A question regarding inverse functions

[christian0710]

Hi I'm reading a book called Calculus lifesaver, and in the book they state that the inverse of a function f(x)= x³ is the same as f^-1(x)=³√x and is the same as f^-1(y)=³√y

So I did a test, with a simpler function and I can't see how this is true

If I have a function f(x)= 2*x

Then the inverse would be y=2x→ x=y/2

So It’s correct then to write f^-1(x)= x/2

And Is it wrong or correct to write?
f^-1(y)= y/2 Because this is not the inverse of y, this is the inverse of X, so the right thing would be to either call it f^-1(x)= x/2 OR f(y)= y/2 right?? If i graph it,
f(y)= y/2 gives us the same graph as f(x)= 2*x and f^-1(x)= x/2 gives us a different graf which is the inverse, so It must be an error to state that f^-1(x)=³√x is the same as f^-1(y)=³√y Or am i wrong?

 

[mfb]

Variable names have no special meaning. As long as you use them consistently, it is fine.
f(r)=2r and f(y)=2y are the same as f(x)=2x.

 

[christian0710]

I see but if we have a xy cordinate system, the have a function f(y)=2y and f(x)=2x we get two different graphs (don't we) because the one is a function of y and the other a function of x? That's my confusion.

 

[mfb]

You get different graphs, but the functions are the same. They are just drawn in a different way.

 

[christian0710]

But how can the functions be the same if they have different graphs? a function only has one inverse, so how can it have two graphs? One of them must be correct and the other wrong? If i graph on a xy axis f(y)=2y and f(x)=2x i get two different graphs.

 

[Mark44]

The coordinate system that you use to graph your functions usually has its horizontal axis labelled as "x" and its vertical axis labelled as "y". When a function is in the form of y = f(x), y is the dependent variable and x is the independent variable.

You know what the graph of y = f(x) = x³ looks like. The equation x = g(y) = $\sqrt[3]{y}$ represents the inverse relationship between x and y, and this equation is equivalent to the equation y = x³, which means that each point (x, y) that is on one graph is also on the other. The only difference is that one function produces a y value for a given input x value, and the other function produces an x value for a given input y value.

Using inverse function notation, these equations say exactly the same thing and have exactly the same graph:

y = f(x) = x³
x = f^-1(y) = $\sqrt[3]{y}$

If I switch the variable names in the function I'm calling g, above, I get y = g(x) = $\sqrt[3]{x}$. Switching x and y has the effect of reflecting each point (x, y) on the original graph across the line y = x. IOW, the point (x, y) gets reflected to (y, x). If that's hard to follow, the point (2, 8) gets reflected to (8, 2).

Our transformed (by reflection) function is now y = g(x) = $\sqrt[3]{x}$. Because of the inverse relationship between f and g, we often write g(x) as f^-1(x).

IMO, this switching gets a lot more attention in Precalc classes and textbooks than it deserves, because it distracts students from the more important concept that a function and its inverse are simply two ways to look at the relationship between x and y on a graph.

A function f and its inverse f^-1 satisfy these relationships:

f (f^-1(y)) = y, and
f^-1(f(x)) = x
as long as y is in the domain of f^-1 and x is in the domain of f.

 

[christian0710]

Man you deserve a Nobel prize for your explanations! This is great. So just to make sure I understand you let me reiterate what you just said:

So the inverse of f(x) y=x³ is obviously x(y)=f^-1(y)=³√y and therefore by switching the variable x and y in the inverse function of y,
f^-1(x)=³√x you get the inverse function of x. And this function is a mirror image of f(x) in the line line y=x.

So equating f-1(x)=3√x with f-1(y)=3√y is not completely correct, because it will give us two different graphs, it would however be correct to say that f-1(x)=3√x has an inverse relationship to y(x)=y³ and gives the same graph.

 

[christian0710]

Here is a small excerpt from the book, i highlighted what i misunderstand.

