I read the answers to What are argument one can give to students on the definition $0^0$? and I noticed some people are mildly bothered that $0^x$ is discontinuous if we define $0^0=1$. However, I think is fairly obvious that if $f(x)=0^x$ then $\text{domain}(f)$ does not contain negative numbers. Surely, if $x <0$ then $$0^x = 0^{-(-x)} = \frac{1}{0^{-x}}=\frac{1}{0}$$ usually an unwelcome expression. My thought here, $0^x$ is not a proper exponential function because the natural domain of the expression $0^x$ is at largest $[0, \infty)$ and, given the convention $0^0=1$, is continuous on $(0,\infty)$. In other words, it is the constant function $g(x)=0$ restricted to $(0,\infty)$. On the other hand, exponential functions with $a>0$ and $a \neq 1$ all have nice properties: they're injective functions with $\text{range}(a^x) = (0, \infty)$ and $\text{domain}(a^x) = (-\infty, \infty)$. Of course, $a=1$ is somewhat boring $a^x=1^x=1$ hence we just have the constant function with an awkward formula.
Ok, enough about the happy place. What then about $f(x)=a^x$ where $a<0$. What is the natural domain for this formula? This leads to my question:
Do you mention the problem of understanding $f(x)=a^x$ for $a<0$ in your teaching?
Is it wise to omit discussion of $a \leq 0$ ?
I don't have a firm opinion on this, I usually find myself talking about the bizarre behavior of such a formula sometime in the first week of calculus. A secondary question, does appreciating the weirdness of negative base "exponential functions" help explain why we shouldn't have much hope for $0^x$ to behave nicely? I put quotes here because I think the definition of an exponential function properly assumes the base is a positive number which is not one.