Range in an Interval I want to find the range of the function, $$f(x) = {{x + 1} \over {{x^2} + 1}}\,\,\;{\rm{when }}\;x \in \left[ { - 1,1} \right].\;\;\;$$
I have isolated $x$ as - $$x = {{1 \pm \sqrt {1 - 4y(y - 1)} } \over {2y}}$$
But I am unable to solve this inequality, $$ - 1 \le {{1 \pm \sqrt {1 - 4y(y - 1)} } \over {2y}} \le 1$$
as it involved $\, \pm \sqrt {} \,$ part. Please tell me how to proceed from here! 
 A: $$\frac{x+1}{x^2+1}\geq0.$$
The equality occurs for $x=-1$, which says that $0$ is a minimum.
Now, by AM-GM for $x\neq-1$ we obtain:
$$\frac{x+1}{x^2+1}=\frac{x+1}{(x+1)^2-2(x+1)+2}=\frac{1}{x+1+\frac{2}{x+1}-2}\leq\frac{1}{2\sqrt2-2}=\frac{1}{2}(\sqrt2+1).$$
The equality occurs for $x+1=\frac{2}{x+1}$, which says that we got a maximal value because for $x=-1$ we get a value $0$.
Id est, the answer is
$$\left[0,\frac{1+\sqrt2}{2}\right]$$ 
A: Here is a method that avoids calculus completely.
Write $y = x+1$, then $\frac{x+1}{x^2+1} = \frac{y}{y^2-2y+2}$.  Then note that $$\frac{y^2-2y+2}{y} = y-2+\frac{2}{y}.$$ Keeping in mind that the range of $y$ is $[0, 2]$, $y+\frac{2}{y}$ approaches $\infty$ as $y \to 0$, and the minimum value over positive $y$ is $2\sqrt{2}$ when $y = \sqrt{2}$ by AM-GM ($y+\frac{2}{y} \geq 2\sqrt{y\cdot\frac{2}{y}}$.)
So $y-2+\frac{2}{y}$ ranges from $[2\sqrt{2}-2, \infty]$ when $y$ is in $[0, 2]$.  Hence the original fraction which is the reciprocal ranges from $[0, \frac{1}{2\sqrt{2}-2}]$, hitting the bounds at $x = -1$ and $x = \sqrt{2}-1$.
