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## Algebra II (2018 edition)

### Unit 1: Lesson 8

Verifying that functions are inverses# Verifying inverse functions by composition

CCSS.Math: ,

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?(8 votes)
- Yes, by definition inverse functions will always be inverses the other way.

Note that f(x) = x^2 is not truly invertible because its inverse is technically not a function.(3 votes)

- Is it always true that if either f(g(x))=x OR g(f(x))=x, then it must also be the case that f(g(x))=g(f(x))=x and the functions are inverses?(4 votes)
- If f and g both have all of ℝ as both their domains and ranges, then yes. But if the domains or ranges are restricted, then not necessarily.

For example, sin(x) has domain ℝ and range [-1, 1]. arcsin(x) has domain [-1, 1] and range [-π/2, π/2].

So if we take a number outside of those ranges, like 3π/4, we have that arcsin(sin(3π/4))=π/4, not 3π/4.(5 votes)

- 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?(4 votes) - As one gets more comfortable with these types of problems, is it okay to use your intuition? I know that simplifying the compounded functions is the only way to make sure you're correct but sometimes you can tell that things are going to cancel out without even having to do the problem. Also, if, say you're given f(x) and h(x), and f(h(x)) simplifies to x, is it reasonably safe to assume that h(f(x)) will equal x too?(4 votes)
- I quote @kubleeka : 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).(2 votes)

- Can someone help me find the inverse of the following function? I've asked some of my friends and they cant figure it out either. F(x)=x^2+3/5(2 votes)
- So basically what Sal is saying at4:20is if f(g(x)) is equal to g(f(x)), they're inverse functions, right?(2 votes)
- Correct. If they are not equal, then the functions are not inverses.(1 vote)

- 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))?(2 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).(2 votes)

- Are every composite functions inverses of each other?(1 vote)
- No. You can compose any two functions you like, so long as their domains and ranges match up right. That is, so long as the outer function is defined for whatever the inner function gives.(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.(2 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)

- 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)

## 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.