How do you solve #((3x-2)(x-4))/(x+4)^2<0# using a sign chart?

1 Answer
Jan 13, 2017

Answer:

The answer is #=x in ] 2/3 , 4 [ #

Explanation:

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

The domain of #f(x)# is #D_f(x)=RR-{-4}#

The denominator is #>0, AA x in D_f(x)#

Let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##2/3##color(white)(aaaaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##3x-2##color(white)(aaaa)##-##color(white)(aa)##color(red)(∥)##color(white)(a)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##3x-2##color(white)(aaaa)##-##color(white)(aa)##color(red)(∥)##color(white)(a)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aa)##color(red)(∥)##color(white)(a)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<0#, when #x in ] 2/3 , 4 [ #