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Current time:0:00Total duration:5:12

AP.CALC:

FUN‑3 (EU)

, FUN‑3.E (LO)

, FUN‑3.E.1 (EK)

let G and H be inverse functions so let's just remind ourselves what it means for them to be inverse functions that means that if I have two sets of numbers let's say one set right over there that's another set right over there and if we view that first set as the domain of G so if you start with some X right over here G is going to map from that X to another value which we would call G of X G of X that's what the function G does now if H is the inverse of G and frankly vice-versa then H could go from that point G of X back to X so H would do this H would get us back to our original value so that's what the function H would do and so we could view this point right over here we could view it as X so that is X but we could also view it as H of G of X so we could also view this as H of G of X and I did all of that so that we can really feel good about this idea if someone tells you that G and H are inverse functions that means that that means that H of G of X is X H of G of X H of G of X is equal to X or you could have gone the other way around you could have you could have started with a well you could have done it in multiple different ways but also G of H of X I could have just swapped these letters here the the letters H and G or somewhat arbitrary so you could have also said that G of H of X is equal to X so G of H of X is equal to X and then they give us some information the following table lists a few values of G H and G Prime all right so they want us to evaluate H prime of 3 they don't even give us H prime of 3 how do we figure it out they gave us G prime and H and G how do we figure this out well here we're going to actually derive something based on chain rule and this isn't the type of problem that you'll see a lot of but it is interesting so we're going to work through it and you there's a chance that you might see it in your in your calculus class so let's start with either one of these expressions up here so let's let's start with the expression well let's start with the let's do it this this one over here so if we have G of H of X is equal to X so let me put that H of X back there which is by definition true if G and H is are inverses well now let's take the derivative of both sides of this so let's take the derivative with respect to X of both sides derivative with respect to X and on the left hand side well we just apply the chain rule this would be G prime of H of X G prime of H of x times H prime of X that's just the chain rule right over there and then that would be equal to what's the derivative with respect to X of X well that's just going to be equal to 1 so now it's interesting we need to figure out what H prime of 3 is we can figure out what H of 3 is and then we can use that to figure out what G prime of whatever H g prime of H of 3 is and so we should be able to figure out H prime of X or we could just rewrite it this way we could rewrite that H prime of X is equal to is equal to 1 over G prime of H of X now in some circles they might encourage you to memorize this and maybe for the sake of doing this exercise on Khan Academy you might want to memorize it but I'll tell you 20 years after I took a calculator made it almost 25 years after I took calculus this is not something that I retain in my long term memory but you I did retain that you can derive this from just what the definition of inverse functions actually are but we can use this now if we want to figure out what H prime of 3 is H prime of 3 is going to be equal to you 1 over 1 over G prime of H of 3 which I'm guessing that they have given us so H of 3 when X is 3 H is 4 so that is H of 3 there so H of 3 is 4 so now we just have to figure out G prime of 4 well lucky for us they have given us when X is equal to 4 G prime G prime is equal to 1/2 so G prime of 4 is equal to 1/2 so H prime of 3 is equal to 1 over 1/2 so 1 over 1/2 1 divided by 1/2 the same thing as 1 times 2 so this is all equal to 2 and we are done

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