Inverse Functions: Why rewrite as y=f(x) ?

[paulb203]

TL;DR

What does this mean? And does it have to be y?

By rewriting, for example, f(x)=2x+3, as y=2x+3, are we simply stating that something = 2x+3; and in the first case we’re calling that something f(x), and in the second case we’re calling it y?

Does the y have anything to do with the y axis as in x,y coordinates axes? Or is just a randomly chosen letter? Could it just as well be z, or a, or b, etc?

 

[Bystander]

@fresh_42

 

[FactChecker]

It doesn't have to be 'y'. In fact, if the result has a meaning, it is better to use a name that reflects that meaning (such as F=mA). The teaching of functions using an X-Y graph tends to make people use 'x' as the input of a function and 'y' as the result. 'x' is an arbitrary member of the domain of the function and 'y' is the resulting value in the range. That is convenient, but sometimes unfortunate.

 

[PeroK]

TL;DR Summary: What does this mean? And does it have to be y?

By rewriting, for example, f(x)=2x+3, as y=2x+3, are we simply stating that something = 2x+3; and in the first case we’re calling that something f(x), and in the second case we’re calling it y?

Does the y have anything to do with the y axis as in x,y coordinates axes? Or is just a randomly chosen letter? Could it just as well be z, or a, or b, etc?

If $f(x)$ is a function, then $y = f(x)$ identifies a dependent variable $y$ as that function of the independent variable $x$.

This defines a functional relationship between $x$ and $y$ that may be useful and, in any case, can be shown as a graph in the x-y plane.

 

[Mark44]

Does the y have anything to do with the y axis as in x,y coordinates axes?

Yes, as already stated.

But the body of your post doesn't seem to have anything to do with the thread title - "Inverse Functions: Why rewrite as y=f(x) ?"

I'll go out on a limb and assume you mean something like this:

If y = f(x) = 2x + 3, what is the inverse function?

In this case, it's pretty simple.
$y = 2x + 3 \Leftrightarrow x = \frac{y - 3}2 = \frac y 2 - \frac 3 2$
So $x = \frac y 2 - \frac 3 2$ is the inverse relationship between x and y.

If you graph y = 2x + 3 its graph will be exactly the same as that of $x = f^{-1}(y) = y/2 - 3/2$.

In response to the implied question in your thread title:

Inverse Functions: Why rewrite as y=f(x) ?

Strictly speaking, we don't. We write the inverse as a function of x, but not as f(x), since the formula for f has already been given.

Many or most precalculus textbooks make a big deal out of writing the inverse function as a function of the independent variable (which is usually x), so we end up with the inverse being $f^{-1}(x) = \frac x 2 - \frac 3 2$. Because we have switched the variables, this graph is now different from the one we started with.

In my view, the whole business of swapping the independent and dependent variables is unfortunate and misguided, because students often get the idea that this switch business is what is important. The important part is that given y in terms of x, can you find a formula for x in terms of y? That is, can you solve for x in the original function definition?

In subsequent classes (e.g., calculus and up) it is often the case that your have a formula for y in terms of x, but you need the equivalent formula for x in terms of y. IOW, you need to solve the given equation for x, with $x = f^{-1}(y)$.

 

[fresh_42]

TL;DR Summary: What does this mean?

It explicitly says that the function $f$ whose function values are noted by $y$ depend on the values of $x,$ short $y=f(x).$

And does it have to be y?

No.

By rewriting, for example, f(x)=2x+3, as y=2x+3, are we simply stating that something = 2x+3; and in the first case we’re calling that something f(x), and in the second case we’re calling it y?

Yes. Some authors write $y(x)=2x+3$ and avoid the double naming with $f.$

I think if we were very picky then we could say that $x$ and $y$ stand for values, whereas $f$ stands for the function, which is the combination of values: $f=\{(x,y)\}=\{(x,2x+3)\}.$ But this is a bit nitpicking to make sense of it all.

Does the y have anything to do with the y axis as in x,y coordinates axes?

