How do you find domain and range for #f(x)=(x+4)/(x^2-4) #?

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Answer 1 · 2017-11-29T17:45:08

The domain is #RR-{-2,2}#.
The range is #y in (-oo,-0.93] uu [-0.67,+oo)#

As we cannot divide by #0#, the denominator is #!=0#

#x^2-4=(x+2)(x-2)!=0#

Therefore,

The domain is #RR-{-2,2}#

To determine the range, proceed as follows

Let, #y=(x+4)/(x^2-4)#

#y(x^2-4)=x+4#

#yx^2-x-4(y+4)=0#

This is a quadratic equation in #x^2# and in order for this equation to have solutions, the discriminant #Delta>=0#

#Delta=b^2-4ac=(-1)^2-4(y)(-4(y+4))>=0#

#1+16y^2+16y>=0#

#16y^2+16y+1>=0#

#y=(-16+-sqrt(16^2-4*16))/(2*16)=(-16+-sqrt(192))/(32)#

#y_1=(-16+sqrt192)/(32)=-0.67#

#y_2=(-16-sqrt192)/(32)=-0.93#

Therefore,

The range is #y in (-oo,-0.93] uu [-0.67,+oo)#

graph{(x+4)/(x^2-4) [-8.89, 8.89, -4.444, 4.445]}