<s>[INST] 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! 
 [/INST] $$\frac{x+1}{x^2+1}\geq0.$$
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
$$\left[0,\frac{1+\sqrt2}{2}\right]$$ 
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