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Question
Find the intervals of monotonicities and hence find the local extremum for the following functions:
f(x) = `x^3/3 - log x`
Solution
f'(x) = `x^2 - 1/x`
f'(x) = 0
⇒ x3 – 1 = 0
⇒ x = 1
The intervals are (0, 1) and `(1, oo)`.
i.e., when x > 0, the function f(x) is defined in the interval (0, 1), f'(x) < 0
∴ f(x) is strictly decreasing in (0, 1) in the interval `(1, oo)`, f'(x) > 0
∴f(x) is strictly increasing in` (1, oo)`
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1
∴ Local minimum
f(1) = `1/3 - log 1 = 1/3 - 0 = 1/3`
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