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

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

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Answer 1 · 2017-08-07T14:57:22

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 $y \in (- \infty, 0] \cup (2, + \infty)$ $y in (-oo,0] uu (2,+oo)$

Explanation:

The function is

$y = \frac{2 x^{2}}{x^{2} - 1}$ $y=(2x^2)/(x^2-1)$

We factorise the denominator

$y = \frac{2 x^{2}}{(x + 1) (x - 1)}$ $y=(2x^2)/((x+1)(x-1))$

Therefore,

$x \neq 1$ $x!=1$ and $x \neq - 1$ $x!=-1$

The domain of y is $x \in (- \infty, - 1) \cup (- 1, 1) \cup (1, + \infty)$ $x in (-oo,-1) uu (-1,1) uu (1,+oo)$

Let's rearrage the function

$y (x^{2} - 1) = 2 x^{2}$ $y(x^2-1)=2x^2$

$y x^{2} - y = 2 x^{2}$ $yx^2-y=2x^2$

$y x^{2} - 2 x^{2} = y$ $yx^2-2x^2=y$

$x^{2} = \frac{y}{y - 2}$ $x^2=y/(y-2)$

$x = \sqrt{\frac{y}{y - 2}}$ $x=sqrt(y/(y-2))$

For $x$ $x$ to a solution, $\frac{y}{y - 2} \geq 0$ $y/(y-2)>=0$

Let $f (y) = \frac{y}{y - 2}$ $f(y)=y/(y-2)$

We need a sign chart

$a a a a$ $color(white)(aaaa)$ $y$ $y$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a a$ $color(white)(aaaaaa)$ $0$ $0$ $a a a a a a a$ $color(white)(aaaaaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $y$ $y$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $y - 2$ $y-2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a$ $color(white)(aa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (y)$ $f(y)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a$ $color(white)(aa)$ $-$ $-$ $a a$ $color(white)(aa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$

Therefore,

$f (y) \geq 0$ $f(y)>=0$ when $y \in (- \infty, 0] \cup (2, + \infty)$ $y in (-oo,0] uu (2,+oo)$

graph{2(x^2)/(x^2-1) [-16.02, 16.02, -8.01, 8.01]}

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

1 Answer

Explanation:

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What is the domain and range of y = (2x^2)/( x^2 - 1)y=2x2x2−1y = (2x^2)/( x^2 - 1)?

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