Let $f: \mathbb{R}\setminus\{0\}\to\mathbb{R}$ be given by $f(x) = x + \frac{1}{x}$. Find $f((0,2))$. So for the question I know it's asking for the interval of the codomain that corresponds to the interval $(0,2)$ of the domain. I know that the upper part of the interval is $(y,2.5)$ but I don't know what $y$ would be. Since $0$ is not even in the domain of $f$, could $y$ be the empty set? This does not seem right to me though. Any help is appreciated.
First, find $f(x)$ at the boundaries.

At the lower bound $x = 0$
$$\lim_{x\to 0^+} \left(x + \frac{1}{x} \right)= \infty $$

At the upper bound $x = 2$
$$ \lim_{x\to 2^-} \left(x + \frac{1}{x} \right)= \infty = 2.5 $$

Second, check to see if the function has any local minima or maxima
$$ f'(x) = 1 - \frac{1}{x^2} = \frac{x^2-1}{x^2}$$

$ f'(x) = 0$ if $x = \pm 1$. One of these points, $x = 1$, lies within the interval of consideration. This is a local minimum as the function is very large and positive near $0^+$. This minimum value is $f(1) = 2$

To conclude, we found that $2 \le f(x) &lt; \infty$ in $(0,2)$, so the codomain is $[2, \infty)$
Hint: $(x-1)^2=x^2-2x+1\geq 0$. What does this imply about 
$$
x+\frac{1}{x}
$$
?