# How do you find f^-1(x) given f(x)=-1/x^2?

Apr 1, 2018

$f {\left(x\right)}^{-} 1 = \pm \sqrt{- \frac{1}{x}}$

#### Explanation:

You substitute the x values for the y values

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

Then we rearrange for y

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

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

$y = \pm \sqrt{- \frac{1}{x}}$

Such a function doesn't exist as you can't have a negative root on the $\mathbb{R}$ plane. Also it fails the function test as you have two x values corresponding to 1 y value.