# How do you find the local max and min for x^5 ln x?

Feb 22, 2017

The function ${x}^{5} \ln x$ has a local minimum for $x = {e}^{- \frac{1}{5}}$

#### Explanation:

Given:

$f \left(x\right) = {x}^{5} \ln x$

we note that the function is defined only for $x > 0$ and that using l'Hospital's rule:

${\lim}_{x \to {0}^{+}} {x}^{5} \ln x = {\lim}_{x \to {0}^{+}} \ln \frac{x}{\frac{1}{x} ^ 5} = {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} \frac{1}{x} ^ 5} = {\lim}_{x \to {0}^{+}} \left(\frac{1}{x}\right) \left(\frac{1}{- \frac{5}{x} ^ 6}\right) = {\lim}_{x \to {0}^{+}} - {x}^{5} / 5 = 0$

evaluate the derivative of the function:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{5} \ln x\right) = \frac{d}{\mathrm{dx}} \left({x}^{5}\right) \ln x + {x}^{5} \frac{d}{\mathrm{dx}} \left(\ln x\right) = 5 {x}^{4} \ln x + {x}^{5} / x = {x}^{4} \left(5 \ln x + 1\right)$

We can then find critical points for the function solving the equation:

${x}^{4} \left(5 \ln x + 1\right) = 0$

since $x > 0$ the only solution is:

$5 \ln x + 1 = 0 \implies x = {e}^{- \frac{1}{5}}$

Now we can evaluate the inequality:

$f ' \left(x\right) > 0$

Since ${x}^{4} > 0$ for $x > 0$ the sign of the derivative is the same as the sign of $5 \ln x + 1$, which means:

$f ' \left(x\right) > 0$ for $x \in \left({e}^{- \frac{1}{5}} , + \infty\right)$

and $f ' \left(x\right) < 0$ for $x \in \left(0 , {e}^{- \frac{1}{5}}\right)$

This shows that the function is strictly decreasing in $\left(0 , {e}^{- \frac{1}{5}}\right)$ and strictly increasing in $\left({e}^{- \frac{1}{5}} , + \infty\right)$, which means that:

$x = {e}^{- \frac{1}{5}} \cong 0.81873 \ldots$

is a local minimum.

graph{x^5lnx [-1.25, 1.25, -0.625, 0.625]}