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Question
f(x) = log(x2 + 2) – log3 in [–1, 1]
Solution
We have, f(x) = log(x2 + 2) – log3
We know that x2 + 2 and logarithmic function are continuous and differentiable
∴ f(x) = log(x2 + 2) – log3 is also continuous and differentiable.
Now f(–1) = f(1) = log3 - log3 = 0
So, conditions of Rolle's theorem are satisfied.
Hence, there exists atleast one c ∈ (–1, 1) such that f'(c) = 0
f(x) = `(2"c")/("c"^2 + 2) - 0` = 0
⇒ c = 0 ∈ (–1, 1)
Hence, Rolle's theorem has been verified.
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