How do you solve: #(x^2-9)/(x^2-1) < 0#?
1 Answer
Explanation:
Your inequality looks like this
#(x^2 - 9)/(x^2 - 1) < 0 #
Right from the start, you know that any solution set that you might come up with cannot include the values of
More specifically, you need to have
#x^2 - 1 != 0 implies x != +- 1#
Now, in order for this inequality to be true, you need to have
#x^2 - 9 < 0" "# and#" "x^2 - 1 > 0#
or
#x^2 - 9 >0" "# and#" "x^2 - 1 < 0#
For the fist set of conditions to be true, you need to have
#{(x^2 - 9 < 0 implies x < +- 3 implies x in (-3, 3)), (x^2 - 1 > 0 implies x > +- 1 implies x in (-oo, -1) uu (1, + oo)) :}#
This means that you need
For the second set of conditions, you need to have
#{(x^2 - 9 > 0 implies x > +- 3 implies x in (-oo, -3) uu (3, + oo)), (x^2 - 1 < 0 implies x < +- 1 implies x in (-1, 1)) :}#
This time, those two intervals will not produce a valid solution set, or
The only option left to you is
graph{(x^2 - 9)/(x^2 - 1) [-18.02, 18.01, -9.01, 9.01]}