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

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.