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Question
Find the maximum and minimum of the following functions : f(x) = x log x
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Solution
f(x) = x log x
∴ f'(x) = `d/dx (x log x)`
= `x.d/dx (log x) + logx.d/dx(x)`
= `x xx (1)/x + (log x) xx 1`
= 1 + log x
and
f"(x) = `d/dx(1 + log x)`
= `0 + (1)/x = (1)/x`
Now, f'(x) = 0, if 1 + log x = 0
i.e. if log x = – 1 = – log e
i.e. if log x = `log(e^-1) = log(1)/e`
i.e. if x = `(1)/e`
When `x = (1)e, f"(x) = (1)/((1/e)` = e > 0
∴ by the second derivative test, f is minimum at x = `(1)/e`.
Minimum value of f at x = `(1)/e`
= `(1)/elog(1/e)`
= `(1)/e.log(e^-1)`
= `(1)/e.(-1)log e`
= `-(1)/e`. ...[∵ log e = 1]
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