Question

f(x) = `sqrt(4 - x^2)` in [– 2, 2]

Answer 1

We have, `sqrt(4 - x^2) = (4 - x^2)^(1/2)`

Since (4 – x²) and square root function are continuous and differentiable in their domain, given function f(x) is also continuous and differentiable in [– 2, 2]

Also f(–2) = f(2) = 0

So, conditions of Rolle's theorem are satisfied.

Hence, there exists a real number c ∈ (–2, 2) such that f'(c) = 0.

Now f'(x) = `1/2(4 - x^2)^((-1)/2)(-2x)`

= `- x/sqrt(4 - x^2)`

So, f'(c) = 0

⇒ `"c"/sqrt(4 - "c"^2)` = 0

⇒ c = 0 ∈ (–2, 2)

Hence Rolle's theorem has been verfired.