Inverse Functions Question; Discrete Maths

[FallonXay]

Heyy, looking over my lecture notes/ doing homework for discrete maths and I'm pretty confused.

The lecture notes say that the inverse of f(x) = 2x - 5 is f^-1(y) = (y + 5) / 2. Is it equivalent/ Would it be correct to say that the inverse of f(x) = 2x - 5 is f^-1(x)= (x + 5) / 2.

So basically is  f^-1(y) = (y + 5) / 2  = f^-1(x)= (x + 5) / 2 . Or what's happening? 

Thanks.

[jamonwindeyer]

Hey! It's way more typical to use \(x\) as far as I know, but the form of the function is the same so I suppose it's practically equivalent! Like, a function is just a box that takes an input and gives an output. The pro numeral doesn't really affect that. So I'd say it's an unusual way to think of an inverse function, but I suppose it is the same

[RuiAce]

Hey! It's way more typical to use \(x\) as far as I know, but the form of the function is the same so I suppose it's practically equivalent! Like, a function is just a box that takes an input and gives an output. The pro numeral doesn't really affect that. So I'd say it's an unusual way to think of an inverse function, but I suppose it is the same

Ah, sorry for the delayed reply but I will have to disagree here Jamon.

In MATH1081, and as far as all university maths courses are concerned, a function is defined usually by f: X->Y. That just means that the domain and codomain are specified. To distinguish between the fact that x is an element in the domain and y is an element in the codomain, by convention we take the pronumerals to be different.

I believe it is crucial to use a different variable for the inverse function f^-1: Y->X