why do inverse functions only exist for one to one functions

rxrx2004

like y=x^2 is a many to one function, but it has an inverse as it can be reflected in the line y=x?

Reply 1

mqb2766

Original post by rxrx2004

like y=x^2 is a many to one function, but it has an inverse as it can be reflected in the line y=x?

At a very simple level you want the property

[br]f^{-1}(f(x)) = x[br]

(edited 2 years ago)

Reply 2

Pangol

Original post by rxrx2004

like y=x^2 is a many to one function, but it has an inverse as it can be reflected in the line y=x?

Part of the definition of a function is that it has to be well-defined, in the sense that if you are given a value to plug in, you get one value out. In your example (but using functional notation), if f(x) = x², then, for example, f(1) = 1 and f(-1) = 1. But what would be f^-1(1)?

(edited 2 years ago)

Reply 3

rxrx2004

Original post by mqb2766

At a very simple level you want the property

[br]f^{-1}(f(x)) = x[br]

but why

Original post by Pangol

1 or -1

Reply 4

mqb2766

Original post by rxrx2004

but why

https://learn.lboro.ac.uk/archive/olmp/olmp_resources/pages/workbooks_1_50_jan2008/Workbook2/2_3_one_2_one_n_inverse_functions.pdf
Is a pretty good overview. A function (forwards or inverse, they're both functions) should evaluate to a single value. The inverse function is defined as per the previous post.

This is why
sqrt(2^2) = 2
i.e. its defined to be the positive root (single valued).

Reply 5

RDKGames

Study Forum Helper

Original post by rxrx2004

1 or -1

Functions only have a single output.

You cannot say that

f^{-1}(1) = \pm 1

because that’s two outputs … so

f^{-1}

is not an inverse function …

Reply 6

Pangol

Original post by rxrx2004

1 or -1

Well that's the problem. If f^-1 were a function, you would be able to tell me the one value of f^-1(1). But there isn't a single value, so it is not a function.

(edited 2 years ago)

Reply 7

rxrx2004

Original post by mqb2766

ok thanks

Original post by RDKGames

Functions only have a single output.

You cannot say that

f^{-1}(1) = \pm 1

because that’s two outputs … so

f^{-1}

is not an inverse function …

thanks

Original post by Pangol

Well that's the problem. If f^-1 were a function, you would be able to tell me the one value of f^-1(1). But there isn't a single value, so it is not a function.

thanks

Reply 8

rxrx2004

Original post by rxrx2004

like y=x^2 is a many to one function, but it has an inverse as it can be reflected in the line y=x?

wait so y=x^2 isnt a function then?

(edited 2 years ago)

Reply 9

mqb2766

Original post by rxrx2004

wait so y=x^2 isnt a function then?

Yes it is, it evaluates to a single value.
its inverse however is only defined on the functions domain x>=0

(edited 2 years ago)

Reply 10

Pangol

Original post by rxrx2004

wait so y=x^2 isnt a function then?

You don't run into the problem with f(x) = x². For any value of x, there is only one value of x².

Reply 11

rxrx2004

Original post by Pangol

You don't run into the problem with f(x) = x². For any value of x, there is only one value of x².

Original post by mqb2766

Yes it is, it evaluates to a single value.
its inverse however is only defined on the functions domain x>=0

oh ok.
so y=root x isnt a function..i think i get it now

Reply 12

mqb2766

Original post by rxrx2004

oh ok.
so y=root x isnt a function..i think i get it now

It is a function, as its single valued. It returns positive values (and 0).

Reply 13

Pangol

Original post by rxrx2004

oh ok.
so y=root x isnt a function..i think i get it now

Well it is, because the definition of square root is to take the positive root. For example, the square root of 9 is 3, it's just the case that the square of negative 3 is also 9.

Reply 14

rxrx2004

Original post by Pangol

Well it is, because the definition of square root is to take the positive root. For example, the square root of 9 is 3, it's just the case that the square of negative 3 is also 9.

but-but i always thought that you add a plus-minus sign at the front of a square root

Original post by mqb2766

It is a function, as its single valued. It returns positive values (and 0).

but if you reflect the graph of y=x^2 in the line y=x you'd get a graph for which it returns two values? :frown:

Reply 15

mqb2766

Original post by rxrx2004

but-but i always thought that you add a plus-minus sign at the front of a square root

but if you reflect the graph of y=x^2 in the line y=x you'd get a graph for which it returns two values? :frown:

The +/- reflects the fact that sqrt() returns a positive value.
You have to multiply it by -1 to get the corresponding negative root as it is defined to return the (single valued) positive value.

If it returned both values (it doesn't) there would be no need to write the +/-.