How do you solve #(16-x^2)/(x^2-9)>=0#?

1 Answer
Jan 24, 2016

Answer:

Identify points and intervals at which this is satisfied to find solution:

#x in [-4, -3) uu (3, 4]#

Explanation:

Let #f(x) = (16-x^2)/(x^2-9) = ((4-x)(4+x))/((x-3)(x+3))#

First note that if #x=+-4# then #16-x^2 = 0# and #x^2-9 != 0#, so these values of #x# are part of the solution set.

Next note that if #x=+-3# then #x^2-9 = 0# and #16-x^2 != 0#, so #f(x)# is not even defined for these values of #x#, which are therefore not part of the solution set.

Apart from these values of #x# the rational expression is non-zero and continuous, so is consistently positive or negative throughout each of the individual intervals: #(-oo, -4)#, #(-4, -3)#, #(-3, 3)#, #(3, 4)#, #(4, oo)#

If #x < -4# or #x > 4# then #16-x^2 < 0# and #x^2-9 > 0#, so #f(x) < 0#.

If #-4 < x < -3# or #3 < x < 4# then #16-x^2 > 0# and #x^2-9 > 0#, so #f(x) > 0#.

If #-3 < x < 3# then #16-x^2 > 0# and #x^2-9 < 0#, so #f(x) < 0#

So #f(x) >= 0# when #x in [-4, -3) uu (3, 4]#