# Let p(x)=x^5-q^2 x-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′(x)=5x^4-q^2 and look for the critical points.

Intervals on which function is increasing and decreasing
Let $p\left(x\right)={x}^{5}-{q}^{2}x-q$, 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 ${p}^{\mathrm{\prime }}\left(x\right)=5{x}^{4}-{q}^{2}$ and look for the critical points.
$5{x}^{4}-{q}^{2}=0⟺x=±\frac{\sqrt{q}}{\sqrt[4]{5}}$
Hence we have to investigate the behavior of p′(x) for each of these intervals $\left(-\mathrm{\infty },-\frac{\sqrt{q}}{\sqrt[4]{5}}\right)$, $\left(-\frac{\sqrt{q}}{\sqrt[4]{5}},\frac{\sqrt{q}}{\sqrt[4]{5}}\right)$ and $\left(\frac{\sqrt{q}}{\sqrt[4]{5}},\mathrm{\infty }\right)$ 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}} and strictly increasing again for $x>\frac{\sqrt{q}}{\sqrt[4]{5}}$.
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indivisast7
Step 1
Assuming we have $q>0$, the derivative is positive whenever $5{x}^{4}>{q}^{2}$, or equivalently $\sqrt{5}{x}^{2}>q$.
This can happen two ways, either with $\sqrt[4]{5}x>\sqrt{q}$ if x is positive, or $-\sqrt[4]{5}x>\sqrt{q}$ if x is negative.
Step 2
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.