# How do you find the critical numbers for f(x) = x^(-2)ln(x) to determine the maximum and minimum?

Apr 15, 2017

$0$ is the minimum

${e}^{\frac{1}{2}}$ is the maximum

#### Explanation:

Take the derivative of $f \left(x\right)$. You will need to use the product rule. You also need to know that the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$:

$f ' \left(x\right) = {x}^{-} 2 \left(\frac{1}{x}\right) + \ln \left(x\right) \left(- 2 {x}^{-} 3\right)$

$f ' \left(x\right) = {x}^{-} 3 - 2 \ln \left(x\right) {x}^{-} 3$

Factor out a ${x}^{-} 3$

$f ' \left(x\right) = {x}^{-} 3 \left(1 - 2 \ln \left(x\right)\right)$

Solve for $x$:

$x = 0 , {e}^{\frac{1}{2}}$

Plug in these numbers into the initial equation:

$f \left(0\right) = 0 \ln \left(0\right) = D N E$.

We'll need to use limits for this:

${\lim}_{x \rightarrow 0} {x}^{-} 2 \ln \left(x\right) = - \infty$. This is definitely a minimum

$f \left({e}^{\frac{1}{2}}\right) = \left({e}^{-} 1\right) \left(\ln \left({e}^{\frac{1}{2}}\right)\right) = \frac{1}{2 e}$. This is a maximum because plugging in anything before or after this will give a value less than this.