Function inverse example 1 | Functions and their graphs | Algebra II | Khan Academy

So we have f(x) = -x + 4. We'd like to figure out the inverse of f. To figure out the inverse, what I like to do is write y = f(x). Now, we solve for x in terms of y. So we have y = -x + 4. Adding x to both sides of the equation, we get y + x = 4. Subtracting y from both sides, we get x = 4 - y. We're always used to writing the dependent variable on the left side. We could rewrite this as x = 4 - y. This is the inverse function right here. We've written it as a function of y, but we can just rename the y as x, so a function of x. So let's do that. Rename y x; we get f^-1(x) = 4 - x.

Let's identify this function. Out of interest, let's graph the inverse function and see how it might relate to this one right over here. So if you look at it, it actually looks fairly identical. -x + 4 is the exact same function. So let's see if we have the y-intercept is 4. It's going to be the exact same thing. The function is its own inverse; it's its own inverse. If we were to graph it, we would put it right on top of this, right on top of this. So there are a couple of ways to think about this. In the first inverse function video, I talked about how function and inverse are the reflection over the line y = x. Where is the line y = x? Here, the line y = x looks like this. The line y = x looks like this. And -x + 4 is actually perpendicular to y = x. If we flip it over, it's its own reflection.

Let's make sure that that actually makes sense when we're dealing with, when we're dealing with I guess the standard function right there. If you input a 2, it gets mapped to a 2. If you input a 4, it gets mapped to 0. What happens if you go the other way? If you input a 2, well, 2 gets mapped to 2. That's what we expect for the regular function. Let's map to zero for the inverse. Regularity explicitly might be obvious to you just in case. Might be helpful. f(5) = 0, or we could say the function f maps 5 to 0. What's f^-1(0)? f^-1(0)? f^-1(0)? f^-1(5)? Again, think these domains and ranges. Dom(f) range(f). f will take us... f takes us... 5. f^-1(5) is just like, just like it's supposed to do.

Let's do one more of these. So here I have g(x) = -2x - 1. Just like the last problem, I like to set y equal to this. So we have y = g(x) = -2x - 1. Solve for x. y + 1 = -2x. Just added 1 to both sides. Now we can divide both sides of this equation by 2, and so you get y/2 - 1/2 = x. We could write x = y/2 - 1/2, or we could write f^-1 as a function of y = y/2 - 1/2. Rename y x. f^-1(x)... f^-1(x)... careful here, that shouldn't be f. The original function was g, so let me be clear: g^-1(y) = y/2 - 1/2. We started with g(x), not f(x). Make sure we get the notation right. We could rename the y and g^-1(x) = x/2 - 1/2. Let's graph it. The y-intercept is -1/2 right over there, and it has a slope of negative 1/2. It has a slope of negative one-half. If we start at -1/2, if we move over to 1 in the positive direction, go down a half. We move over 1 again, go down a half again. We're back... I'll try my best to draw that. Keep in mind the directions. Let's see if it's reflected over y = x. y = x, y = x. It looks like that reflection. Reflect, reflect. Yes, inverse function. Yes.