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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

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  • $\begingroup$ 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, $I_i$. $\endgroup$ – Dmoreno Aug 15 '14 at 12:12
  • $\begingroup$ That is exactly what I am asking for. How to determine whether the derivative will be positive or negative on these intervals? $\endgroup$ – user124471 Aug 15 '14 at 12:14
  • $\begingroup$ Are there any restrictions on the value of $q$? $\endgroup$ – Dmoreno Aug 15 '14 at 12:15
  • $\begingroup$ nope, it just says, $q$ is a prime number $\endgroup$ – user124471 Aug 15 '14 at 12:16
  • $\begingroup$ 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 $I_2$. $\endgroup$ – Dmoreno Aug 15 '14 at 12:20
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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.

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  • $\begingroup$ Thank you, I understand it now. $\endgroup$ – user124471 Aug 15 '14 at 13:00

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