 In this video, I want to talk about graphing radical functions. And I want to do that through the lens that radicals are the inverses to monomials. But even before that, I want to show us how can we actually compute the inverse of a one-to-one polynomial function? Now, not every polynomial function's gonna be one-to-one. We've seen many of them. They're like, wow, like that. There's so many wiggles and turning points that, you know, there's gonna be many places where it fails, you know, the horizontal line test, something like that. So not every polynomial function's gonna be one-to-one. Therefore, it's not invertible, but there will be many that are. So for example, consider the function f of x equals three x to the fifth plus one, which you can see that right here, three x to the fifth plus one. So looking at that graph, f, this would be the standard y equals x to the fifth, which looks like an odd monomial. You have this type of shape right here. But there's been some transformations done to it, right? We've stretched it vertically by a factor of three. So it's three times a steep. So as you went from the vertex to the next point, you actually go three steps up instead of one step. You see that. And then also it was shifted up by one. So instead of the vertex being at zero zero, it's instead at the point zero comma one. So we have transformed this graph right here. And I do wanna summarize what we have here. So f of x equals three x to the fifth plus one. So we took our, as our basic graph, as the base, you take y equals x to the fifth. And then in terms of transformations, you stretch, I was gonna write it this way, you're gonna stretch it vertically by a factor of three and you're gonna shift it up by one unit. So that's the basic graph of this function right here. That's an important thing to keep track of. Now we can see visually that the yellow graph is a one-to-one function. Like I said, it passes this horizontal line test at all locations in its domain. Since it's one-to-one, it should have an inverse function. How do we compute this inverse function? Well, to compute the inverse function, we're gonna use the technique that we developed previously. So we have our function y, well, f of x, we'll say that first, f of x equals three x to the fifth plus one. When you're trying to solve for the inverse function, what I want you to do is replace the function notation with y, right? So y equals three x to the fifth plus one. And then I want you to switch the roles of x and y. So x becomes y and y becomes x. This then gives you that x is equal to three y to the fifth, excuse me, plus one. And then you need to proceed to solve for y. So to do that, I'd subtract one from both sides so that the one cancels on the left, on the right, excuse me. Then we get three x to the fifth is equal to x minus one. We then perform the next inverse operation, which will be division by three, division by three. So we end up with y to the fifth is equal to x plus one over three, or if you prefer, we're gonna do one third x minus a third. Either one is perfectly fine, it doesn't matter. It doesn't matter which one you approach. I'm gonna stick with the first one. And then, and so notice what we did each time, right? So in the first example, we had this plus one. So we looked at all these numbers that were attached to the y. So there's a five that's attached to the y. There's a three that's attached to the y. There's a one that's attached to the y. And so in the morning, when you get ready to go to class, you put your socks on, then your shoes. But then when you're done with the day, you take your shoes off, then your socks. The order of operations gets reversed. So you were first taught in grade school and such, please excuse my dear Aunt Sally that when it comes to order of operations, we do whatever in parentheses first then exponents, multiplication, division, addition, subtraction, right? There's this order of operation. When we do inverse operations, we do them backwards. So when you're reversing the process, you take your shoes off before your socks, addition becomes the first operation. And so you undo addition by subtracting. So the one is attached to the y via addition. So we undo it by subtracting one on both sides. What's good for the goose is good for the candor. And then you come down to the next one over here. Whoops. We had this three x to the fifth, right? Three was attached to the y. How is it attached to the y? It's attached by multiplication. So to get rid of the three, we're gonna divide both sides by three. What's good for the goose is good for the candor. We have to do it to the same thing, the both sides of the equation to keep equality. We divide by three to cancel out the multiplication by three. And so that gets us to the point we're at right now. This y to the fifth equals x minus one over three. This is the important step here because in order to get rid of the five, how do we get rid of the five? We wanna move the five to the other side of the equation, just like we did with the one and just like we did with the three. Well, how do you move the five to the other side? Well, how is the five attached to the y there? We have y exponent five. That's the operation here, this x one of this power. So we get rid of the power of five by taking the power of the one fifth. That is, we're gonna take the fifth root of both sides. That's the inverse operation in play here. So we need to take the fifth root of y and the fifth root on the right-hand side. This then gives us that y will equal the fifth root of x minus one over three. And thus, this is our inverse function here. F inverse of x is equal to the fifth root of x minus one over three. So the strategy is the same like we saw previously. In order to find the inverse function, you are going to swap the x and y and solve for the new y. And so the graph of this function you can see now illustrated here on the screen. And as these are inverse functions, you should see with this reflection across the diagonal line y equals x. Because all of these points will swap into, will swap their order. So you have this point right here of zero comma one. It's the y intercept of the function. Y intercepts will translate to become x intercepts. You're gonna switch the order around one comma zero like that. Another point we mentioned on the graph right here, you have the point one comma four. Well, when you switch to the inverse function, one comma four becomes the point four comma one. All of the x and y coordinates get swapped around. Look