Question

Verify Rolle's theorem for the function f(d): x² + x - 1
in [-2, 1]​

Answer 1

$\large\underline{\bold{Solution-}}$

Definition of Rolle's Theorem :-

Let us consider a real valued function f(x) defined on [a, b] such that

• f(x) is continuous on [a, b]

• f(x) is differentiable on (a, b).

• f(a) = f(b)

This implies, Rolle's Theorem is applicable.

• Therefore, there exist atleast one real number c belongs to (a, b) such that f'(c) = 0.

Let's solve the problem now!!

Given,

• f(x) = x² + x - 1 in [- 2, 1].

1. Since, f(x) is a polynomial function

$\rm :\implies\:f(x) \: is \: continuous \: on \: [ - 2, 1]$

2. f'(x) = 2x + 1

$\rm :\implies\:f(x) \: is \: differentiable \: on \: ( - 2, 1)$

3. Now,

$\rm :\longmapsto\:f( - 2) = {( - 2)}^{2} - 2 - 1 = 4 - 3 = 1$

$\rm :\longmapsto\:f(1) = {(1)}^{2} + 1 - 1 = 1$

$\bf\implies \:f( - 2) = f(1)$

• This implies, Rolle's Theorem is applicable.

• Therefore, there exist atleast one real number c belongs to (- 2, 1) such that f'(c) = 0.

$\rm :\implies\:2c + 1 = 0$

$\bf :\implies\:c \: = \: - \: \dfrac{1}{2} \: \in \: ( - 2, 1)$

More information :-

Geometrical Meaning of Rolle's Theorem :-

• This theorem help us to find a point 'c' on the curve y = f(x) at which tangent is always parallel to x - axis.

• Every polynomial function is always continuous as well as differentiable.

• Every sine function is always continuous as well as differentiable on its domain.

• Every cosine function is always continuous as well as differentiable on its domain.

• Every Exponential functions are always continuous as well as differentiable