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]}