Verifying inverse functions by composition

- [Voiceover] Let's say that f of x is equal to x plus 7 to the third power, minus one. And let's say that g of x g of x is equal to the cube root of x plus one the cube root of x plus one, minus seven. Now, what I wanna do now is evaluate f of g of x I wanna evaluate f of g of x and I also wanna evaluate g of f of x g of f of x, and see what I get I encourage you, like always, pause the video and try it out Let's first evaluate f of g of x. That means, g of x, this expression is going to be our input So, everywhere we see an x in the definition for f of x, we would replace it with all of g of x so, f of g of x is going to be equal to so it's going to be equal to, well I see an x right over there so I'll write all of g of x there so that's the cube root of x plus one minus seven, and then I have plus seven plus seven to the third power, minus one notice, whenever I saw the x, since I'm taking f of g of x, I replace it with what g of x is, so, that is, the cube root of x plus one minus seven. Alright, I'll see if we can simplify this. Well, we have a minus seven plus seven so that simplifies nicely. So, this just becomes, this is equal to, I can do a neutral color now, this is equal to the cube root of x plus one to the third power, minus one. Well, if I take the cube root of x plus one and then I raise it to the third power, well, that's just gonna give me x plus one. So, this part this part just simplifies to x plus one, and then I subtract one, so it all simplified out to just being equal to x. So we're just left with an x. So, f of g of x is just x. So now, let's try what g of f of x is. So, g of f of x is going to be equal to, I'll do it right over here, this is going to be equal to the cube root of actually, let me write it out, wherever I see an x, I can write f of x instead, I didn't do it that last time, I went directly and replaced with the definition of f of x but just to make it clear what I'm doing everywhere I'm seeing an x, I replace it with an f of x. So, the cube root of f of x plus one, minus seven. Well, that's going to be equal to the cube root of cube root of f of x, which is all of this business over here so that is x plus seven to the third power, minus one, and then we add one and we add one, and then we subtract the seven lucky for us, subtracting one and adding one, those cancel out. Next, we're gonna take the cube root of x plus seven to the third power. Well, the cube root of x plus seven to the third power is just going to be x plus seven so, this is going to be x plus seven, for all of this business simplifies to x plus seven, and then we do subtract seven and these two cancel out, or they negate each other and we are just left with x. So, we see something very interesting. f of g of x is just x and g of f of x is x. So, in this case, if we start with an x if we start with an x, we input it into the function g and we get g of x we get g of x and then we input that into the function f, then we input that into the function f, f of g of x gets us back to x. It gets us back to x. So we kind of did a round-trip. And the same thing is happening over here. If I put x into f of x... I'm sorry, if I put x into the function f, and I get f of x, the output is f of x, and then I input that into g, into the function g into the function g, once again I do this round-trip and I get back to x. Another way to think about it, these are both composite functions, one way to think about it is, if these are the set of all possible inputs into either of these composite functions, and then, these are the outputs, so you're starting with an x, I'll do this case first, so, g is a mapping, let me write down, so, g is going to be a mapping from x to g of x so, this is what g is doing so, the function g maps from x to some value, g of x and then if you'd apply f to this value right over here if you apply f to this value, the g of x, you get all the way back to x. So, that is f of g of x. And vice versa. If you start with x and apply f of x first, so, if you start with f, if you apply f of x first, let me do that, so, if you apply f of x first, you see you get to this value so, that is f of x, so you applied the function f when you apply the function g to that, you apply the function g to that, you get back. So this g of f of x, I should say, or g of f, we're applying the function g to the value f of x and so, since we get a round-trip either way, we know that the functions g and f are inverses of each other in fact, we can write that f of x is equal to the inverse of g of x, inverse of g of x, and vice versa, g of x is equal to the inverse of f of x inverse of f of x. Hope you enjoyed that.