Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would I find the f intervals for the following two functions.


using the chain rule I got $(x-2)^2(2)(x+1)(1)+(2)(x-2)(1)(x+1)^2$

then I got f decrease $(-\infty,-1]$

and f increase $[2,\infty)$

but the area between -1 and 2 in confusing me.

My second function is


differentiating I got $\frac{x^3-8}{x^3}$

so I got f increase $(\infty,0)$ and $[2,\infty)$

f decrease $(0,2]$

but did I do this correctly.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Your first function is decreasing $(-\infty, -1),$ and increasing on $(2, \infty)$. And it is increasing, then decreasing on $(-1, 2)$. It increases on $(-1, 1/2)$ and decreases on $(1/2, 2).$

This can be seen if you draw the graph:

enter image description here

It helps in cases like this to find the zeros of the derivative: where $f'(x) = 0$. It also helps to expand, and factor you derivative after computing it:

$$f'(x) = 4x^3 - 6x^2 - 6x +4 = 2(x-2)(x+1)(2x - 1)$$ $$f'x)=0 \implies x = 2, x = -1, x = 1/2$$

For your section function: $$f(x) = x+\frac 4{x^2}$$

Again, we graph the function to get an intuitive idea about what's happening (note that the function is not defined at $x = 0$):

enter image description here

$$f'(x) = 1 - \frac{8}{x^3}=\frac{x^3 - 8}{x^3} = \frac{(x-2)(x^2 + 2x + 4)}{x^3}$$

Note the derivative has a sole zero at $x = 2$.; another critical point is $x = 0$. Note the function has a vertical asymptote at $x = 0$. (Why?). We see that the function is strictly increasing on the interval $(-\infty, 0)$ and strictly decreasing on the interval $(0, 2)$, and strictly increasing on the interval $(2, \infty)$. So your conclusions here are correct, except that you do not want to include $2$ in your intervals at which the function is increasing or decreasing, since at $x = 2$, $f'(x) = 0$, hence not increasing nor decreasing.

share|improve this answer
my question is how would I describe that it increases and decrease in the (-1,2) interval –  Fernando Martinez Mar 2 '13 at 18:27
for the zeroes I got-1 and 2 maybe there is another one? –  Fernando Martinez Mar 2 '13 at 18:28
I graphed the wrong function. Now I've got it right. There's an additional zero at x = 1/2, check the factoring above. –  amWhy Mar 2 '13 at 18:31
I have a quick question how did you get (2x-1) that part when taking the derivative. –  Fernando Martinez Mar 2 '13 at 18:35
I expanded out the derivative (which you computed correctly), and factored. Sometimes it helps to factor so you can see ALL the zeros. –  amWhy Mar 2 '13 at 18:37

As you did $f'(x)$ correctly, we have $$f'(x)=2(x-2)(x+1)(2x-1)$$ When:

  • $x\le -1~~$, $(x-2)\leq0$ and then two terms $(x+1)$ and $(2x-1)$ are negative. So $f'(x)$ is negative.

  • $x\le\frac{1}2~~$, $(x-2)\leq0$ and $(x+1)>0$ and $(2x-1)<0$ is negative, so $f'(x)$ is positive.

  • $x\le2~~$, $(x-2)\leq0$ and then two terms $(x+1)$ and $(2x-1)$ are positive, so $f'(x)$ is nagative again.

  • And if $x>2$ all terms are positive and so $f'(x)$ is positive again.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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