Given:

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

Firstly, we have to show that f(x) is continuous.

Here, f(x) is continuous because f(x) has a unique value for each x ∈ [-2,2]

Condition 2:

Now, we have to show that f(x) is differentiable

[using chain rule]

∴ f’(x) exists for all x ∈ (-2, 2)

So, f(x) is differentiable on (-2,2)

Hence, Condition 2 is satisfied.

Condition 3:

Now, we have to show that f(a) = f(b)

so, f(a) = f(-2)

and f(b) = f(2)

∴ f(-2) = f(2) = 0

Hence, condition 3 is satisfied

Now, let us show that c ∈ (0,1) such that f’(c) = 0

On differentiating above with respect to x, we get

Put x = c in above equation, we get

Thus, all the three conditions of Rolle’s theorem is satisfied. Now we have to see that there exist c ∈ (-2,2) such that

f’(c) = 0

⇒ c = 0

∵ c = 0 ∈ (-2, 2)

Hence, Rolle’s theorem is verified