Calculate domain $f(x)=x^{\frac{x+1}{x+2}}$ I have the following function:
$$f(x)=x^{\frac{x+1}{x+2}}$$
I tried to calculate the domain, which seems easy, and my result is: $D(f)=(0,\infty)$.
When I tried to calculate it, by using Wolfram-Alpha, I obtain: $D(f)=[0,\infty)$.
Can someone explain me the reason, or if it is just a Wolfram's error?

I proceed in this way:
$$f(x)=x^{\frac{x+1}{x+2}} = e^{\frac{x+1}{x+2} \log(x)}$$
$$
\left\{ 
\begin{array}{c}
x+2\ne0 \ \Rightarrow\ x\ne -2 \\ 
x>0
\end{array}
\right. 
$$
Hence: $D(f)=(0,\infty)$.

I think Wolfram is wrong and you are right.
If we write $(f(x))^{g(x)}$ then the domain it's
$$D(g)\cap\{x|f(x)>0\},$$
where $d(g)$ it's the domain of $g$.
I think it's better to define such that even $0^{\frac{1}{2}}$ does not exist, but $\sqrt0=0.$
