Please help with finding the inverse of a one to one function

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How can I find the inverse of a function?

I am struggling with functions, finding the domain and range, reading and graphing functions and finding the inverse of a function. Just an FYI of my level of understanding if someone were to try and explain to me how to find the inverse of a function my understanding is very low level.

This is a problem straight from my lesson:

An inverse of a one-to-one function \(\displaystyle f(x),\), which we write as \(\displaystyle f^{-1}(x),\) is a function where the composition \(\displaystyle f\left(f^{-1}(x)\right)\, =\, f^{-1}\left(f(x)\right)\, =\, x.\)

What does this say? I don't know what this means:(. How do I find the inverse of this equation? Please help, thank you!!!
 
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How can I find the inverse of a function. I am struggling with functions, finding the domain and range, reading and graphing functions and finding the inverse of a function. Just an fyi of my level of understanding if someone were to try and explain to me how to find the inverse of a function my understanding is very low level. This is a problem straight from my lesson: An inverse of a one-to-one function f(x), which we write as
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, is a function where the composition
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. What does this say? I don't know what this means
:(. How do I find the inverse of this equation? Please help, thank you!!!

What it means is that an inverse function "undoes" what the original function does.

When they say, \(\displaystyle f^{-1}(f(x))=x\), that means that if you start with a number x, put it into the function f, then put the result into the inverse function, you end up with the same x you started with: the inverse "undid" f.

The other equality, \(\displaystyle f(f^{-1}(x))=x\), says the same thing with the functions reversed: either undoes the other.

But what you quoted is not a problem in which you have to find an inverse; it is a definition of what an inverse is.

In fact, though, it is how you would check whether one function is the inverse of another. As a simple example, suppose I have the function \(\displaystyle f(x) = x + 2\). That says to add 2 to the input. Undoing that would, clearly, be subtracting 2 from the number: \(\displaystyle f^{-1}(x) = x - 2\). To verify that this really is the inverse function, you calculate this:

\(\displaystyle f^{-1}(f(x))=f^{-1}(x+2)=(x+2)-2=x\)

So subtracting 2 does undo adding 2, and we have the right inverse function.

Now, your question about how to find an inverse function is a separate matter. There are actually a couple slightly different ways to do it; I'd like you to look at any examples you were given of the process and try to do it yourself. Then either show us your work for that, or at least show us an example you were given and ask specific questions about it. That way we can make sure we are working with your textbook rather than against it!
 
Rather than jump straight in to the idea of inverses, it seems to me like it would be helpful to go back and review everything leading up to the topic, starting from "what is a function?" A function is a mathematical process that takes in an input (represented as a variable) and does something, then returns the output. Typically, the notation f(x) is used if the input variable is x, but you might also see functions of other variables like f(t) or g(w) or P(r). As an example, consider the function f(x) = x + 3. This function takes in some value x, adds 3 to it, and returns the result. If we "feed" x = 4 into f(x), we'll get the value 7 out.

The domain of a function is defined as the set of all inputs for which there is a valid output. Often this ends up being all real numbers, but not always. Consider the function \(\displaystyle g(w) = \sqrt{w}\). This function takes an input w and returns its square root. For negative values of w, this process is undefined, so the domain is all positive real numbers. This might be written as Domain: \(\displaystyle w \ge 0\), or Domain: \(\displaystyle [0, \infty)\), among other possibilities. The range of a function, then, is the set of all possible outputs. Using the same g(w) as before, we see that any positive real number has a unique square root, so the range is also all positive real numbers. In this specific case, the domain and range are the same, but that's not always the case.

Consider the function P(r) = \(\displaystyle \dfrac{1}{r-5}\). Here, "feeding" r = 5 into P(r) produces an error, because we tried to divide by zero. But any other real number will produce a valid result. So the domain might be written as \(\displaystyle r \ne 5\) or \(\displaystyle (-\infty, 5) \cup (5, \infty)\). As for the range, the function can produce pretty much any output you desire by "feeding" in an appropriate value of r. But, no matter what we "feed" in, the output will never be 0. Can you see why this is the case? Because of this, the range is \(\displaystyle r \ne 0\) or \(\displaystyle (-\infty, 0) \cup (0, \infty)\).

