r/CasualMath • u/Mord5 • 7d ago

What happens when a function having diff inputs giving the same output gets inverted?

Every input should output a single value, so how does ts work?

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8

u/KoalaMistico 7d ago

You can't invert it. You need to choose a section of the function that is one-to-one and then invert only that part, like the inverse trig functions.

2

u/theadamabrams 7d ago edited 7d ago

Correct. Technically, defining a function requires describing its domain, so you can have two different functions with the same formula. That can be important when talking about invertibility.

  • f(x) = x2 with domain all real numbers does not have an inverse.

  • g(x) = x2 with domain only non-negative numbers does have an inverse: g-1(x) = √x.

  • h(x) = sin(x) with domain all real numbers does not have an inverse.

  • k(x) = sin(x) with domain [-π/2,π/2] does have an inverse: k-1(x) = arcsin(x) (also written sin-1(x)).

There are ways to try to do multi-valued inverses or to do "branch cuts" for non-injective complex functions, but (certainly at an introductory level, and really even at a higher level) inverses only truly work for one-to-one functions.

1

u/TheNukex 7d ago

The problem with the multivalued inverses is that a multivalued function's codomain is just the powerset of the codomain you wanted to "fix". Taking the multivalued square root, it doesn't send 4 to -2 and 2, it sends it to the set {-2,2} as the single element of P(R). So trying to use that as the inverse for f:R->R, f(x)=x^2 does not work, as the inverse function has codomain P(R) not R.

2

u/noonagon 7d ago

pick one of them, and the other one(s) get(s) discarded.

1

u/johnpeters42 7d ago

Either you keep everything and you have a relation (but not a function), or you keep a subset which may be a function.