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]}

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