How do you graph and find the discontinuities of $f(x)=1/x^2$ ?

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Answer 1 · 2015-10-03T21:58:12

$f(x)$ is defined and $f(x) > 0$ for all $x in (-oo, 0) uu (0, oo)$ and is continuous on this domain.

$f(x)$ has a horizontal asymptote $y = 0$ and vertical asymptote $x = 0$ .

$f(0)$ is undefined, so $0$ is not part of the domain. For all other Real values of $x$ , $f(x)$ is well defined.

$f(-1) = f(1) = 1/1 = 1$

As $x -> +-oo$ , $f(x) -> 0$

As $x -> 0$ , $f(x) -> oo$

For all $c in (-oo, 0) uu (0, oo)$ we find $lim_(x->c) f(x)$ exists and is equal to $f(c)$ .

The graph looks like this...

graph{1/x^2 [-9.96, 10.04, -2, 8]}

How do you graph and find the discontinuities of f(x)=1/x^2f(x)=1/x^2?