Exponential Functions where $-1 < x < 1$

Question

In $f(x) = a^x$, I understand why $a > 0$ and $a \neq 1$. But how is $f(x) = a^x$ guaranteed to be a function when $-1 < x < 1$? I realize there are an infinite number of values in that range, and an infinite number of them will be fine (and an infinite number of them won't), so that is a hand-wavey restriction I've set, but let me clarify with an example. If $f(x) = 4^x$, then $f(1/2) = \{-2, 2\}$. This clearly can't be within the domain a function, since it breaks the rule that every input gives a single output. What makes those values okay to be considered within the exponential function domain?

Answer 1

$a^{\frac{1}{2}}$ is not a set of numbers. Usually $a^b$ is defined to be $\exp(\ln_π(a)b)$ when $a,b \in \mathbb{C}$, where $\ln_π$ is the principal branch of the complex logarithm. In more basic mathematics $a^b$ can be defined as $a^b = \sup(\{ a^r : r \in \mathbb{Q} \})$ but only when $a > 0$ and $b \in \mathbb{R}$. Also, high-school mathematics might consider $(-a)^\frac{1}{n} = -(a^\frac{1}{n})$ for $a > 0$ and odd integer $n$, but that actually conflicts with the more general definition for complex numbers, so you cannot have both.