# Let s>1,s in mathbb{R}, and let f be a function defined by f(x)=ln(x)/x^s, x>0 Determine the monotone intervals of f.

Show the function is decreasing in an interval
Let $s>1,s\in \mathbb{R}$, and let f be a function defined by
$f\left(x\right)=\frac{ln\left(x\right)}{{x}^{s}},x>0$
Determine the monotone intervals of f.
I note that f(x)'s domain is $\left(0,\mathrm{\infty }\right)$. I then find the derivative of f(x).
${f}^{\prime }\left(x\right)={x}^{-s-1}\left(1-s\cdot ln\left(x\right)\right)$
${f}^{\prime }\left(x\right)=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x={e}^{\frac{1}{s}}$
So ${e}^{\frac{1}{s}}$ is a critical point for f.
I'm uncertain how to evaluate the function around the critical point. I'm thinking to evaluate at the points ${e}^{\frac{0}{s}}$ and ${e}^{\frac{2}{s}}$?
So ${f}^{\prime }\left({e}^{\frac{0}{s}}\right)={e}^{{\frac{0}{s}}^{-s-1}}\left(1-s\cdot ln\left({e}^{\frac{0}{s}}\right)\right)$.
I then need to figure out if the expression above is greater, less or equal to 0. And likewise with the other critical point.
I'm not sure if I've done this correctly and I'm not sure how to make the expression for the evaluated critical point any nicer.
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Luna Wells
Explanation:
To find intervals of monotonicity you only have to known when ${f}^{\prime }>0$ and when ${f}^{\prime }<0$. No need to evaluate f at any point. Note that ${x}^{-s-1}\left(1-s\mathrm{ln}x\right)>0$ iff $\mathrm{ln}x<\frac{1}{s}$. The function is increasing in $\left(0,\frac{1}{s}\right)$ and decreasing in $\left(\frac{1}{s},\mathrm{\infty }\right)$.