Calculate #x# in #3/(x^2-6x+8)-6/(x^2-16)<=0# ?

2 Answers
Apr 17, 2017

Answer:

See below.

Explanation:

If #3/(x^2-6x+8) le 6/(x^2-16)# then

# 6/(x^2-16)-3/(x^2-6x+8) ge 0# or

#6/((x-4)(x+4))-3/((x-2)(x-4)) ge 0# or

#1/(x-4)(6/(x+4)-3/(x-2))ge 0# or

#(x-4)(3(x-8))/((x+4)(x-2))) ge 0 #

Now we have the sign dependence

#(-oo)("+++")(-4)("----")(2)("+++")(4)("----")(8)("++++")(oo)#

Then the inequality is true for

#(-oo,-4]uu(2,4)uu[8,oo)#

Apr 17, 2017

Answer:

#(-oo,-4)uu(2,4)uu[8,+oo)#

Explanation:

#"Express the rational functions as a single rational function"#

#3/(x^2-6x+8)-6/(x^2-16)<=0#

#"factorise the denominators"#

#3/((x-2)(x-4))-6/((x-4)(x+4))<=0#

#rArr(3(x+4))/((x-2)(x-4)(x+4))-(6(x-2))/((x-2)(x-4)(x+4))<=0#

#rArr(-3(x-8))/((x-2)(x-4)(x+4))<=0#

#"the zero's of the numerator/denominator are"#

#" numerator " x=8, "denominator " x=2,x=4,x=-4#

These indicate where the rational function may change sign and which values x cannot be on the denominator.

#"the intervals for consideration are"#

#x<-4,color(white)(x)-4 < x < 2,color(white)(x)2 < x < 4,4 < x <=8,x > 8#

#" Consider " color(blue)" a test point " " in each interval"#

We want to find where the function is negative, that is < 0

Substitute each test point into the function and consider it's sign.

#color(red)(x=-6)to (+)/(-)to color(blue)" negative"#

#color(red)(x=-2)to(+)/(+)tocolor(red)" positive"#

#color(red)(x=3)to(+)/(-)tocolor(blue)" negative"#

#color(red)(x=6)to(+)/(+)tocolor(red)" positive"#

#color(red)(x=8)to0/(+)tocolor(magenta)" x-intercept"#

#color(red)(x=10)to(-)/(+)tocolor(blue)" negative"#

#rArr(-oo,-4)uu(2,4)uu[8,+oo)#
graph{(-3(x-8))/((x-2)(x-4)(x+4)) [-10, 10, -5, 5]}