<s>[INST] Why do I get different results when testing for increasing/decreasing intervals of a function here? I have the function $f(x)=6+\frac{6}{x}+\frac{6}{x^2}$ and I want to find the intervals where it increases or decreases. The problem is that when I find $f'(x)=0$, which becomes $x=-2$. Once I put $-2$ on a number line, and find whether the numbers higher and lower than $-2$ produce positive or negative numbers when plugged into $f'(x)$, I get both positives and negatives for different numbers when $x > -2$. For example, $f'(-1) > 0$ and $f'(5) < 0$.
I am not sure if this is because the function has a vertical asymptote at $x=0$ and a horizontal attribute at $y=6$. If so, I'd like to know what I need to do to handle them.
How I found the first derivative:
$\frac{d}{dx} \frac{6}{x}=\frac{[0] - [6\cdot1]}{x^2}=\frac{-6}{x^2}$
$\frac{d}{dx}\frac{6}{x^2}=\frac{[0]-[6 \cdot 2x]}{(x^2)^2} =\frac{-12}{x^3}$
$f'(x)=\frac{-6}{x^2}-\frac{-12}{x^3}$
How I found the asymptotes:
When you combine the fractions of $f(x)$, $f(x)=\frac{6x^2+6x+6}{x^2}$. When set equal to zero, $x^2$ has an x-value of zero. Therefore, $f(x)$ has a vertical asymptote at $x=0$.
Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, $f(x)=\frac{6}{1}$. So $f(x)$ has a horizontal asymptote at $y=6$.
How I found the value of $f'(x)=0$:
$\frac{-6}{x^2}-\frac{12}{x^3}=0$
Add $\frac{12}{x^3}$ to both sides
$\frac{6}{x^2}=\frac{-12}{x^3}$
Multiply both sides by $x^3$.
$6x=-12$
Divide both sides by $6$.
$x = \frac{-12}{6}=-2$
 [/INST] Notice that
$$ f^\prime(x)=-\frac{6}{x^2} -\frac{12}{x^3}=-\frac{6}{x^3}(x+2)$$
So $f^\prime$ is undefined at $x=0$ and $f$ has slope $0$ at $x=-2$.
You should therefore check for increasing/decreasing on the intervals $(-\infty,-2),\,(-2,0)$ and $(0,\infty)$
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