# What intervals is this function: f(x)= x^5(lnx) increasing and decreasing?

Dec 29, 2017

$f$ strictly decreasing in $\left(0 , {e}^{- \frac{1}{5}}\right]$ & $f$ strictly increasing in $\left[{e}^{- \frac{1}{5}} , + \infty\right)$

#### Explanation:

$f \left(x\right) = {x}^{5} \ln x$ , ${D}_{f} = \left(0 , + \infty\right)$

For $x$$\in$${D}_{f}$,
$f ' \left(x\right) = \left({x}^{5} \ln x\right) ' =$

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

$5 {x}^{4} \ln x + {x}^{5} / x$ $=$

$5 {x}^{4} \ln x + {x}^{4}$ $=$

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

• $f ' \left(x\right) = 0$ $\iff$ ${x}^{4} = 0$ $\iff$ $x = 0$ -> Impossible because $x > 0$

$5 \ln x + 1 = 0$ $\iff$ $5 \ln x = - 1$ $\iff$ $\ln x = - \frac{1}{5}$ $\iff$ $x = {e}^{- \frac{1}{5}}$ $\cong$ $0.81$

For $x = \frac{1}{2} = 0.5$ ---> $f ' \left(\frac{1}{2}\right) = 5 \cdot \frac{1}{2} ^ 4 \ln \left(\frac{1}{2}\right) + \frac{1}{2} ^ 4$ $=$
$\frac{5}{16} \left(\ln 1 - \ln 2\right) + \frac{1}{16}$ $=$ $- \ln 2 \cdot \frac{5}{16} + \frac{1}{16}$ $< 0$

because $\ln 2 \cong 0.7$
$\frac{1}{16} = 0.0625$

For $x = 1$ ----> $f ' \left(1\right) = 1 > 0$

As a result we have,

• $f$ continuous in $\left(0 , {e}^{- \frac{1}{5}}\right]$ with $f ' \left(x\right) < 0$ when $x$$\in$$\left(0 , {e}^{- \frac{1}{5}}\right)$ so $f$ strictly decreasing in $\left(0 , {e}^{- \frac{1}{5}}\right]$

• $f$ continuous in $\left[{e}^{- \frac{1}{5}} , + \infty\right)$ with $f ' \left(x\right) > 0$ when $x$$\in$$\left({e}^{- \frac{1}{5}} , + \infty\right)$ so $f$ strictly increasing in $\left[{e}^{- \frac{1}{5}} , + \infty\right)$