How to determine monotonicity intervals for a given function?

Question

Given the function $$ f(x) = \frac{\ln(x)}{x^s}, \ \ \ x > 0 $$ where $s$ is real number $> 1$ I have to find the monotony intervals for the function. In high school I used to find $f'(x)$, solve $f'(x) = 0$ and pick values of x less and greater than the x value found solving $f'(x) = 0$ to find where $f'(x) > 0 $ and $f'(x) < 0 $. What do I do now? I have found that $$ f'(x) = \frac{1}{x^{s+1}} -sx^{s-1} \ln(x) $$ using both the Product and the Chain Rule but it is not possible to solve $$ f'(x) = 0 $$ unless I can pick a $s>1$. Am I allowed to do that? Or do I maybe have to use the fact that I know that $\frac{1}{x^s}$ converges faster towards zero than $\ln(x)$ does to infinity.

Thank you in advance for your help.

Answer 1

Note that $f'(x) = \frac{1-s \ln(x) }{x^{s+1}}$. It suffices to check when $f'(x) > 0$ and $f'(x) < 0$, which reduces to $1 - s \ln(x) > 0$ and $1-s\ln(x) < 0$ respectively. Performing the calculation yields that $x \in (0,e^{\frac{1}{s}}) \implies f'(x) > 0 $ and $x \in (e^{\frac{1}{s}}, \infty) \implies f'(x) < 0$. So the monotonic intervals of $f$ are: $$ \begin{cases} (e^{\frac{1}{s}}, \infty) & \text{ Monotonic Decreasing } \\ (0,e^{\frac{1}{s}}) & \text { Monotonic Increasing } \end{cases} $$

Answer 2

Solving for $f'(x) = 0$ is easier from the correct derivative. \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \frac{\ln x}{x^s} &= \frac{\mathrm{d}}{\mathrm{d}x} x^{-s} \ln x \\ &= x^{-s} \left( \frac{\mathrm{d}}{\mathrm{d}x} \ln x \right) + \left( \frac{\mathrm{d}}{\mathrm{d}x} x^{-s}\right) \ln x \\ &= x^{-s} \frac{1}{x} + (-s)x^{-s-1}\ln x \\ &= x^{-s-1} + (-s)x^{-s-1}\ln x \\ &= x^{-s-1}(1 - s \ln x) \text{.} \\ \end{align*}

The product of two (or more) things is zero when one (or more) of them is zero, so we solve $x^{-s-1} = 0$ and $1 - s \ln x = 0$. Since $s > 1$, the power $-s-1 < -2$, so the former has no solution. The latter has the solution $x = \mathrm{e}^{1/s}$.