Question

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

Answer 1

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)`

Answer 2

`(-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]}