What is the domain and range of $f(x)=(x^2+1)/(x+1)$ ?

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Answer 1 · 2017-09-04T17:42:53

The domain is $x in (-oo,-1) uu(-1,+oo)$
The range is $y in (-oo, -2-sqrt8] uu [-2+sqrt8, +oo)$

As we cannot divide by $0$ , $x!=-1$

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

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

So,

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

$x^2+yx+1-y=0$

In order for this equation to have solutions, the discriminant is

$Delta<=0$

$Delta=y^2-4(1-y)=y^2+4y-4>=0$

$y=(-4+-(16-4*(-4)))/(2)$

$y=(-4+-sqrt32)/2=(-2+-sqrt8)$

$y_1=-2-sqrt8$

$y_2=-2+sqrt8$

Therefore the range is

$y in (-oo, -2-sqrt8] uu [-2+sqrt8, +oo)$

graph{(x^2+1)/(x+1) [-25.65, 25.66, -12.83, 12.84]}

What is the domain and range of f(x)=(x^2+1)/(x+1)f(x)=(x^2+1)/(x+1)?