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

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Answer 1 · 2018-05-06T08:02:54

The domain is $x in (-oo, -1)uu(-1,1)uu(1,+oo)$ . The range is $y in (-oo,-1]uu(0,+oo)$

The denominator is $!=0$

$x^2-1!=0$

$(x+1)(x-1)!=0$

$x!=-1$ and $x!=1$

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

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

Therefore,

$yx^2-y=1$

$yx^2-(y+1)=0$

This is a quadratic equation in $x$

The real solutions are when the discriminant is

$Delta>=0$

$0-4*y(-(y+1))>=0$

$4y(y+1)>=0$

The solutions to this equation is obtained with a sign chart.

$y in (-oo,-1]uu(0,+oo)$

The range is $y in (-oo,-1]uu(0,+oo)$

graph{1/(x^2-1) [-7.02, 7.024, -3.51, 3.51]}

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