How to find the range of $\frac{x^{2}}{1 - x^{2}}$ $x^2/(1-x^2)$ ?

Question

I thought that the answer is $R : (- \infty, - 1) \cup (- 1, + \infty)$ $R:(-oo,-1)uu(-1,+oo)$ but the book says that the correct answer is $R : (- \infty, - 1) \cup [0, + \infty)$ $R:(-oo,-1)uu[0,+oo)$
Can anyone help please?

15767 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-10T21:12:05

The range of $\frac{x^{2}}{1 - x^{2}}$ $x^2/(1-x^2)$ is $(- \infty, - 1) \cup [0, \infty)$ $(-oo, -1) uu [0, oo)$

Let:

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

and solve for $x$ $x$ ...

Multiplying both sides by $1 - x^{2}$ $1-x^2$ , we get:

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

Adding $y x^{2}$ $yx^2$ to both sides, this becomes:

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

Then dividing both sides by $(y + 1)$ $(y+1)$ we get:

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

This has solutions if and only if:

$\frac{y}{y + 1} \geq 0$ $y/(y+1) >= 0$

That is, if either of the following:

$y \geq 0$ $y >= 0" "$ and $y + 1 > 0$ $" "y + 1 > 0$ . That is $y \geq 0$ $y >= 0$
$y \leq 0$ $y <= 0" "$ and $y + 1 < 0$ $" "y + 1 < 0$ . That is $y < - 1$ $y < -1$

So the range of $\frac{x^{2}}{1 - x^{2}}$ $x^2/(1-x^2)$ is $(- \infty, - 1) \cup [0, \infty)$ $(-oo, -1) uu [0, oo)$

graph{x^2/(1-x^2) [-10, 10, -5, 5]}