# What are the critical values of f(x) = xlnx ?

Mar 19, 2017

$f \left(x\right)$ is defined for $x > 0$
$f \left(x\right)$ has a zero at $x = 1$
$f \left(x\right)$ has an absolute minimum of$- \frac{1}{e}$ at $x = \frac{1}{e}$

#### Explanation:

$f \left(x\right) = x \ln x$

Since $\ln x$ is defined for $x > 0 \to f \left(x\right)$ is defined for $x > 0$

To find the zero: $f \left(x\right) = 0$

$x \ln x = 0 \to \ln x = 0$ Since $x > 0$

$\ln x = 0 \to x = 1$

$f ' \left(x\right) = x \cdot \frac{1}{x} + \ln x$ (product rule and standard differential)

$f ' \left(x\right) = 0 \to \ln x = - 1$

$x = {e}^{-} 1$

$x = \frac{1}{e} \cong 0.3679$

$f ' ' \left(x\right) = \frac{1}{x}$ which is $> 0 \forall x > 0$

Hence: ${f}_{\min} = f \left({e}^{-} 1\right) = - {e}^{-} 1 \cong - 0.3679$

These points can be seen on the graph of $f \left(x\right)$ below:

graph{xlnx [-10, 10, -5, 5]}