A calculus problem by Ankit Nigam

Let's avoid having to use derivatives for this one.

Since $x^2+1$ is always greater than zero, $f(x)$ is always defined. $f(x)=1-\frac2{x^2+1}$ , meaning we have to maximise $\frac2{x^2+1}$ meaning we have to minimise $x^2+1$ . Hence we need $x=0$ , and so $f(0)=\frac{0^2-1}{0^2+1}=\boxed{-1}$ .

Find the asymptotes of the graph, along with the axis - intercepts which can be done as follows: $lim_{x \rightarrow\pm \infty } \frac{x^2 - 1}{x^2 +1} = lim_{x \rightarrow\pm \infty } \frac{1-1/x^2}{1+1/x^2} = 1 \ (from \ below)$ with intercepts (0,-1), (1,0) and (-1,0). Now if you draw the graph you should get a curved V-shape with a minimum point of -1.

$y=\frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 }$

$x=\sqrt {\frac{y+1}{1-y}}$ Hence we get $-1 \le y <1$