How do solve #(x^2-x-12)/(x^2+4)<=0# and write the answer as a inequality and interval notation?

1 Answer
Jan 21, 2017

Answer:

The answer is #x in [-3, 4 ]# or #-3<=x<=4#

Explanation:

Let's factorise the numerator

#x^2-x-12=(x+3)(x-4)#

Let #f(x)=(x^2-x-12)/(x^2+4)#

The domain of #f(x)# is #RR#

The denominator is #>0#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##4##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<=0# when #x in [-3, 4 ]#

#-3<=x<=4#