Yes, because as a function is the set of pairs $f=\{(x,y)\}$, and those pairs are drawn in the $(x,y)$-coordinate system. $y$ is just the name of one axis where we look for the second coordinate of the point $(x,y)=(x,f(x)).$

Or is just a randomly chosen letter?

Not randomly. It is a convention because $y$ comes after $x$ in the alphabet.

Could it just as well be z, or a, or b, etc?

Whatever symbol you like, not even necessarily a Latin letter. But make sure that you will be understood. Naming it $y$ is recognized by everybody and allows you to omit any further explanation. If you use $\xi$ instead, i.e. write $\xi=f(x)$ and $f=\{(x,\xi)\}$ then you should explain this beforehand. You would also need a $(x,\xi)$-coordinate system to draw the function.

 

[paulb203]

Thanks, guys.
Just to clarify; although y does have something to do with x,y graphs, I don't need to be thinking about graphs (for now) if I'm simply asked to find the inverse function of the example given, f(x)=2x+3?

For now all they want us to do, it seems, is to convert this to f^-1(x)=x-3/(2), doing the thing that you, Mark44, said was misguided (swapping the input and output around).

Reminder; I'm only at GCSE level so a lot of this, at this stage, looks impossibly complex.

 

[FactChecker]

Thanks, guys.
Just to clarify; although y does have something to do with x,y graphs, I don't need to be thinking about graphs (for now) if I'm simply asked to find the inverse function of the example given, f(x)=2x+3?

For now all they want us to do, it seems, is to convert this to f^-1(x)=x-3/(2), doing the thing that you, Mark44, said was misguided (swapping the input and output around).

Reminder; I'm only at GCSE level so a lot of this, at this stage, looks impossibly complex.

$f^{-1}(x)$ does NOT equal $x-3/(2)$. When you divide the entire side of an equality by 2, consider enclosing the entire side with parenthesis to make it clear. That should be $f^{-1}(x) = (x-3)/2$.

 

[PeroK]

For now all they want us to do, it seems, is to convert this to f^-1(x)=x-3/(2), doing the thing that you, Mark44, said was misguided (swapping the input and output around).

It's clear that if $y = 2x +3$, then $x = (y-3)/2$. We can then identify these as inverse functions:
$$y = f(x), \ x = f^{-1}(y)$$However, a function doesn't depend on what variable you use. So, you are free to write:
$$f(x) = 2x +3, \ f^{-1}(x) = (x-3)/2$$This last step may confuse some students, because they associate a different variable with each function. But, a function is a function and whether you use $x$ or $y$ or something else as a variable, the functions $f$ and $f^{-1}$ are what they are.

 

[FactChecker]

If the variables have a defined meaning, then the variables CAN NOT be switched. The inverse of $F=mA$ is $A=F/m$. The $F$ and $A$ can not be switched. It is only when the variables $x$ and $y$ are generic variables in a function definition, $y = f(x)$, that they can be switched to say $y = f^{-1}(x)$.

 

[PeroK]

If the variables have a defined meaning, then the variables CAN NOT be switched. The inverse of $F=mA$ is $A=F/m$. The $F$ and $A$ can not be switched. It is only when the variables $x$ and $y$ are generic variables in a function definition, $y = f(x)$, that they can be switched to say $y = f^{-1}(x)$.

A function is a function. Mathematically at least.

 

[Mark44]

It is only when the variables $x$ and $y$ are generic variables in a function definition, $y = f(x)$, that they can be switched to say $y = f^{-1}(x)$.

A function is a function. Mathematically at least.

I agree in principle with both of the above, but in many cases that involve formulas that relate two physical quantities, it doesn't make sense to switch variables.
Simple example, the conversion from Celsius to Fahrenheit: $F = \frac 9 5 C + 32$. The inverse relationship (a function) is given by $C = \frac 5 9 (F - 32)$.
In the first equation F is a function of C ($F = f(C)$ in function notation); in the second, C is a function of F ($C = f^{-1}(F)$). It would make no sense to switch variables in the second equation to get $F = \frac 5 9 (C - 32)$.