x.

f'(x) = 2x + 2

f'(c) = 2c + 2

To find value of c, solve the equation f '(c) = 0.

2c + 2 = 0

2c = - 2

c = (- 2)/2

c = - 1

The value of c lies in [-3, 1].

Solution c = - 1.