How to find range of a rational function? How to find range of $$f(x) = x^2 + \frac{1}{x^2+1} \quad ?$$
Its domain is all real numbers. If I use calculus it is very lengthy but if I put RHS of equation equal to y I get a quadratic equation in $x^2$ but I cannot find range of y by imposing condition on discriminant as even if let us say $x$ was complex or imaginary but there is the  possibility that when it is raised to power 2 or 4 it becomes purely real?
 A: First notice that the function is even, that is $f(x) = f(-x)$.
I'd like to show that $1$ is either the upper or lower bound of the function. We have $f(0) = 1$, and solving the equation $f(x) = 1$ gives only one solution. Assume there are two real numbers $a$ and $b$ such that $f(a) < 1 < f(b)$. If any of them are negative, simply use $f(x) = f(-x)$ and change their sign, thereby assuming WLOG that they are both positive. By the intermediate value theorem there must be a $c$ between $a$ and $b$ such that $f(c) = 1$, but since $0 < c$ this is a contradiction. Therefore all real numbers in the range must be to the same side of $1$.
Since $x^2 < f(x)$ and $x^2 \to \infty$  as $x \to \infty$ we must have $f(x) \to \infty$ as  $x \to \infty$.
From this we can conclude that the range of the function is $[1;\infty[$.
A: Hint: write it as $\,(x^2+1)+\cfrac{1}{x^2+1}-1\,$ and use $\,a+\cfrac{1}{a}\ge 2\,$ for $\,a \ge 1\,$.
A: Clearly the range is all positive as both terms are positive.  As $x$ gets large it goes to $+\infty$ so you just need to find the minimum.  Take the derivative, set to zero, find the minimum, and you are done.
