What is the range of #8/(x^2+2)#?

1 Answer
Jun 15, 2015

Answer:

#x^2+2# has range #[2, oo)#, so #8/(x^2+2)# has range #(0,4]#

Explanation:

#f(x) = 8/(x^2+2)#

#f(0) = 8/2 = 4#

#f(-x) = f(x)#

As #x->oo# we have #f(x)->0#

#f(x) > 0# for all #x in RR#

So the range of #f(x)# is at least a subset of #(0, 4]#

If #y in (0, 4]# then #8/y >= 2# and #8/y - 2 >= 0#

so #x_1 = sqrt(8/y - 2)# is defined and #f(x_1) = y#.

So the range of #f(x)# is the whole of #(0, 4]#