# Function f(x)=6+6/x+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.

Why do I get different results when testing for increasing/decreasing intervals of a function here?
I have the function $f\left(x\right)=6+\frac{6}{x}+\frac{6}{{x}^{2}}$ and I want to find the intervals where it increases or decreases. The problem is that when I find ${f}^{\prime }\left(x\right)=0$, 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, ${f}^{\prime }\left(-1\right)>0$ and ${f}^{\prime }\left(5\right)<0$.
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:
$\frac{d}{dx}\frac{6}{x}=\frac{\left[0\right]-\left[6\cdot 1\right]}{{x}^{2}}=\frac{-6}{{x}^{2}}$
$\frac{d}{dx}\frac{6}{{x}^{2}}=\frac{\left[0\right]-\left[6\cdot 2x\right]}{\left({x}^{2}{\right)}^{2}}=\frac{-12}{{x}^{3}}$
${f}^{\prime }\left(x\right)=\frac{-6}{{x}^{2}}-\frac{-12}{{x}^{3}}$
How I found the asymptotes:
When you combine the fractions of f(x), $f\left(x\right)=\frac{6{x}^{2}+6x+6}{{x}^{2}}$. 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, $f\left(x\right)=\frac{6}{1}$. So f(x) has a horizontal asymptote at $y=6$.
How I found the value of ${f}^{\prime }\left(x\right)=0$:
$\frac{-6}{{x}^{2}}-\frac{12}{{x}^{3}}=0$
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.
$x=\frac{-12}{6}=-2$
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Step 1
Notice that ${f}^{\mathrm{\prime }}\left(x\right)=-\frac{6}{{x}^{2}}-\frac{12}{{x}^{3}}=-\frac{6}{{x}^{3}}\left(x+2\right)$.
So f′ is undefined at $x=0$ and f has slope 0 at $x=-2$.
Step 2
You should therefore check for increasing/decreasing on the intervals $\left(-\mathrm{\infty },-2\right),\phantom{\rule{thinmathspace}{0ex}}\left(-2,0\right)$ and $\left(0,\mathrm{\infty }\right)$.