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Uni Stuff => Science => Faculties => Mathematics => Topic started by: FallonXay on March 09, 2017, 07:56:07 pm

Title: Inverse Functions Question; Discrete Maths
Post by: FallonXay on March 09, 2017, 07:56:07 pm
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?  :o

Thanks.
Title: Re: Inverse Functions Question; Discrete Maths
Post by: jamonwindeyer on March 09, 2017, 08:08:33 pm
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 ;D
Title: Re: Inverse Functions Question; Discrete Maths
Post by: RuiAce on March 23, 2017, 06:41:55 pm
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 ;D
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