 Hello, everyone, and welcome to this lesson on derivatives of algebraic inverse functions. We are not talking here of inverse trig functions, nor are we talking about logarithmic functions or exponential functions. We are talking about finding inverses of purely algebraic functions. So let's begin with a review of some things that hopefully sound familiar to you from Algebra 2, starting with the definition of a function. Recall that a function is a set of ordered pairs of relation in which each element of the domain is paired with one and only one element of the range. When you learned about functions, perhaps you recall that there are three representations of a function. The first would be that as a relation, a set of ordered pairs. So for example, we have here the function that is given by the relation of points 5, 3, 2, 3, negative 1, 0, and negative 3, negative 1. The second representation would be that of a graph. And what we have here is a plotting of those four ordered pairs that were in the relation I just provided for you. Starting over on the far left, we have the point negative 3, negative 1, the point negative 1, 0, then 2, 3, and 5, 3. This would be an example of a function because notice each x is paired with only one y. Perhaps you remember the vertical line test. If you were to take a vertical line seen here in red and run it along your graph from the left to the right, if you find that vertical line passes through the curve or the set of points only one place at a time, then you know that you have a function because each x is paired with only one y value. The last of the three representations of a function is that as an equation. So for example, suppose we have the function denoted by f of x to be equal to 2x squared minus 5. Now obviously this is not the equation that models those four ordered pairs in the relation I gave you before. But this is indeed a function because for each x that you substitute into this function, you will obtain only one answer, only one y value. However it is a function. Of course we all know we can graph functions on our graphing calculator. Here we have the graph of that 2x squared minus 5. Think also, since we're looking at the graph, of the vertical line test. If you were to run a vertical line across the graph from the left to the right, that vertical line is passing through the curve only one place at a time, therefore a nice visual for the fact that we do indeed have a function. You would have also learned how to find the inverse of a function from each of these three representations. So let's begin with the representation as a relation. Here we have the same relation that I provided you with before. If we want to find the inverse of this, we simply flip-flop the x and the y values. You will also hopefully remember the notation for the inverse of a function which is f negative one of x. Therefore we obtain the inverse here, 3,5, 3,2, 0, negative 1, negative 1, negative 3. Now the big question is whether or not the inverse you find is itself a function or not. If you take a look at this relation that we have as the inverse of f of x, notice how this x value of 3 is paired with two different y values. It's paired with 5 and then it's paired with 2. Therefore this relationship that is the inverse of f of x is not a function itself. Let's look at finding an inverse graphically. What we have here is the plotting of those ordered pairs we obtained by finding the inverse of that relation we just looked at. So going from the left to the right, that point that was originally negative 3, negative 1, when you invert or flip-flop the x and y values, you now have the point negative 1, negative 3. The point that was originally at 1,0 when you flip-flop the x and y values is now at 0, negative 1. The point that was originally at 2,3 becomes 3,2 and the point that was originally 5,3 becomes the point 3,5. Once again you have a great visual, if you think of the vertical line test, of the fact that this does not represent a function. It fails the vertical line test because you would have the vertical line passing at x equals 3 through two different points. So once again we have a great visual of the fact that this is not a function. Finally, perhaps the most common type of problem you encountered in finding inverses before is when you are working with an equation. So here we once again have our function 2x minus 5. We utilize the same basic idea of flip-flopping the x's and the y's. Now remember f of x is really y and we're going to start by literally exchanging or flip-flopping the x for the y and vice versa. So when we do that, we have x equals 2y squared minus 5. We want to now solve for y. So we have to add over the 5, divide by 2, that isolates the y squared. So when we solve for y, obviously we're going to have to take the square root. And we do need to remember that it's both the positive and negative square root. Our calculators will nicely graph the inverse of a function. So let's go take a look at that and how to do that. So here we have the graph of that parabola 2x squared minus 5. If you want to have the calculator draw the inverse for you. Go back to your working screen and you'll notice over the program button it says draw. So we're going to do second program to bring up our draw menu. And you will notice that number 8 is draw inverse. So simply hit 8. We need to tell the calculator we want it to draw the inverse of that equation we have under y1. So we need to bring up y1. Simply hit your vars button, go across the top to yvars, into function, the first option and voila there is your y1. So simply hit enter once again. So now you'll see it says draw the inverse of that which we have under y1. Hit enter again and you'll notice the view pops to your graph and it draws that sideways parabola that is the inverse of the original f of x function. This also confirms how when we were writing the equation of the inverse algebraically and we took the square root remember of both sides we said it had to be the positive and negative square root. Well this makes sense now because for any of these x values except for negative 5 there are two y values both the positive and the negative of the square root. It also confirms once again if you think of the vertical line test that this sideways parabola is obviously not a function because except for that point at the vertex over at x equals negative 5 the vertical line you would have would cut through the sideways parabola in two places every place you were to move it. So when we rearrange this equation this which we ended up with is our equation for the inverse. So as we've been working through these and recalling how in order to find an inverse we flip-flop the x and the y that