Verify Rolle’s Theorem for the function: f(x) = x3 – 7x2 + 16x

1.	Verify Rolle’s Theorem for the function: f(x) = x3 – 7x2 + 16x – 12 in [2, 3]
Answer» We know that (i) f (x) = x³ – 7x² + 16x – 12 is a polynomial which is continuous for all x ϵ R Hence, f (x) = x³ – 7x² + 16x – 12 is continuous on [2, 3] (ii) f’(x) = 3x² – 14x + 16 exist in [2, 3] Hence, f (x) = x³ – 7x² + 16x – 12 is differentiable on (2, 3) (iii) We know that f (2) = (2)³ – 7 (2)² + 16(2) – 12 = 0 Similarly f (3) = 3³ – 7(3)² + 16 (3) – 12 = 0 Here f (2) = f(3) The conditions of Rolle’s Theorem are satisfied. There exist at least one c ϵ (2, 3) where f’(c) = 0 3c² – 14c + 16 = 0 It can be written as (c – 2) (3c – 7) = 0 which gives c = 7/3 We know that value of c = 7/3 ϵ (2, 3) Therefore, Rolle’s Theorem is satisfied.

Verify Rolle’s Theorem for the function: f(x) = x3 – 7x2 + 16x – 12 in [2, 3]

Answer»

We know that

(i) f (x) = x³ – 7x² + 16x – 12 is a polynomial which is continuous for all x ϵ R

Hence, f (x) = x³ – 7x² + 16x – 12 is continuous on [2, 3]

(ii) f’(x) = 3x² – 14x + 16 exist in [2, 3]

Hence, f (x) = x³ – 7x² + 16x – 12 is differentiable on (2, 3)

(iii) We know that

f (2) = (2)³ – 7 (2)² + 16(2) – 12 = 0

Similarly f (3) = 3³ – 7(3)² + 16 (3) – 12 = 0

Here f (2) = f(3)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (2, 3) where f’(c) = 0

3c² – 14c + 16 = 0

It can be written as

(c – 2) (3c – 7) = 0 which gives c = 7/3

We know that value of c = 7/3 ϵ (2, 3)

Therefore, Rolle’s Theorem is satisfied.