How do you find the domain and range of #y= x^2 / (x^2-16)#?

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Answer 1 · 2017-11-17T06:56:24

The domain is #x in (-oo,-4) uu(-4,4) uu (4,+oo)#. The range is #y in (-oo,1) uu (1,+oo)#

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

#x^2-16!=0#

#x^2!=16#

#x!=4# and #x!=-4#

The domain is #x in (-oo,-4) uu(-4,4) uu (4,+oo)#

Calculation of the range :

#y=x^2/(x^2-16)#

#y(x^2-16)=x^2#

#yx^2-16y=x^2#

#yx^2-x^2=16y#

#x^2(y-1)=16y#

#x^2=(16y)/(y-1)#

#x=+-sqrt((16y)/(y-1))#

Therefore,

#y!=1#

The range is #y in (-oo,1) uu (1,+oo)#

graph{x^2/(x^2-16) [-10, 10, -5, 5]}