Question

What is the range of the function `f(x)=(3x^2+3x-6)/(x^2-x-12)`?

Answer 1

The range is `yin (-oo,0.614]uu[2.692,+oo)`

Explanation:

Let `y=(3x^2+3x-6)/(x^2-x-12)`

To find the range, proceed as follows

`y(x^2-x-12)=3x^2+3x-6`

`yx^2-3x^2-yx-3x-12y+6=0`

`x^2(y-3)-x(y+3)-(12y-6)=0`

This is a quadratic equation in `x` and in order for this equation to have solutions, the discriminant `Delta>=0`

`Delta=b^2-4ac=(-(y+3))^2-4(y-3)(-(12y-6))>=0`

`y^2+6y+9+4(y-3)(12y-6)>=0`

`y^2+6y+9+4(12y^2-42y+18)>=0`

`y^2+6y+9+48y^2-168y+72>=0`

`49y^2-162y+81>=0`

`y=(162+-sqrt(162^2-4*49*81))/(2*49)`

`=(162+-101.8)/(98)`

Therefore,

The range is `yin (-oo,0.614]uu[2.692,+oo)`

graph{(3x^2+3x-6)/(x^2-x-12) [-14.24, 14.23, -7.12, 7.12]}

Answer 2

Range: `f(x)in RR or (-oo,oo)`

Explanation:

`f(x) =(3x^2+3x-6)/(x^2-x-12)` or

`f(x) =(3(x+2)(x-1))/((x-4)(x+3))`

`f(x)=0` for `(x=1, x=-2)`

`f(x)` is undefined for `(x=-3, x=4)`

`f(x)= oo and f(x) = -oo` when `x` approaches `-3 and 4`

Therefore range is any real value ,i.e`f(x)in RR or (-oo,oo)`

Range: `f(x)in RR or (-oo,oo)`

graph{(3x^2+3x-6)/(x^2-x-12) [-40, 40, -20, 20]} [Ans]