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### Course: Precalculus > Unit 1

Lesson 5: Verifying inverse functions by composition- Verifying inverse functions from tables
- Using specific values to test for inverses
- Verifying inverse functions by composition
- Verifying inverse functions by composition: not inverse
- Verifying inverse functions by composition
- Verify inverse functions
- Composite and inverse functions: FAQ

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# Verifying inverse functions by composition

Sal composes f(x)=(x+7)³-1 and g(x)=∛(x+1)-7, and finds that f(g(x))=g(f(x))=x, which means the functions are inverses!

## Want to join the conversation?

- Would it be possible to have two functions where f(g(x)) maps back to x but g(f(x)) doesn't? In other words: if two functions are inverses one way, will they always be inverses the other way?(7 votes)
- Yes a function and its inverse are both inverses of each other.(14 votes)

- So basically what Sal is saying at4:20is if f(g(x)) is equal to g(f(x)), they're inverse functions, right?(7 votes)
- Well, the point isn't that their equal. The main point is that they take us back to x. That is considered the main reason (At least to me).(0 votes)

- Is this true for all functions?(4 votes)
- The method shown in the video is a common way to check if two functions are inverses of each other. If

f(g(x)) = x and

g(f(x)) = x for all

x in the domain of the functions, then

f(x) and

g(x) are inverses of each other. If this isn't true, then they're not inverses. Just find out if f(g(x)) is equal to g(f(x)). This is a handy tool for proving or disproving if functions are inverses!(3 votes)

- Do we know that g(x) is the inverse of f(x) because f(g(x)) and g(f(x)) equal x, or is it because f(g(x)) = g(f(x))?

For example, if it were the first, whenever we solve for a function of a function and get just x, that would mean they are inverses.

However, if it were the second, we would have to solve both ways to compare answers to make sure its the same.(3 votes)- They are inverses of each other because f(g(x)) = x = g(f(x)). Or looking at it another way, g(x) = f^-1(x) "g undoes f". Or f(x) = g^-1(x) and "f undoes g". I suspect that you only have to solve it one way, find you got x and there is your inverse.(1 vote)

- How do I solve for this video's set of equations?

I know the two functions and graphed them on desmos.

https://www.desmos.com/calculator/8eunthihxq

The solution for this set of functions appears to be (-9,-9) but how do I show this algebraically?(2 votes) - If I am verifying inverse functions by compostion and I do f(g(x)) and get x as a result, do I also need to do g(f(x))?(0 votes)
- If you know that f has an inverse (nevermind what it is), and you see that f(g(x))=x, then apply f ⁻¹ to both sides to get

f ⁻¹(f(g(x))=f ⁻¹(x)

g(x)=f ⁻¹(x)

So if you know one function to be invertible, it's not necessary to check both f(g(x)) and g(f(x)). Showing just one proves that f and g are inverses.

You know a function is invertible if it doesn't hit the same value twice (e.g. if the functions is strictly increasing or decreasing).(5 votes)

- I know that this seems pretty obvious, but ¿Is always a function the inverse of its inverse?(1 vote)
- yes it is, but only if the inverse exists(2 votes)

- Do you always have to check 2 cases? I mean f(g(x)) must be equal to x and g(f(x)) must be equal to x? Or it's enough to check only one composite function?(1 vote)
- There is no need to check the functions both ways. If you think about it in terms of the function f(x) "mapping" to the result
*y_ and the inverse f^-1(x) "mapping" back to _x*in the opposite direction, one always gives you the result of the other. Therefore, once you have proven the functions to be inverses one way, there is no way that they could not be inverses the other way.

Similarly, if you look at a graph, inverses will always mirror each other over the line y=x. There is no way you can alter a reflection to let it be true one way and not the other!(2 votes)

- But why does this work? How does showing f(h(X))=h(f(x))=x prove that the two functions are inverses? It doesn't seem to be clicking for me.(1 vote)
- Think of the definition for the inverse of a function. The inverse for a function y=x is x=y. Thus this should logically make sense if you say that f(x) is x and h(x) is y. Thus the composition of the 2 functions would prove they are inverses.(2 votes)

- but g(-2) is non real so why is f(g(-2))=-2?(1 vote)
- If you are referring to the functions in the video, g(-2) is a real number. You can do cube roots of negative numbers. g(-2) = ∛(-2+1)-7 = ∛(-1)-7 = -1-7 = -8.

Notice: ∛(-1) = -1 because (-1)^3 = -1

A square root of a negative number is not a real number. But, cube roots of negative numbers are real numbers.

Hope this helps.(2 votes)

## Video transcript

- [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.