How do you solve #x/(x^2 - 16)>0#?

1 Answer
Jun 17, 2016

Answer:

#x/(x^2-16) > 0 iff x in (-4,0)uu(4,oo)#

Explanation:

An expression #f(x)/g(x)# will be greater than #0# when either both #f(x)# and #g(x)# are positive or both #f(x)# and #g(x)# are negative.

Looking at the signs of the given functions, we have

#x < 0 if x in (-oo,0)#
#x>0 if x in (0,oo)#

and

#x^2-16 < 0 if x in (-4,4)#
#x^2-16 > 0 if x in (-oo,-4)uu(4,oo)#

If we compare the signs, then, we have both being negative on #(-4,0)# and both being positive on #(4,oo)#.

Thus, we have #x/(x^2-16) > 0 iff x in (-4,0)uu(4,oo)#