brings to mind one of the major characteristics of a function and its inverse. The fact that the domain of f of x becomes the range of the inverse f negative 1 of x and the range of the original function f of x becomes the domain of f negative 1 of x and remember of course the domain is our set of x's the range is the set of y's so this simply is confirming how we know the process works to find the inverse of flip-flopping the x and the y. As we have seen though the big question becomes if the inverse we find is a function itself it does not necessarily have to be the only types of functions that are going to have inverses that are also functions are what we call one-to-one functions. I don't know if you've heard those before a one-to-one function is a function whereby each x is paired with only one y which we know it has to be by definition of a function but at the same time each y comes from only one x it's a one-to-one relationship between the x's and the y's hence the name one-to-one function. So how do you know if a function is one-to-one? You can simply look at the values of the original function and see if each x has only one y as is required by definition of a function but also that each y comes from only one x. Now another way to do this since you are in calculus now you can use the derivative to help you determine if the original function f of x is a one-to-one function. You would find f prime of x the first derivative. You would graph that first derivative and if it is the case that the entire derivative graph is greater than zero meaning it lies above the x-axis for all values of x or if that first derivative graph is less than zero meaning it lies below the x-axis for all values of x then you know the original f of x function is a one-to-one function and therefore you know that when you find the inverse that inverse is guaranteed to also be a function itself. There are two theorems related to functions and their inverses. The first is that if we know the original function f of x is continuous on its domain then the inverse of f of x will also be continuous on its domain and remember the domain of the inverse of f of x would have been the range of the original function. Secondly if f of x is differentiable at the point c comma d meaning you can find the derivative at the point c comma d and if f prime of c is not equal to zero then the inverse of f of x will be differentiable at the point d comma c. Now take a little notice of this and remember what we just discussed about ordered pairs on the graph of the original function versus on the graph of its inverse. If the point on the original function is c comma d remember that on the inverse you need to flip-flop the x's and y's so the point that would then lie on the inverse graph is going to be d comma c. That becomes incredibly important when you're doing some of these problems. So let's now turn to the matter at hand and talk about how to find the derivative of an algebraic inverse function. We're going to start out knowing that we have a function f that is differentiable on an interval i and the function f is going to have an inverse that we are going to denote as f negative one. Then the derivative of that inverse is simply one over the derivative of the original. Now take note of the notation that you see here. These are two of the common notations you're going to see. d dx is obviously derivative and we're finding the derivative of f negative one of x the inverse. Another notation you might see is f negative one prime just like we typically would have f prime. This is f negative one prime of x. Yes it's this simple. You simply do one over the derivative of the original and that is going to always be the derivative of the inverse equation. Most of the time however you are concerned with the derivative at a certain point. So for instance if you wanted to find the derivative at a certain point and we know the point c comma d is on the graph of the original function f. Then we would have the derivative of the inverse function at d is going to be one over f prime of c. Now think again about what we know about ordered pairs on the original graph f of x versus on the inverse graph. c comma d is on the original graph so that means the point on the inverse would be d comma c. If we're finding the derivative of the inverse we all know that number inside the parenthesis is an x value d is the correct x for the point on the inverse graph. That's going to be equal to one over f prime of c because c is the correct x on the graph of the original. That's typically what people will mess up is using the correct x value in the correct place. So let's take a look at an example to serve as a proof that this actually works and we're just going to consider an easy linear equation 2x plus 3 and we want to determine the derivative of its inverse function. So we want to find the derivative of the inverse and we're not looking at this point for a certain location on the graph, we're just looking for the derivative in general. Remember how the rule goes, it's simply one over f prime of x. So we'd have one over the derivative of 2x plus 3 is simply 2. So this is saying that the derivative of the inverse is a half. So to prove it to you in case you're not convinced, let's do it the old fashioned way. Let's actually find the equation of this inverse of the given f of x and then we're actually going to take the derivative of it. So if we want to find the inverse of this equation, we're going to have to flip-flop the x and the y. We're going to have to solve for y so we'd have x minus 3 over 2 equal to y. That is our inverse. Now to make this easy, let's perhaps rewrite it. So our inverse, if we think of x over 2, well that's one half x minus 3 halves. Geez, what's the derivative of that? One half. It's amazing it works, but it will always work and that's the really neat thing about taking derivatives of algebraic inverse functions. It can save you a lot of time, especially when you have more complicated functions. You don't have to actually find the equation of the inverse. You simply invoke this theorem that we know it's going to the derivative of the inverse is going to be 1 over the first derivative of the original and then if by chance you are asked for the derivative at certain points, you just have to be careful to use the correct x with the correct function. That's what you're going to see demonstrated in the example videos that will be provided for you. These typically can be pretty easy. Don't make them harder than they have to be. You do need to keep your thoughts organized. The hardest part is making sure you think through them correctly. But if you do that, I think you will see a good bit of success with these.