How do you solve #(x^2+x-6)/(x^2-3x-4)<=0#?

1 Answer
Jul 9, 2017

Answer:

#x in [-3, -1) uu [2, 4)#

Explanation:

Note that:

#x^2+x-6 = (x+3)(x-2)#

#x^2-3x-4 = (x-4)(x+1)#

Each of the linear factors occurs precisely once, so the sign of the given rational expression will change at each of the points where one of the linear factors is zero. That is at:

#x = -3, -1, 2, 4#

Note that when #x# is large, the #x^2# terms will dominate the values of the numerator and denominator, making both positive.

Hence the sign of the value of the rational expression in each of the intervals #(-oo, -3)#, #(-3, -1)#, #(-1, 2)#, #(2, 4)# and #(4, oo)# follows the pattern #+ - + - +#. Hence the intervals #(-3, -1)# and #(2, 4)# are both part of the solution set.

When #x = -1# or #x = 4#, the denominator is zero so the rational expression is undefined. Since the numerator is non-zero at those values, the function will have vertical asymptotes at those points (and not satisfy the inequality).

When #x = -3# or #x = 2#, the numerator is zero and the denominator is non-zero. So the function will be zero and satisfy the inequality at those points.

Hence the solution is:

#x in [-3, -1) uu [2, 4)#

graph{(x^2+x-6)/(x^2-3x-4) [-10, 10, -5, 5]}