Verify Rolle’s Theorem for the function: f(x) = x2 – 3x – 18

1.	Verify Rolle’s Theorem for the function: f(x) = x2 – 3x – 18 in [-3, 6]
Answer» We know that (i) f(x) = x² – 3x – 18 is a polynomial which is continuous for all x ϵ R Hence, f(x) = x² – 3x – 18 is continuous on [-3, 6] (ii) f’(x) = 2x – 3 exist in [-3, 6] Hence, f(x) = x² – 3x – 18 is differentiable on (- 3, 6) (iii) We know that f(-3) = (-3)² – 3(-3) – 18 = 0 Similarly f(6) = 6² – 5(6) – 18 = 0 Here f(-3) = f(6) The conditions of Rolle’s Theorem are satisfied. There exist at least one c ϵ (- 3, 6) where f’(c) = 0 2c – 3 = 0 which gives c = 3/2 We know that value of c = 3/2 ϵ (-3, 6) Therefore, Rolle’s Theorem is satisfied.

Verify Rolle’s Theorem for the function: f(x) = x2 – 3x – 18 in [-3, 6]

Answer»

We know that

(i) f(x) = x² – 3x – 18 is a polynomial which is continuous for all x ϵ R

Hence, f(x) = x² – 3x – 18 is continuous on [-3, 6]

(ii) f’(x) = 2x – 3 exist in [-3, 6]

Hence, f(x) = x² – 3x – 18 is differentiable on (- 3, 6)

(iii) We know that

f(-3) = (-3)² – 3(-3) – 18 = 0

Similarly f(6) = 6² – 5(6) – 18 = 0

Here f(-3) = f(6)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (- 3, 6) where f’(c) = 0

2c – 3 = 0 which gives c = 3/2

We know that value of c = 3/2 ϵ (-3, 6)

Therefore, Rolle’s Theorem is satisfied.