How do you graph $f(x)=2/(x^2+1)$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Answer 1 · 2017-02-01T11:55:20

$f(x)$ has an absolute maximum of 2 at $x=0$
$f(x) -> 0$ as $x-> +- oo$

$f(x) = 2/(x^2+1)$

Since $x^2 + 1 > 0 forall x in RR$ there exists no holes in $f(x)$

Also, $lim_"x-> +-oo" f(x) = 0$

$f'(x) = (-4x)/(x^2+1)^2$

For a maximum or minimum value; $f'(x) = 0$

$:. (-4x)/(x^2+1)^2 = 0 -> x=0$

$f(0) = 2/(0+1) = 2$

Since $f''(0) < 0$ $f(0) = 2$ is a maximum of $f(x)$

The critical points of $f(x)$ can be seen on the graph below:
graph{2/(x^2+1) [-5.55, 5.55, -2.772, 2.778]}

How do you graph f(x)=2/(x^2+1)f(x)=2/(x^2+1) using holes, vertical and horizontal asymptotes, x and y intercepts?