# What are the values and types of the critical points, if any, of f(x)=xlnx?

Apr 6, 2018

Local minimum: $\left(\frac{1}{e} , - \frac{1}{e}\right)$

#### Explanation:

Take the first derivative, noting that the domain of the original function is $\left(0 , \infty\right) .$

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

$f ' \left(x\right) = 1 + \ln x$

The domain of the first derivative is also $\left(0 , \infty\right) ,$ so there won't be any critical points where the first derivative does not exist.

Set to zero and solve for $x .$

$1 + \ln x = 0$

$\ln x = - 1$

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

Recalling that ${e}^{\ln} x = x :$

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

This is not a very complex function, so we can take the second derivative and apply the Second Derivative Test, which tells us that if $x = a$ is a critical value and $f ' ' \left(a\right) > 0 ,$ then it is a minimum, and if $f ' ' \left(a\right) < 0 ,$ then it is a maximum.

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

$f ' ' \left(\frac{1}{e}\right) = \frac{1}{\frac{1}{e}} = e > 0$

Thus, we have a local minimum at $x = \frac{1}{e} .$ Find the $y -$coordinate:

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

Local minimum: $\left(\frac{1}{e} , - \frac{1}{e}\right)$