Inverse of a Power Function with Natural Exponent
Understanding $n$-th roots as inverse functions of $f(x) = x^n$
1. Starting Point — The Power Function
We begin with the power function with a natural exponent $n$:
$$ f(x) = x^n, \quad n \in \mathbb{N} $$
Recall that a function has an inverse if and only if it is one-to-one (injective). Whether $f(x) = x^n$ is invertible depends on whether $n$ is odd or even.
Odd $n$ (e.g., $n = 1, 3, 5, 7, \ldots$)
$f(x) = x^n$ is strictly increasing on all of $\mathbb{R}$, so it is one-to-one and therefore invertible on $\mathbb{R}$.
Its inverse is:
$g(x) = \sqrt[n]{x}$
called the $n$-th root of $x$.
Even $n$ (e.g., $n = 2, 4, 6, \ldots$)
$f(x) = x^n$ is not one-to-one on $\mathbb{R}$ (e.g., $(-2)^2 = 2^2 = 4$). We must restrict the domain to $[0, +\infty)$ to make it invertible.
Its inverse (on $[0, +\infty)$) is:
$g(x) = \sqrt[n]{x}, \quad x \geq 0$