Question

f(x) = `sqrt(25 - x^2)` in [1, 5]

Answer 1

We have, f(x) = `sqrt(25 - x^2)` in [1, 5]

Since 25 – x² and square root function are continuous and differentiable in their domain, given function f(x) is also continuous and differentiable.

So, condition of mean value theorem are satisfied.

Hence, there exists atleast one c ∈ (1, 5) such that,

f'(c) = `("f"(5) - "f"(1))/(5 - 1)`

⇒ `(-"c")/sqrt(25 - "c"^2) = (0 - sqrt(24))/4`

⇒ 16c² = 24(25 – c²)

⇒ 40c² = 600

⇒ c² = 15

⇒ c = `sqrt(15) ∈ (1, 5)`

Hence, mean value theorem has been verified.