# Finding f intervals

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

$f(x)=(x-2)^2(x+1)^2$

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

$f(x)=x+\frac{4}{x^2}$

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.

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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:

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 $x = 0$):

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

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.

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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.

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