# How do you find the range of f(x)= (x^3+1)^-1?

Sep 28, 2015

$\left\{y | y \ne 0\right\}$

#### Explanation:

Do find the range, get the domain of the function's inverse.

$y = {\left({x}^{3} + 1\right)}^{-} 1$

Flip $x$ and $y$.

$x = {\left({y}^{3} + 1\right)}^{-} 1$

Now isolate $y$.

$x = {\left({y}^{3} + 1\right)}^{-} 1$
$x = \frac{1}{{y}^{3} + 1}$
$\left({y}^{3} + 1\right) x = \left({y}^{3} + 1\right) \frac{1}{{y}^{3} + 1}$
$\left({y}^{3} + 1\right) x = 1$
${y}^{3} x + x = 1$
${y}^{3} x + x - x = 1 - x$
${y}^{3} x = 1 - x$
$\left(\frac{1}{x}\right) {y}^{3} x = \left(\frac{1}{x}\right) \left(1 - x\right)$
${y}^{3} = \frac{1 - x}{x}$
$\sqrt{{y}^{3}} = \sqrt{\frac{1 - x}{x}}$
$y = \sqrt{\frac{1 - x}{x}}$
$\textcolor{b l u e}{{f}^{-} 1 \left(x\right) = \sqrt{\frac{1 - x}{x}}}$

Now looking for the domain of the inverse function, we will find that it will only be undefined at $x = 0$. So its domain is $\left\{x | x \ne 0\right\}$.

Therefore, the range of $f \left(x\right) = {\left({x}^{3} + 1\right)}^{-} 1$ is $\left\{y | y \ne 0\right\}$.