## Precalculus (10th Edition)

Domain: $[0,\infty)$, inverse:$f^{-1}(x)=\sqrt[4] x$.
We restrict the domain to $[0,\infty)$, (such that for every $y=f(x)$, there is a unique $x$, because we want it to pass the horizontal line test, because we can only find the inverse of a one-to-one function). If I want to find the inverse of a function, then I first must express $x$ from the function. Then, switching $x$ to $f^{−1}(x)$ and $y$ to $x$ in the function basically gives the inverse. Hence: $y=x^4\\\sqrt[4] y=x$ Therefore $f^{-1}(x)=\sqrt[4] x.$