How do you find the range of $f(x)= x^2/(1-x^2)$ ?

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Answer 1 · 2018-03-21T21:53:14

$(-oo, -1) uu [0, oo)$

Given:

$f(x) = x^2/(1-x^2)$

Let $y = f(x)$ and attempt to solve for $x$ ...

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

Add $1$ to both ends to get:

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

Multiply both sides by $(1-x^2)/(y+1)$ to get:

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

Add $x^2-1/(y+1)$ to both sides to get:

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

In order for this to have a real valued solution, we require:

$1-1/(y+1) >= 0$

That is:

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

Hence we require one of:

$(y >= 0 ^^ y+1 > 0) rarr y in [0, oo)$

$(y < 0 ^^ y+1 < 0) rarr y in (-oo, -1)$

So the range of $f(x)$ is $(-oo, -1) uu [0, oo)$

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

How do you find the range of f(x)= x^2/(1-x^2)f(x)= x^2/(1-x^2)?