Same expressions, but different functions

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The discussion clarifies that the functions f(x) = x/x and g(x) = 1 are indeed different due to their domains. While f(x) is undefined at x = 0, g(x) is defined for all real numbers. Algebraic transformations can simplify expressions, but one must maintain awareness of the domain restrictions, particularly when dealing with functions. The graph of f(x) has a "hole" at (0,1), distinguishing it from g(x).

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f(x) = x/x

g(x) = 1

so... those are different functions? i don't know how to treat expressions like that.


When expression is not defined for some x=a, but after some algebraic transformations i get expresion defined for x=a ... what happends?

Can i make algebraic transformations on expressions with an unknown number without knowing it's value?
I know can.
But can i do this with epression that is part of a function?
 
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You talk about "functions", but then it becomes very relevant to mention the domain. Assuming you mean the "maximal domain", i.e. all real numbers for which the expression is meaningful, then both functions are indeed different. The first expression is not defined for x = 0, since division by zero is undefined. The second expression of course also exists for x = 0; so yes: they are different.
The graph of f would be exactly the same as the graph of g, except for a "small hole" (or "perforation") at (0,1). The algebraic manipulation you would perform to go from x/x to 1, is only allowed for x nonzero so if you start with x/x, you can only simplify it to "1" if you keep track of the condition that this only holds for all nonzero x.
 
And even if you wrote f(x) = g(x) = x, they might still have different codomains and so be different functions.
 

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