What is the domain and range of $f (x) = \frac{3 x}{x^{2} - 1}$ $f(x)= (3x) /(x^2-1)$ ?

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Answer 1 · 2018-07-02T16:01:45

The domain is $x \in (- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$ $x in (-oo,-1)uu(-1,1)uu(1,oo)$ . The range is $y in RR$ .

As you cannot divide by $0$ , the denominator is $!=0$

Therefore,

$x^2-1!=0$

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

So,

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

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

To calculate the range, let

$y=(3x)/(x^2-1)$

$=>$ , $y(x^2-1)=3x$

$=>$ , $yx^2-y=3x$

$=>$ . $yx^2-3x-y=0$

This ia a quadratic equation in $x$ and in order to have solutions, the discriminant must be $>=0$

Therefore,

$Delta=(-3)^2-4(y)(-y)>=0$

$9+4y^2>=0$

So,

$AA y in RR$ , $9+4y^2>=0$

The range is $y in RR$

graph{3x/(x^2-1) [-18.02, 18.02, -9.01, 9.02]}

What is the domain and range of f(x)= (3x) /(x^2-1)f(x)=3xx2−1 f(x)= (3x) /(x^2-1)?