Question

Determine the maximum and minimum value of the following function.

f(x) = x log x

Answer 1

f(x) = x log x

∴ f'(x) =${\displaystyle \text{x}\frac{\text{d}}{\text{dx}}\left(\log \text{x}\right)+\log \text{x}\frac{\text{d}}{\text{dx}}\left(\text{x}\right)}$

${\displaystyle =\text{x}\times \frac{1}{\text{x}}+\log \text{x}\times 1=1+\log \text{x}}$

and f''(x) = ${\displaystyle 0+\frac{1}{\text{x}}=\frac{1}{\text{x}}}$

Consider, f'(x) = 0

∴ 1 + log x = 0

∴ log x = - 1

∴ log x = - log e = log e^-1 = log ${\displaystyle \left(\frac{1}{\text{e}}\right)}$

∴ x = ${\displaystyle \frac{1}{\text{e}}}$

For x = ${\displaystyle \frac{1}{\text{e}}}$

${\displaystyle f\prime \prime \left(\frac{1}{\text{e}}\right)=\frac{1}{\frac{1}{\text{e}}}=\text{e}>0}$

∴ f(x) attains minimum value at x = ${\displaystyle \frac{1}{\text{e}}}$.

∴ Minimum value = ${\displaystyle \text{f}\left(\frac{1}{\text{e}}\right)=\frac{1}{\text{e}}\log \left(\frac{1}{\text{e}}\right)=\frac{1}{\text{e}}{\log \text{e}}^{-1}}$

${\displaystyle =\left(\frac{-1}{\text{e}}\right)\left(1\right)=\left(\frac{-1}{\text{e}}\right)}$

∴ The function f(x) has minimum value ${\displaystyle \frac{-1}{\text{e}}}$ at x = ${\displaystyle \frac{1}{\text{e}}}$.