Verify Rolle’s Theorem for the function: f(x) = x2 – x – 12 i

1.	Verify Rolle’s Theorem for the function: f(x) = x2 – x – 12 in [-3, 4]
Answer» We know that (i) f(x) = x² – x – 12 is a polynomial which is continuous for all x ϵ R Hence, f(x) = x² – x – 12 is continuous on [-3, 4] (ii) f’(x) = 2x – 1 exist in [-3, 4] Hence, f(x) = x² – x – 12 is differentiable on (-3, 4) (iii) We know that f(- 3) = (- 3)² – 3 – 12= 0 Similarly f(4) = 4² – 4 – 12 = 0 Here f(-3) = f(4) The conditions of Rolle’s Theorem are satisfied. There exist at least one c ϵ (-3, 4) where f’(c) = 0 2c – 1 = 0 which gives c = 1/2 We know that value of c = 1/2 ϵ (-3, 4) Therefore, Rolle’s Theorem is satisfied. f(x)= x² - X -12 f(x) being a polynomial function is continuous for all x €[-3,4] f'(x) = 2x-1 Where f(x) is clearly differentiable for all x€[-3,4]. f(-3)= (-3)² -(-3)-12 f(-3)= 0 f(4)= (4)²-4-12 f(4)= 0 f(-3)=f(4) There must exist a real number c such that f'(c)= 0 Or. 2c-1=0 c=½ €[-3,4] Hence Rolle's theorem is verified Hope it helps...

Verify Rolle’s Theorem for the function: f(x) = x2 – x – 12 in [-3, 4]

Answer»

We know that

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

Hence, f(x) = x² – x – 12 is continuous on [-3, 4]

(ii) f’(x) = 2x – 1 exist in [-3, 4]

Hence, f(x) = x² – x – 12 is differentiable on (-3, 4)

(iii) We know that

f(- 3) = (- 3)² – 3 – 12= 0

Similarly f(4) = 4² – 4 – 12 = 0

Here f(-3) = f(4)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-3, 4) where f’(c) = 0

2c – 1 = 0 which gives c = 1/2

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

Therefore, Rolle’s Theorem is satisfied.

f(x)= x² - X -12

f(x) being a polynomial function is continuous for all x €[-3,4]

f'(x) = 2x-1

Where f(x) is clearly differentiable for all x€[-3,4].

f(-3)= (-3)² -(-3)-12

f(-3)= 0

f(4)= (4)²-4-12

f(4)= 0

f(-3)=f(4)

There must exist a real number c such that

f'(c)= 0

Or. 2c-1=0

c=½ €[-3,4]

Hence Rolle's theorem is verified

Hope it helps...