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

Question

995 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-29T12:23:13

The domain of #f(x)# is #d_f(x)=RR-{-1}#
The range is #f(x) in RR#

Our function is

#f(x)=(x^2+2x)/(x+1)#

As we cannot divide by #0#, #x+1!=0#

Therefore,

The domain of #f(x)# is #D_f(x)=RR-{-1}#

Let,

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

So,

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

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

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

In order for this quadratic equations to have solutions, the discriminant

#Delta>=0#, #=>#, #b^2-4ac>=0#

#(2-y)^2-4*1+(-y)>=0#

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

#y^2+4>=0#

#AA y in RR#, #y^2+4>=0#

The range is #y in RR#

graph{(x^2+2x)/(x+1) [-12.66, 12.65, -6.33, 6.33]}