Calculate domain $f(x)=x^{\frac{x+1}{x+2}}$

Question

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?

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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: $D(f)=(0,\infty)$.

Answer 1

The problem is on the first line of your proof. The functions $log$ and $e$ are inverse to one another, but are not both defined on all of the real line. $$ log : (0, \infty) \rightarrow \mathbb{R} $$ $$ e : \mathbb{R} \rightarrow (0 , \infty) $$ and so it is only true that, $$ e(\log(x)) = x \mbox{ , }\forall x \in (0, \infty) $$

The function $f$ itself can be defined at zero, and also at $-1$, $$ f(-1) = (-1)^{\frac{(-1)+1}{(-1) + 2}} = (-1)^{0} = 1$$

Answer 2

However, if you input $f(0)$ into your function, it returns $0$, and there is no problem with this.

$$f(0)=0^{\frac{0+1}{0+2}}=f(0)=0^{\frac{1}{2}}=0$$

Answer 3

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 $\sqrt0=0.$

Answer 4

You can write the function into exponential form only if $f(x) >0$, for example consider that $x=e^{\ln x} $ only when $x>0$

Answer 5

The domain and range functionality in WolframAlpha was designed specifically to answer questions for students in algebra, precalculus, and calculus. In that context, the convention that $0^{1/2}=0$ is quite reasonable. As you might hope, it's also completely consistent with the way that Mathematica treats things. For example:

In[1]:= Log[0]
Out[1]= -Infinity

In[2]:= Exp[-Infinity]
Out[2]= 0

Now, as others have argued, there are other reasonable ways to interpret $0^{1/2}$ and that's fine. If your objective is to understand WolframAlpha's response, however, then it's a simple matter of design choice.