# Find the f intervals for the following two functions. f(x)=(x-2)^2(x+1)^2, f(x)=x+4/x^2.

Finding f intervals
How would I find the f intervals for the following two functions.
$f\left(x\right)=\left(x-2{\right)}^{2}\left(x+1{\right)}^{2}$
using the chain rule I got $\left(x-2{\right)}^{2}\left(2\right)\left(x+1\right)\left(1\right)+\left(2\right)\left(x-2\right)\left(1\right)\left(x+1{\right)}^{2}$
then I got f decrease $\left(-\mathrm{\infty },-1\right]$
and f increase $\left[2,\mathrm{\infty }\right)$
but the area between -1 and 2 in confusing me.
My second function is $f\left(x\right)=x+\frac{4}{{x}^{2}}$
differentiating I got $\frac{{x}^{3}-8}{{x}^{3}}$
so I got f increase $\left(\mathrm{\infty },0\right)$ and $\left[2,\mathrm{\infty }\right)$
f decrease (0, 2]
but did I do this correctly.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Bianca Chung
Step 1
As you did f′(x) correctly, we have ${f}^{\prime }\left(x\right)=2\left(x-2\right)\left(x+1\right)\left(2x-1\right)$
Step 2
When:
- , $\left(x-2\right)\le 0$ and then two terms $\left(x+1\right)$ and $\left(2x-1\right)$ are negative. So f′(x) is negative.
- , $\left(x-2\right)\le 0$ and $\left(x+1\right)>0$ and $\left(2x-1\right)<0$ is negative, so f′(x) is positive.
- , $\left(x-2\right)\le 0$ and then two terms $\left(x+1\right)$ and $\left(2x-1\right)$ are positive, so f′(x) is nagative again.
- And if $x>2$ all terms are positive and so f′(x) is positive again.