It is not clear what you mean by "we say f is the mapping: x -> Y".

@AnneBauval It's the definition of a function in the linked article.

No it is not. "the mapping: X -> Y" makes no sense. Can you please clarify your question? A cleaner formatting might help, but won't suffice.

Note that the function is a map from the domain into the range. The domain isn't a variable $x$. You're likely wanting to say $ f: [ -1, 5] \rightarrow Y$?

@CalvinLin Have updated the question.

Your notation is all over the place. Distinguish between capital and lower case, please.

$f(x+1)=x$ is an equation with respect to $x$. You would be asking which $x$ in the domain of $f$ satisfy said equality.

How do you know that "the domain X is still [-1, 5]"? Can you explain your reasoning? EG What is $f(-1)$ equal to, and how did you obtain that from the condition?

@CalvinLin I have updated the question, and given a distinguishable notation to avoid the ambiguous.

Can you clarify what is the set "X+1"? EG What is the set "$[-2, 2] +1$"? $\quad$ Also, please answer my previous question where I asked you to explain the reasoning (You stated "its form is different from the first function", which isn't an explanation for "why that is the domain")

@CalvinLin Sorry, that's the wrong expression. I don't know how to express the mapping relationship for the function g(x+1) = x.

When its domain is $X$ and its codomain is $Y,$ $f$ is a (not the) "mapping $X\to Y$". Now, again (as AlvinL already asked you) when $f$ is defined by "$f(x+1) = x, x\in[-1,5]$", what is the domain of $f?$ i.e. what is the set of elements $t$ such that $f(t)$ is defined?

@AnneBauval Now, when 𝑓 is defined by "𝑓(𝑥+1)=𝑥,𝑥∈[−1,5], what is the domain of 𝑓? Isn't the [-1,5] is the domain of this f?

No. Please think it over.what are the $t$'s for which $f(t)$ is defined, i.e. the $t$'s for which $t=x+1$ for some $x\in[-1,5]$?

@xmh0511 As far as i understand from the discussion above in the comments, all you want to know is what is the domain of $g$ where $g(x+1)=x \hspace{0.1cm} \forall x \in [-1,5]$. See how $g(x)=x-1$ and this holds for every $x$ in $[0,6]$ and hence your domain of $g$ would be $[0,6]$.

@OmJoglekar Isn't [-1,5] is the domain of g(x+1)?

Please answer Calvin Lin's main question, which he insisted upon and which I reformulated twice.

@CalvinLin How do you know that "the domain X is still [-1, 5]"? because x∈[-1,5] in g(x+1) = x, where the range of x is the domain of g? Maybe, my understanding of the domain of a function is wrong. Please feel free to point out my wrong.

If the domain is $[-1, 5]$ like you said, then what is the value of $g(-1)$? How did you figure that out?

@CalvinLin if you passed -1 as the argument, that means, x+1 = -1, which makes the x outside its range, so you cannot pass -1 as the argument.

Great! So how can we figure out the actual domain?

@CalvinLin the actual domain is x∈[-1,5] and the passed argument t should be t ∈[0,6] for f(x+1).

Great! More accurately, $g(t) = t-1$ for $ t \in [0, 6]$. The domain is $[0, 6]$.