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

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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]}

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