Domain and Range of a Function and Its Inverse- Polynomials

[opus]

Homework Statement

Consider the function $f\left(x\right)=\sqrt {x+2}$. Determine if the function is a one-to-one function, If so, find $f^{-1}\left(x\right)$ and state the domain and range of $f\left(x\right)$ and $f^{-1}\left(x\right)$

Homework Equations

N/A

The Attempt at a Solution

I start by noticing that it is a one to one function, by the horizontal line test.

I then find the inverse function of $f\left(x\right)$:
$$f^{-1}\left(x\right)=x^2-2$$

I then note that the $D_{f^{-1}}$ = $R_{f}$ and $R_{f^{-1}}$ = $D_f$

I then get:
$$D_{f^{-1}} = [0,∞]$$
$$R_{f^{-1}} = [-2,∞]$$
$$D_f = [-2,∞]$$
$$R_f = [0,∞]$$

My question:
The inverse function is a polynomial, and to my understanding, all polynomials have a domain of [-∞,∞]. Yet, the domain of the inverse function is also equal to the range of the function. The range of the function is not all real numbers. So these are two contradictory statements in this case, and I don't understand why.

 

[Mark44]

Homework Statement

Consider the function $f\left(x\right)=\sqrt {x+2}$. Determine if the function is a one-to-one function, If so, find $f^{-1}\left(x\right)$ and state the domain and range of $f\left(x\right)$ and $f^{-1}\left(x\right)$

Homework Equations

N/A

The Attempt at a Solution

I start by noticing that it is a one to one function, by the horizontal line test.

I then find the inverse function of $f\left(x\right)$:
$$f^{-1}\left(x\right)=x^2-2$$

I then note that the $D_{f^{-1}}$ = $R_{f}$ and $R_{f^{-1}}$ = $D_f$

I then get:
$$D_{f^{-1}} = [0,∞]$$
$$R_{f^{-1}} = [-2,∞]$$
$$D_f = [-2,∞]$$
$$R_f = [0,∞]$$

My question:
The inverse function is a polynomial, and to my understanding, all polynomials have a domain of [-∞,∞]. Yet, the domain of the inverse function is also equal to the range of the function. The range of the function is not all real numbers. So these are two contradictory statements in this case, and I don't understand why.

Your inverse function is a polynomial, but it is one with a restricted domain, because the range of $y = \sqrt{x + 2}$ is $[0, \infty)$. Since f is one-to-one, its inverse $f^{-1}$ will also be one-to-one.

The equations $y = f(x) = \sqrt{x + 2}$ and $x = f^{-1}(y) = y^2 - 2, y \ge 0$ are equivalent, meaning that any ordered pair (x, y) that satisfies one equation also satisfies the other. Without the restriction that I showed for $f^{-1}$ we could evaluate $f^{-1}(-3) = (-3)^2 - 2 = 7$. I.e., x = 7, y = -3. The problem is that the point (7, -3) is not on the graph of f.

 

[opus]

So the inverse function's domain and range is explicit in that although there may be more possibilities, we state acceptable values before hand due to the first function?

And you said that the two equations are equivalent, but it was my understanding that they were opposite. That is, whatever the first function does, the inverse function reverses it. Is the domain that you specified making them equal? A little confused on the last paragraph.

Had to make this post a little wordy, as I am posting from my phone.

 

[Mark44]

So the inverse function's domain and range is explicit in that although there may be more possibilities, we state acceptable values before hand due to the first function?

Yes.

And you said that the two equations are equivalent, but it was my understanding that they were opposite. That is, whatever the first function does, the inverse function reverses it. Is the domain that you specified making them equal? A little confused on the last paragraph.

Notice that I wrote $y = f(x) = \sqrt{x + 2}$ and $x = f^{-1}(y) = y^2 - 2, y \ge 0$. These two equations, $y = \sqrt{x + 2}$ and $x = f^{-1}(y) = y^2 - 2$ (including the restriction) are equivalent, meaning that they are expressing exactly the same relationship between x and y, although from two different perspectives.

If you graph these two equations, you get a single graph (in part that's what equivalence means). The two functions involved, f and f^-1, are inverses, which is a bit more precise than saying opposites. The two relationships are $f(f^{-1}(y)) = y$ and $f^{-1}(f(x)) = x$.
Where students get confused is switching the variables from $x = f^{-1}(y)$ to $y = f^{-1}(x)$. When you do that, the graph is reflected across the diagonal line y = x. Now, when you graph y = f(x) and y = f^-1(x), one graph is the reflection of the other across the line y = x.

IMO, this whole swapping x for y and y for x is dumb, being done only because when we graph functions, we like them to be functions of x, not y. In later courses, e.g., calculus, you'll be working with inverses, and it will be counterproductive or even foolish to swap variables. The main point of function inverses is that if you have an equation y = g(x) and you need to solve for x, you use the inverse, providing that it exists. If g has an inverse, then the solution is x = g^-1(y).

 

[opus]

Yes.

Notice that I wrote $y = f(x) = \sqrt{x + 2}$ and $x = f^{-1}(y) = y^2 - 2, y \ge 0$. These two equations, $y = \sqrt{x + 2}$ and $x = f^{-1}(y) = y^2 - 2$ (including the restriction) are equivalent, meaning that they are expressing exactly the same relationship between x and y, although from two different perspectives.

If you graph these two equations, you get a single graph (in part that's what equivalence means). The two functions involved, f and f^-1, are inverses, which is a bit more precise than saying opposites. The two relationships are $f(f^{-1}(y)) = y$ and $f^{-1}(f(x)) = x$.
Where students get confused is switching the variables from $x = f^{-1}(y)$ to $y = f^{-1}(x)$. When you do that, the graph is reflected across the diagonal line y = x. Now, when you graph y = f(x) and y = f^-1(x), one graph is the reflection of the other across the line y = x.

