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

2 Answers
Dec 30, 2017

Answer:

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]}

Dec 30, 2017

Answer:

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]