Question

f(x) = `1/(4x - 1)` in [1, 4]

Answer 1

We have, f(x) = `1/(4x - 1)` in [1, 4]

Clearly f(x) is continuous in [1, 4]

Also, f'(x) = `-4/(4x - 1)^2`,  which exists in  (1, 4)

So, it is differentiable in (1, 4)

Thus conditions of mean value theorem are satisfied.

Hence, there exists a real number c ∈ (1, 4) such that

f'(c) = `("f"(4) - f(1))/(4 - 1)`

⇒ `(-4)/(4"c" - 1)^2 = (1/(16 - 1) - 1/(4 - 1))/(4 - 1)`

= `(1/15 - 1/3)/3`

⇒ `(-4)/(4"" - 1)^2 = (-4)/45`

⇒ `(4"c" - 1)^2` = 45

⇒ 4c – 1 = `+- 3  sqrt(5)`

⇒ c = `(3sqrt(5) + 1)/4 ∈ (1, 4)`

Hence mean value theorem has been verified.