Question

Prove the Rolle’s theorem for the following :

(a) f(x) = (x - 2)√x, x ∈ [0, 2]

(b) f(x) = (x - 1)(x - 3), x∈ [1, 3]

Answer 1

(a) Given

f(x) = (x - 2)√x, x ∈ [0, 2]

Her, f(x) is continuous in interval [0, 2] and f'(x) = (3x - 2)/2√x

which is defined and exist at every point in interval (0, 2).
So, function is differentiable in interval (0, 2).
∵ f(0) = 0 = f(2)
⇒ f(0) = f(2)
Here,f(x) satisfies Rolle’s theorem in given interval.

(b) Given f(x) = (x – 1) (x – 3), x ∈ [1, 3]

Here, f (x) is continuous in interval [1, 3] and f'(x) =2x – 4, which is defined and exist at every point in interval (1, 3). 

So,f(x) is differentiable in interval (1, 3).

∵ f(1) = 0 = f(3)

⇒ f(1) = f(3)

Here,f(x) satisfies Rolle’s theorem in given interval.

Hence f'(c) = 0

f'(c) = 2c – 4 = 0

⇒ 2c = 4

⇒ c = 2

∴ c = 2 ∈ (1,3)

such that

f'(c) = 0

Hence, Rolle’s theorem verified for c = 2.