Question

True of false: for a one-to-one function, y=f(x), then x=f^-1(y). Explain your answer.

I have no idea how to even start doing this problem. HELP!!!

Answer 1

Well, the hard part is not figuring true or false, but the "Explain your answer". But ok.. let's try.
Do you know the difference between a relation and a function?

Answer 2

No, not really

Answer 3

Ok, well relations are actually somewhat simpler than functions in some way.
A relation just ties the possible values of variables together. A relation can also be a function, but I'll go on that later.
First, let's see an example of a relation:
$$
y = 2x
$$This relation ties the values of \(x\) and \(y\) together, because if \(x=5\) Then \(y\) has to be double than that and therefore \(y=10\).
And it also goes the opposite direction, if \(y=30\) then \(x\) has to be half then that and therefore \(x=15\)
We can therefore rewrite that relation as
$$
\frac{y}{2} = x
$$It is equivalent, the same relation. It ties the variables in the same way
Makes sense so far?

Answer 4

yeah, kind of

Answer 5

The thing is that relations are not really restricted. A relation could allow more than one possible value for a variable. For example:
$$
y = x^2
$$This relation ties a single value for \(y\) for every value of \(x\), so if \(x=3\) then \(y = 3^2 = 9\).
But, it ties two values of \(x\) for a single value of \(y\), because if we say \(y=9\) then both \(x=3\) and \(x=-3\) work, in both cases \(x^2 = 9\)

Answer 6

okay, I understand that

Answer 7

Ok, good =)
So now to functions. A function is defined to assign a single output for every possible input.
For example, the function \(f(x) = 2x\) will just double every input and that doubled value is a single output.
The following is also a function \(f(x) = 5\)
Although the output is the same for all inputs (no matter what x is), it is a still a single value for every input. No matter what x is, the output is just 5.

The following is NOT a function: \(f(x) = \color{red}{\pm} x\)
Because there is no single output.

We can now look at relations and see when can it be functions.
For example, the first relation:
$$
y = 2x
$$can be a function of \(x\), because for every \(x\) we pick there is a single possible output \(y\)
It can also be a function of \(y\) because for every \(y\) we pick there is a single possible \(x\)
In both cases we say it is a one-to-one function, because the output value is guaranteed to be unique for every single input. You wouldn't encounter the same output for different inputs in this relation.
That is what allows us to define it as a function either for \(x\) or \(y\).

In our second example, however:
$$
y = x^2
$$We can only define this as a function of \(x\). Because for every given input \(x\) there is a single possible output value \(y\)
It cannot be defined as a function of \(y\), because a single value of \(y\) could lead to multiple possible values for \(x\).
We can set it as a function of \(x\), but that function is not one-to-one function, because there are two possible inputs (values of \(x\)) that could lead to the same output (value of \(y\)).
So we cannot set it to be function for both variables, only for \(x\).

Makes sense? =)

Answer 8

yes, it's finally starting to make sense

Answer 9

Ok, so now we can face the question itself.
They say that \(y = f(x)\) is a one to one function, and ask if we can say \(x = f^{-1}(y)\)
Right?

Answer 10

yeah, and \[x=f^{-1}(y)\] would be a one-to-one function because there is only one output variable....right?

Answer 11

hmm not exactly. First of all a function has one possible output for every given input. A one to one function also says that each possible output value can be generated by a single input value.
You can say that in \(y= x^2\) we also have a single output variable. If we assign \(y\) to be our input then \(x\) will be our output.
The problem is that \(x\) can have more than one possible value for every value of \(y\), so it cannot be a function. For example if we set \(y = 4\) then the possible values for \(x\) are both \(2\) and \(-2\), in both cases \(x^2 = y\)

Answer 12

But we don't really care if x=f^{-1}(y) is a one-to-one function (though it has to be), because we just want to figure out if it is possible given what we know

Answer 13

The expression \(y = f(x)\) says that \(y\) is a function of \(x\). Why? because we take \(x\), we operate some function on it and \(y\) is tied to be that function's output.
Now we ask ourselves, can \(y\) have multiple possible values for a given \(x\)?
No, because the output of the function \(f()\) has to be single output, and that is the only thing we allow \(y\) to be.
So we don't really care about what \(f()\) really is, we just say that if we make sure that our operation on \(x\) results in a single possible output, then \(y\) is a function of \(x\)

Answer 14

Now, they are asking of we can say \(x = f^{-1}(y)\)
That means, is \(x\) a function of \(y\)? Is there a single possible value for \(x\) for any given \(y\)?
Well in that form it seems there is, we do some operation on \(y\)... that leads to a single output, and that is what \(x\) is.
But the question is if it is the same relation as \(y = f(x)\).
If it is the same relation, then we can say \(x\) is a function of \(y\) and \(y\) is a function of \(x\).
For every value of \(x\) there is one possible \(y\), and for every value of \(y\) there is one possible \(x\).
You see what I mean?

Answer 15

yeah

Answer 16

So, what do we need in order to define the relation as a function of \(y\)? what is the requirement?

Answer 17

I'm stuck again...i'm not sure

Answer 18

Well, just remember what a function is. It is like a blackbox that takes an input and converts it into an output.
One input -> one output. that's it.
If we say that something is a 'function of \(y\)' for example, it means that the input is \(y\). Set any value you want for \(y\), the function can take it and generate a single output out of it.
Functions are consistent, they set a well-defined output for every possible input you can give. You won't get a different result for the same input. One input -> One output

Answer 19

With that in mind, if we want \(x\) to be a function of \(y\).. Then \(y\) is our input and \(x\) our output of the function.. and in order for that to be a function, what restriction do we have on the output?

Answer 20

true, since our function is 1-1 we have that \(f(x) = f(y) \implies x=y\) we never have two different x's mapping to the same y. thus there is only one choice for \(f^{-1}(y)\) and that is \(x\).