Verify Rolle’s Theorem for the function: f(x) = (x – 1) (x – 2

1.	Verify Rolle’s Theorem for the function: f(x) = (x – 1) (x – 2)2 in [1, 2]
Answer» We know that (i) f (x) = (x – 1) (x – 2)² is a polynomial which is continuous for all x ϵ R Hence, f (x) = (x – 1) (x – 2)² is continuous on [1, 2] (ii) f’(x) = (x – 2)² + 2 (x – 1) (x – 2) exist in [1, 2] Hence, f (x) = (x – 1) (x – 2)² is differentiable on (1, 2) (iii) We know that f (1) = (1 – 1) (1 – 2)² = 0 Similarly f (2) = (2 – 1) (2 – 2)² = 0 Here f (1) = f (2) The conditions of Rolle’s Theorem are satisfied. There exist at least one c ϵ (1, 2) where f’(c) = 0 So we get (c – 2)² + 2 (c – 1) (c – 2) = 0 We know that (3c – 4) (c – 2) = 0 Which gives c = 2 or 4/3 We know that value of c = 4/3 ϵ (1, 2) Therefore, Rolle’s Theorem is satisfied.

Verify Rolle’s Theorem for the function: f(x) = (x – 1) (x – 2)2 in [1, 2]

Answer»

We know that

(i) f (x) = (x – 1) (x – 2)² is a polynomial which is continuous for all x ϵ R

Hence, f (x) = (x – 1) (x – 2)² is continuous on [1, 2]

(ii) f’(x) = (x – 2)² + 2 (x – 1) (x – 2) exist in [1, 2]

Hence, f (x) = (x – 1) (x – 2)² is differentiable on (1, 2)

(iii) We know that

f (1) = (1 – 1) (1 – 2)² = 0

Similarly f (2) = (2 – 1) (2 – 2)² = 0

Here f (1) = f (2)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (1, 2) where f’(c) = 0

So we get

(c – 2)² + 2 (c – 1) (c – 2) = 0

We know that

(3c – 4) (c – 2) = 0

Which gives c = 2 or 4/3

We know that value of c = 4/3 ϵ (1, 2)

Therefore, Rolle’s Theorem is satisfied.