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Question
Find the maximum and minimum of the following functions : f(x) = `logx/x`
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Solution
f(x) = `logx/x`
∴ f'(x) = `d/dx(logx/x)`
= `(xd/x(logx) - logxd/dx(x))/x^2`
= `(x(1/x) - (logx)(1))/x^2`
= `(1 - logx)/x^2`
and
f"(x) = `d/dx((1 - logx)/x^2)`
= `(x^2d/dx(1 - logx) - (1 - log x)d/dx(x^2))/x^4`
= `(x^2(0 - 1/x) - (1 - logx) xx 2x)/x^4`
= `(- x - 2x + 2x log x)/x^4`
= `(x(2 log x - 3))/x^4`
∴ f"(x) = `(2 log x - 3)/x^3`
Now, f'(x) = 0 if `(1 - log x)/x^2` = 0
i.e. if 1 – log x = 0, i.e. if log x = 1 = log e
i.e. if x = e
and
f"(e) = `(2log e - 3)/e^3`
= `(-1)/e^3 < 0` ...[∵ log e = 1]
∴ by the second derivative test, f(x) is maximum at x = e.
Maximum value of f at x = e
= `loge/e`
= `(1)/e`. ...[∵ log e = 1]
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