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