# Intervals on which function is increasing and decreasing

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

• You are asked to determine wether $p'(I_i)$ is less than or greater than zero for each one of the intervals you have determined, $I_i$. – Dmoreno Aug 15 '14 at 12:12
• That is exactly what I am asking for. How to determine whether the derivative will be positive or negative on these intervals? – user124471 Aug 15 '14 at 12:14
• Are there any restrictions on the value of $q$? – Dmoreno Aug 15 '14 at 12:15
• nope, it just says, $q$ is a prime number – user124471 Aug 15 '14 at 12:16
• Oh, you're right. I see it now in the question. Well, we know that $x=0 \in I_2 \equiv (-\sqrt{q}/5^{1/4},\sqrt{q}/5^{1/4})$ and we have $p'(0) = -q^2 < 0$ and therefore $p(x)$ is decreasing on $I_2$. – Dmoreno Aug 15 '14 at 12:20

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.