at this point right here. That's not quite negative one, negative one. But this is the points, the points on the diagonal, this diagonal y equals x. That's where these graphs are gonna intersect each other. Because this is a place where the x and y corner are the same. So when you swap them around, nothing gets moved around. Another observation I want you to make here is let's think of this function in terms of transformations. What would the transformations be here? Well, in this situation, the base function, instead of being the monomial y equals x to the fifth, it's gonna be y equals the fifth root of x. So the base functions are inverses of each other. But now notice all of the numbers are inside of the square root. And so if you're inside the fifth root, excuse me, if you're inside the radical, you're actually in the horizontal zone. And therefore everything in the horizontal zone will affect the x coordinates, but it also will always affect things backwards. So if you look at this three, for example, dividing by three actually stretches the function horizontally by a factor of three. And subtracting one in the horizontal zone actually moves the graph in the positive direction, which would of course be the right if we're talking about the positive horizontal direction. And so I want you to compare these things. What were the transformations? We stretched it vertically by three versus we stretched it horizontally by three. It's the same thing, except vertical turned into horizontal here. And then the original graph we shifted in the positive direction by one vertically, which is what we mean by up. And then here we shifted in the positive direction by one horizontally, which of course what we mean by right. In this context, transformations are the same, except we switched everything from vertical to horizontal. And the other would also be true as we switched from a function to its inverse, horizontal transformations turn into vertical transformations because vertical and horizontal are inverses of each other. Let's look at a slightly more involved example here. Take the function f of x equals four x cube plus one divided by two x cube minus three. This would be a rational function. I'm not gonna graph this one, but you could put this in like desmos.com or something. You could see that this rational function would in fact be a one to one function. What we're gonna do is I wanna compute the inverse operation, the inverse function, which you're gonna see is gonna involve radicals. It's gonna actually be the cube root of a rational function. The strategy for doing that is the exact same. So we start off with our function f of x equals four x cube plus one over two x cubed minus three. I want you to realize that the f of x here is just the name of the function, but it represents the y-coordinate. So we get y equals four x cube plus one over two x cube minus three. Then the step that we switched from f of x to f inverse of x, we have to swap the x's with the y's. So the y will become an x, and then the x's will become y's. So we get x equals four y cube plus one over two y cubed minus three. Now, unlike the previous example, this one had multiple y's inside of the formula. So before we can start ripping away numbers attached to y's, we have to reunite the y's. They were separated by war. We need to combine together the family. They miss each other. Now, the y cubed in the denominator is separated by the y cubed kind of like by this Berlin wall here, which we call the fraction bar, the vinculum. It's the fancy word for it. We need to get the y cubed out of the denominator. So to do that, we wanna clear the denominators. On the right hand side, if we were to times both sides by two y cubed minus three, they would cancel out, but what's good for the goose is good for the gander. We have to do it to both sides of the equation. So we get x times two y cubed minus three is equal to four y cube plus one. But like our heroes often face peril when they get past one obstacle, there's another one, right? So now these parentheses are separating our lost family members here. So in order to get the y cubed out of the parentheses, we can accomplish that by distributing the x. This would give us two x y cubed minus three x. This is equal to four y cubed plus one. And so now we're in a situation that the terms are only separated by the equal sign, which is not really a separation whatsoever. We can move the four y cubed to the left hand side by subtraction so that those cancel. And then we'll add the three x to both sides to cancel out and out right there. And so then we'll see that the left hand side has only multiples of y cubed on the left. So we have the two x y cubed that was already there. Then we move the four y cubed to the other side. So they're on the left hand side. The right hand side now consists of three x plus one. You'll now notice that on the left hand side, we gathered every multiple of y. And in particular, everything on the left hand side is a multiple of y cubed. Let's factor it out of the expression here. So we get y cubed times two x minus four is equal to three x plus one. In which case then to start solving the y for y here, what's attached to the y? There's this multiplication of two x minus four to get rid of the multiplication we do division. So we divide both sides by two x minus four so that these cancel out right here. And then let's make that look more like a four. And then we have what's here, y cubed, which is equal to three x plus one over two x minus four. Well, we saw this previously. How do you get rid of the power of three? We need to take the cube root of both sides, like so. And so this then gives us y equals the cube root of three x plus one over two x minus four, which then would be our inverse function. F inverse of x equals the cube root of three x plus one over two x minus four. This would then be the inverse function of that original rational function. Now in these two examples, I looked at situations where the powers of x were odd powers like five and three. This same principle would also work for even powers, but there's just a few caveats we have to worry about. We do have to restrict the domain and range so that they are one-to-one functions. And if ever you have to take a plus or minus, guys, if you have to take a square root, because we've restricted the domain, we'll only focus on the positive case. But as we'll see in the next video, when you start solving equations, we have to make sure we include both the positive and negative possibilities.