Intervals on which function is increasing and decreasing

Question

Let $p(x)=x^5-q^2x-q$ , where $q$ is a prime number. I want to understand how to determine when the function will be decreasing and increasing on the intervals given below.

We compute $p^{\prime}(x)=5x^4-q^2$ and look for the critical points.

$5x^4-q^2=0\Longleftrightarrow x=\pm \frac{\sqrt{q}}{\sqrt[4]{5}}$

Hence we have to investigate the behavior of $p^{\prime}(x)$ for each of these intervals $(-\infty,-\frac{\sqrt{q}}{\sqrt[4]{5}})$, $(-\frac{\sqrt{q}}{\sqrt[4]{5}},\frac{\sqrt{q}}{\sqrt[4]{5}})$ and $(\frac{\sqrt{q}}{\sqrt[4]{5}},\infty)$ this will indicate when the function will be increasing and decreasing. How can this be determined when the expression $\frac{\sqrt{q}}{\sqrt[4]{5}}$ contains a prime number???

The answer should be : the function will be increasing for $x<\frac{\sqrt{q}}{\sqrt[4]{5}}$ and strictly decreasing for $-\frac{\sqrt{q}}{\sqrt[4]{5}}<x<\frac{\sqrt{q}}{\sqrt[4]{5}}$ and strictly increasing again for $x>\frac{\sqrt{q}}{\sqrt[4]{5}}$

Can someone explain this last part? Thank you

Answer 1

Assuming we have $q\gt 0$, the derivative is positive whenever $5x^4\gt q^2$, or equivalently $\sqrt 5 x^2\gt q$

This can happen two ways, either with $\sqrt[4]5x\gt \sqrt q$ if $x$ is positive, or $-\sqrt[4]5x\gt \sqrt q$ if $x$ is negative.

For a negative derivative, the inequalities are reversed.

I am not sure what you mean by "contains a prime number" - the function is presumably being taken over the real numbers, over which the [positive real] square and fourth roots of non-negative real numbers are well-defined.