# What is the absolute minimum of f(x)=xlnx?

Minimum Point at $\left(\frac{1}{e} , - \frac{1}{e}\right)$

#### Explanation:

the given $f \left(x\right) = x \cdot \ln x$

obtain the first derivative $f ' \left(x\right)$ then equate to zero.

$f ' \left(x\right) = x \cdot \left(\frac{1}{x}\right) + \ln x \cdot 1 = 0$

$1 + \ln x = 0$

$\ln x = - 1$

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

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

Solving for $f \left(x\right)$ at $x = \frac{1}{e}$

$f \left(x\right) = \left(\frac{1}{e}\right) \cdot \ln \left(\frac{1}{e}\right)$

$f \left(x\right) = \left(\frac{1}{e}\right) \cdot \left(- 1\right)$

$f \left(x\right) = - \frac{1}{e}$

so the point $\left(\frac{1}{e} , - \frac{1}{e}\right)$ is located at the 4th quadrant which is a minimum point.