Discuss the applicability for Rolle’s theorem, when:f(x) = (x - 1)(2

1.	Discuss the applicability for Rolle’s theorem, when:f(x) = (x - 1)(2x - 3), where 1 ≤ x ≤ 3
Answer» Condition (1): Since, f(x) = (x - 1)(2x - 3) is a polynomial and we know every polynomial function is continuous for all x ϵ R. ⇒ f(x) = (x -1)(2x - 3) is continuous on [1,3]. Condition (2): Here, f’(x) = (2x - 3)+ 2(x - 1) which exist in [1,3]. So, f(x) = (x -1)(2x - 3) is differentiable on (1,3). Condition (3): Here, f(1) = [1 - 1][2(1) - 3] = 0 And f(5) = [3 -1][2(3) - 3] = 6 i.e. f(1) ≠ f(3) Condition (3) of Rolle’s theorem is not satisfied. So, Rolle’s theorem is not applicable.

Discuss the applicability for Rolle’s theorem, when:f(x) = (x - 1)(2x - 3), where 1 ≤ x ≤ 3

Answer»

Condition (1):

Since, f(x) = (x - 1)(2x - 3) is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = (x -1)(2x - 3) is continuous on [1,3].

Condition (2):

Here, f’(x) = (2x - 3)+ 2(x - 1) which exist in [1,3].

So, f(x) = (x -1)(2x - 3) is differentiable on (1,3).

Condition (3):

Here, f(1) = [1 - 1][2(1) - 3] = 0

And f(5) = [3 -1][2(3) - 3] = 6

i.e. f(1) ≠ f(3)

Condition (3) of Rolle’s theorem is not satisfied.

So, Rolle’s theorem is not applicable.