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]}