Same expressions, but different functions

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The discussion clarifies that f(x) = x/x and g(x) = 1 represent different functions due to their domains, specifically that f(x) is undefined at x = 0 while g(x) is defined for all x. Algebraic transformations can be performed on expressions with unknown values, but caution is necessary when these expressions are part of a function, as the domain must be considered. The graph of f(x) resembles g(x) but has a "hole" at (0,1), indicating the point of discontinuity. Simplifying x/x to 1 is valid only for nonzero x, highlighting the importance of tracking conditions during manipulation. Ultimately, even if two functions appear similar, their domains and codomains can establish them as distinct.
hellbike
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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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