# Math Help - What is the inverse function of...

1. ## What is the inverse function of...

...f(x)= (1 - x)^2 on [-1, 2]?

2. The inverse of $f(x) = y = (1-x)^2$ is found as follows

Swap x and y and solve for y.

$x = (1-y)^2$

$\sqrt{x} = 1-y$

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

$1-\sqrt{x} =y$

$y=f^{-1}(x) = 1-\sqrt{x}$

but the inverse only exists for a function that is one to one. Does your interval imply this?