 

[Vargo]

Howdy,

In what your are writing, you are identifying a function too much with its graph. Or, rather, you are putting too much meaning in the specific variables x and y. This question is about functions, not variables. Think of a function as something numerical. It takes in an input and returns an output. The given function takes an input x and returns an output x^3. In other words, the output is the cube of the input (the letter we use to label the input is irrelevant).

The inverse operation "undoes" what the original operation did. So the inverse of the previous function would take a numerical input and return the cube root. Again, the letter we use to label the input is irrelevant. So whether we write the inverse function in terms of an input x or an input y is irrelevant as long as the output is the cube root of the input.

 

[christian0710]

The inverse operation "undoes" what the original operation did. So the inverse of the previous function would take a numerical input and return the cube root. Again, the letter we use to label the input is irrelevant. So whether we write the inverse function in terms of an input x or an input y is irrelevant as long as the output is the cube root of the input.

Thank you, You are right I get too caught up in definitions because I fear that I misunderstand something, I see now that the output of any variable will still be a value plotted on the y axis, even f^-1(y)=y² :)

 

[Mark44]

Man you deserve a Nobel prize for your explanations! This is great. So just to make sure I understand you let me reiterate what you just said:

So the inverse of f(x) y=x³ is obviously x(y)=f^-1(y)=³√y and therefore by switching the variable x and y in the inverse function of y,
f^-1(x)=³√x you get the inverse function [STRIKE]of x[/STRIKE] as a function of x. And the graph of this function is a mirror image of the graph of f(x) in the line line y=x.

So equating f-1(x)=3√x with f-1(y)=3√y is not completely correct,

It's not even partially correct.

because it will give us two different graphs, it would however be correct to say that f-1(x)=3√x has an inverse relationship to y(x)=y³ and gives the same graph.

Here's the corrected version. y = f(x) = x³ is equivalent to x = f^-1(y) = $\sqrt[3](y)$ - same graph.
At each point on this graph, the first equation tells you how to get the y value if you know the x value. The second equation tells you how to get the x value if you know the y value.

Here's a different example that shows how switching the variables can be a major distraction. The example involves converting Celsius temperatures to Fahrenheit and vice versa.

F = (9/5)C + 32

If we solve the equation above for C, we get
F - 32 = (9/5)C
(5/9)(F - 32) = C
so C = (5/9)(F - 32)

The first equation above gives you the temp in °F if you know the Celsius temp, so we have F = f(C) in function notation.

The last equatio above gives you the temp in °C if you know the Fahrenheit temp, so we have C = f^-1(F)

It would be extremely foolish to switch C and F in either formula...

 

[christian0710]

Perfect then i understand:

Inverse Relationship between x and y:
f(x)=y=x³
x(y)=f^-1(y)=³√y

Inverse function of f(x)
f(x)=y=x³
f^-1(x)=³√x

And thank you for verifying that the definition from the book was incorrect.

 

[Mark44]

Are you talking about the part that you highlighted? If so, there's nothing wrong with it. They are saying that f^-1(y) = $\sqrt[3]{y}$, but if you switch letters, you get f^-1(x) = $\sqrt[3]{x}$.

 

[christian0710]

In the last part you mean f^-1(x) = ³√x instead of √y right? :)

Okay, so they suggest this f^-1(x) = ³√x instead of f^-1(y) = ³√y because it might fool you to think that the input would be read on the y-axis and the output f(y)=x would be read on the x-axis.

 

[Mark44]

In the last part you mean f^-1(x) = ³√x instead of √y right? :)

Yep, you got me. Copy/paste error.

Okay, so they suggest this f^-1(x) = ³√x instead of f^-1(y) = ³√y because it might fool you to think that the input would be read on the y-axis and the output f(y)=x would be read on the x-axis.

It's probably more because people think of functions being "of x".

 

[christian0710]

Thanks again. I appreciate your help!