Question

The Rolle’s constant for the function y = x² on [-2, 2] is ...

(a) 2√3/3 

(b) 0

(c) 2

(d) -2

Answer 1

(b) 0

Let f(x) = y = x² 

f'(x) = 2x = 0 

By Rolle’s theorem there exists a ‘c’ such that 

f'(c) = 0 ⇒ 2c = 0 ⇒ c = 0 ∈ (-2, 2)