How do you find domain and range for #f(x)=(x^2+2x)/(x+1) #?

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Answer 1 · 2018-06-06T09:09:40

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

The denominator must be #!=0#.

Therefore,

#x+1!=0#

#=>#, #x!=-1#

The domain is #x in (-oo,-1)uu(-1,+oo)#

To calculate the range, proceed as follows :

Let #y=(x^2+2x)/(x+1)#

#y(x+1)=x^2+2x#

#x^2+2x-yx-y=0#

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

In order for this quadratic equation in #x# to have solutions, the discriminant #Delta>=0#

The discriminant is

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

#=4+y^2-4y+4y#

#=y^2+4#

Therefore,

#Delta=y^2+4>=0#

So,

#AA y in RR, Delta>=0#

The range is # y in RR#

graph{(x^2+2x)/(x+1) [-16.02, 16.01, -8.01, 8.01]}