IMO, this whole swapping x for y and y for x is dumb, being done only because when we graph functions, we like them to be functions of x, not y. In later courses, e.g., calculus, you'll be working with inverses, and it will be counterproductive or even foolish to swap variables. The main point of function inverses is that if you have an equation y = g(x) and you need to solve for x, you use the inverse, providing that it exists. If g has an inverse, then the solution is x = g^-1(y).

Man I feel like I've got a brick wall preventing me from seeing this. I graphed the two equations on my calculator and they aren't the same at all. I'm thinking that the reason is because the calculator can only set things equal to Y and the second equation should be set equal to X? Are you saying that the same thing is happening to Y in both equations, and the same thing is happening to X in both equations?
What you're saying, about switching the variables, is exactly what I was taught in class. I don't see how it works or what it does, just that finds the inverse and that's just not okay to me. Let me read through this a bit longer. I don't know quite what to ask yet.

 

[Mark44]

Man I feel like I've got a brick wall preventing me from seeing this. I graphed the two equations on my calculator and they aren't the same at all. I'm thinking that the reason is because the calculator can only set things equal to Y and the second equation should be set equal to X?

Yes, because both of your equations are of the form y = <some expression in terms of x>.

Are you saying that the same thing is happening to Y in both equations, and the same thing is happening to X in both equations?

Yes. Any ordered pair of numbers (x, y) satisfies both equations, which means that both equations that I wrote have exactly the same graph. Once you swap variables, though, you get a different graph.

What you're saying, about switching the variables, is exactly what I was taught in class.

At the precalc level, that's the way things are usually done, something like this:
1. Start with an equation y = <something involving x>
2. Swap x for y and y for x so that you now have x = <same something only now involving y>
3. Solve the equation in step 2 for y so that you now have y = f^-1(x)

IMO, this is silly, and many students think that the act of switching variables somehow magically gives them the inverse - it doesn't.

Here's a picture to help illustrate what I'm talking about.

Here's a crude graph of $y = e^x = f(x)$. If you want to find the y coordinate when x = 1, you use the function $f(1) = e^1 = e$.
On the other hand, if you know the y value and need to find the x value, you need the inverse.
$y = e^x \Leftrightarrow x = \ln(y) = f^{-1}(y)$, so $f^{-1}(4) = \ln(4) \approx 1.386$
This is an example of how inverses are really used in calculus courses.

Here's an example that shows why swapping variables is silly.
The conversion formula for Celsius to Fahrenheit temperatures is $F = f(C) = \frac 9 5 C + 32$. To find the inverse function, follow this procedure

1. Solve for C

What could be simpler than a procedure with one step?

$F = \frac 9 5 C + 32 \Leftrightarrow F - 32 = \frac 9 5 C \Leftrightarrow C = \frac 5 9 (F - 32)$
It would be silly to swap variables.
In the first equation I have $F = f(C)$. In the last equation I have $C = f^{-1}(F - 32)$
Note that f means one thing and F means something else.

 

[opus]

So I graphed the two equations on Desmos, where I can set one of them equal to x and saw exactly what you were saying. With y≥0, they are exactly the same.
So the point of inverse functions, isn't necessarily to "undo" what functions do, but rather to find the x when we have y, or find the y when we have x?

 

[Mark44]

So the point of inverse functions, isn't necessarily to "undo" what functions do, but rather to find the x when we have y, or find the y when we have x?

It's not one or the other. Undoing the actions of one function is what allows you to solve for the other variable.

Suppose f is a function that has an inverse, and that y = f(x).
Applying the inverse function to both sides of this equation yields $f^{-1}(y) = f^{-1}(f(x))$.
Since the composition of a function and its inverse, in either order, results in no change in the argument, the expression on the right just above is equal to x.
So we have $f^{-1}(y) = f^{-1}(f(x)) = x$, or $x = f^{-1}(y)$.

In my pictured example with the exponential function, to find the x-coordinate at (?, 4) we start with this:
$4 = e^x$
Apply the natural log function to both sides.
$\ln(4) = \ln(e^x) = x$, since ln and the exponential function are inverses. So the point in question is (1.386, 4), with the x-value rounded to three dec. places.

 

[opus]

Sorry for the late response. I've been having to ping pong back and forth between classes.
Your last post makes more sense now. However, after thinking longer about it, and completing my homework assignments on this topic, I've come up with more questions.
First, how would we go about finding a more complicated inverse, say $f\left(x\right)=\frac {2x^2-2} {x+1}$? In this example, we can't really look at it and say "the function is taking this, and doing that, then doing this" because there are two x values. So how would we find the inverse without swapping variables?

My second question: My book says that $f\left(x\right)$ and its inverse function are symmetric about the line y=x. I'm not sure what this means, as I thought they were the same graph?

Apologies for the strung out questions. I went to the tutor lab and they didn't have the answers I was looking for. I asked about the variable swapping thing and was told "to not worry about it until Calculus".

 

[Alloymouse]

Sorry for the late response. I've been having to ping pong back and forth between classes.
My second question: My book says that $f\left(x\right)$ and its inverse function are symmetric about the line y=x. I'm not sure what this means, as I thought they were the same graph?

try plotting e^x and ln(x) on the same axes, and draw the line y=x. You should be able to understand :)

Sorry for the late response. I've been having to ping pong back and forth between classes.
First, how would we go about finding a more complicated inverse, say $f\left(x\right)=\frac {2x^2-2} {x+1}$? In this example, we can't really look at it and say "the function is taking this, and doing that, then doing this" because there are two x values. So how would we find the inverse without swapping variables?

Again, solve for x (make x the subject of the formula). This way of finding inverse is generally applicable to most equations you will encounter at this time.