Next, we can move to function composition. You've probably seen the notation f(g(x)) or similar before. This means to let the function g act on x, and take whatever number it outputs and feed that output into f(x). Consider the functions \(\displaystyle f(x) = \dfrac{1}{x - 2}\) and \(\displaystyle g(x) = \sqrt{x}\). The composite function f(g(x)) says to take some input x, take its square root, and then return the result of 1 divided by that output minus 2. When determining the domain of composite functions, we have to be careful because if either function would cause an error, that input is not part of the domain. The range is a bit easier, but care is still required because even if g(x) can only produce some limited set of outputs, those outputs might produce a different set of possible outputs for f(x).

Let's work with this f(x) and g(x) and see what the domain and range are. We begin with g(x), and we already know it can only take in positive real numbers and only output positive real numbers. Our domain is thus already restricted to the range of g(x). But, what happens as we "feed" these values into f(x)? An error due to division by zero occurs if x = 2, so that's no good. But any other positive real number is fine. In the end, the domain is all positive real numbers except 2, and the range is all positive real numbers except 0 (because remember that we can never get zero from the division in f(x), but we can get any other positive real number).

Finally, we can get to the topic we're actually interested in: "What is an inverse of a function?" Using the composite function notation, the definition of an inverse function is exactly what you quoted from your textbook. We need to find a function \(\displaystyle f^{-1}(x)\) such that:

\(\displaystyle f(f^{-1}(x)) = x\) and \(\displaystyle f^{-1}(f(x)) = x\)

Recall what a composite function means. If we start by applying the inverse function first, then "feeding" its output into f(x), we will always get back x. In other words, no matter what input we start with (so long as it's in the domain of \(\displaystyle f^{-1}(x)\) that is), if we apply the inverse function and then the regular function, we always end up exactly back where we started, as if we'd done nothing at all. Similarly, if we start by first applying the function and then apply the inverse function, we also end up back where we started.

If you think about what it means to invert something, this actually makes a lot of sense. Suppose you had a regular six-sided dice, oriented such that the number 1 is on top. You have a friend look at this dice, and then leave the room. While they're gone, you flip it over so the number 6 is on top. We can think of this as applying the function. Then you flip it over again and you end up with the number 1 on top again. We can think of this as applying the inverse function. Then your friend returns to the room. Would they be able to tell you'd even done anything to the dice?

As for how to find the inverse of a function, that's often not easy to do. If a function is one-to-one, so that every input produces exactly one output and every output is produced by exactly one input, then we can guarantee that function has an inverse, although we may not know what the inverse is. For many simple functions, such as \(\displaystyle f(x) = \dfrac{x}{3}\), this page from CoolMathhttp://www.coolmath.com/algebra/16-inverse-functions/05-how-to-find-the-inverse-of-a-function-01 has a list of steps you can follow to find the inverse. For more complicated functions like \(\displaystyle g(x) = x^5 + x^4 + 1\), finding the inverse is far more difficult.

This provides at least a basic overview of all of the ideas, although I'm obviously not set-up to provide an in-depth instruction. But for that, you can always use your textbook, class notes, or even just Google various webpages.
 
I'd like to thank both Dr. Peterson and ksdhart for their pertinent and extended answers.
 
How can I find the inverse of a function?

I am struggling with functions, finding the domain and range, reading and graphing functions and finding the inverse of a function. Just an FYI of my level of understanding if someone were to try and explain to me how to find the inverse of a function my understanding is very low level.

This is a problem straight from my lesson:

An inverse of a one-to-one function \(\displaystyle f(x),\), which we write as \(\displaystyle f^{-1}(x),\) is a function where the composition \(\displaystyle f\left(f^{-1}(x)\right)\, =\, f^{-1}\left(f(x)\right)\, =\, x.\)

What does this say? I don't know what this means:(. How do I find the inverse of this equation?
This isn't a "problem"; it's a definition. That is, this statement is not a homework exercise, and you are not being asked to find the inverse of any function; indeed, no function is even mentioned. Instead, this defines what an inverse function does. The original function takes "x" and spits out some value of "y". The inverse function takes that "y" and spits out the original "x".

To learn how to find the inverse of a given function, try here. ;)
 
Thank You ?

Thank you all so much for your thorough explanation and advice, much appreciated! I'll be checking out those links, too!
 
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