Does $1 + \frac{1}{x} + \frac{1}{x^2}$ have a global minimum, where $x \in \mathbb{R}$? Does the function 
$$f(x) = 1 + \frac{1}{x} + \frac{1}{x^2},$$
where $x \in {\mathbb{R} \setminus \{0\}}$, have a global minimum?
I tried asking WolframAlpha, but it appears to give an inconsistent result.
 A: Considered as a polynomial in $\frac1x$,
$$
1+\frac1x+\frac1{x^2}=\left(\frac1x+\frac12\right)^2+\frac34
$$
has a global minimum of $\frac34$ at $x=-2$.
A: $g(x)=1+\frac{1}{x}+\frac{1}{x^2},x\ne 0$
$g'(x)=-\frac{1}{x^2}-2\frac{1}{x^3}=0\Rightarrow \frac{-2}{x^3}=\frac{1}{x^2}\Rightarrow x=-2$
$g''(x)=\frac{2}{x^3}+\frac{6}{x^4}|_{x=-2}> 0$
$\Rightarrow g(-2)$ is the minimum.
Edit:This shows that $g$ has a local minimum at $x=-2$ but this is also the global minimum because 
$g(x)-g(-2)=1+\frac{1}{x}+\frac{1}{x^2}-1-(\frac{-1}{2})-\frac{1}{2^2}=\frac{x+2}{2x}+\frac{(2-x)(2+x)}{4x^2}=(x+2)\frac{2x+2-x}{4x^2}=\frac{(x+2)^2}{4x^2}\ge0$
so $g(x)\ge g(-2)$ so $g(-2)$ is a global minimum.
A: Hint: Take the derivative and find where it's zero.
A: Methodically in my opinion it would be as follows. Let $ A  = \mathbb{R}\backslash \{0\} $ be the domain of definition of the function $f$. $f$ is continous and differentiable in $A$. $A$ is an open unbounded set. Since we have no boundary, we don't have to check if the boundary points are local extrema. So it remains to see if the function derivative becomes zero for some point $x\in A$. 
$$
f'(x) = -\frac{1}{x^{2}} - \frac{2}{x^3} = -\frac{x+2}{x^3}
$$
which is actually zero for $x=-2$. In this point we don't know if the function attains a maximum or a minimum, so we check with the second derivative and we evaluate it for $x=-2$, thus
$$
f''(x)
=
\frac{2(x+3)}{x^4}
$$
and we get $f''(-2) >0$ and therefore we have found a minimum at $x=-2$. I hope it helped to see it without a graph ;)
