Main content

### 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Verifying inverse functions by composition: not inverse

Sal composes f(x)=2x-3 and g(x)=½x+3, and finds that f(g(x)) ≠ g(f(x)) ≠ x, which means the functions are

*inverses.***not**## Want to join the conversation?

- Is it possible to have two functions such that f(x) = g^-1(x), but g(x) does not = f^-1(x)?(26 votes)
- That's a really good question. I think if one function equals the inverse of the other function then it must work vice versa.(5 votes)

- can you have one function = x but not the other?(9 votes)
- No, if either f(g(x)) = x or g(f(x)) = x then they have to be inverse functions, since the functions cancel each other out to result in the unchanged x.(7 votes)

- What if f(g(x)) = x + 3 but g(f(x)) = x + 3? Would they still be equal? How about if one was x + 3 and the other was x -3? How would that be different?(5 votes)
- They have to
**both**equal x to be inverse.(4 votes)

- If in the place of x there may be ( x+a) or (x-b) such that f(g(x))=g(f(x)), then are they considered as inverse functions?(2 votes)
- As long as both functions give the same answer, yes.(2 votes)

- How did this happen?

f(g(x))=3(1/3x^2-9)^2+9

=3(1/9x^4-6x^2+81)+9

This is from Khan Academy (in the hint).

From where they brought (-6x^2)?

Thanks in advance(1 vote)- I've got you:

the*x^2*in*f(x)*is replaced with the function**g(x)* which is *(1/3x^2-9)*, which makes *(1/3x^2-9)*^_2_, which can be expanded as *(1/3x^2-9)(1/3x^2-9)*, which equals *1/9x^4-9/3x^2-9/3^2+81**. Add the two like terms, *(-9/3x^2-9/3x^2)* and you get *(-18/3x^2)*, which is simplified as *(-6x^2)*, which gives you**1/9x^4-6x^2+81**, then you can solve for the rest of f(g(x)), then do the same for g(f(x)).

To simplify,

f(g(x)) = 3(1/3x^2-9)^2+9

multiply (1/3x^2-9)(1/3x^2-9)

simplify (1/9x^4-9/3x^2-9/3^2+81)

f(g(x))=3(1/9x^4-6x^2+81)+9

distribute (3[1/9x^4]) (3[-6x^2]) (3[81]) + 9

add (243+9)

f(g(x)) = 1/3x^4-18x^2+252

I hope that helps!(4 votes)

- Hi

If we work out that f(g(x)) = 0, does that automatically mean that g(f(x)) = 0?(2 votes)- If 𝑓 and 𝑔 are inverses, then the answer is always yes. Because:

𝑓(𝑔(𝑥)) = 𝑔(𝑓(𝑥)) = 𝑥

So in your case, if 𝑓 and 𝑔 were inverses, then yes it would be possible. (This also implies that 𝑥 = 0).

However, if 𝑓 and 𝑔 are arbitrary functions, then this is not necessarily true. One can easily construct a counter example. Try to do so yourself! Comment if you have questions!(2 votes)

- Is it possible that f(g(x)) = x but g(f(x)) is not equal to X.. I mean that why do we prove that both the equations have a final value of x in order to show that both the equations are inverses of each other. ?(2 votes)
- Well, according to wikipedia on inverse functions (https://en.wikipedia.org/wiki/Inverse_function), once you have proved f(g(x))=x then you know f(x) = g^-1(x). In other words, f is the inverse of g. As stated on the wiki page, there is a symmetry of inverse functions. In other words, if g^-1(g(x)) = x then g(g^-1(x)) =x is true. Once you have proved it one way, you don't have to prove it the other way. You are done. However, I think Sal was just illustrating both ways to better explain the issue.(1 vote)

- f(x)=(1)/(x^(3))+3 y(x)=\root(3)((1)/(x-3)) functions are inverses or of each other(2 votes)
- Yes, f(x) = (1/x^3) + 3 and y(x) = cube_root(1/(x-3)) are inverses of each other(1 vote)

- Can you solve this problem please:If f:N tends to N and h: N tends to N be defined by f(x)=3x , g(x)=2x+3 , h(x)=cosx , verify that ho(gof)=(hog)of.(2 votes)
- Why solve for g(f(x)) if f(g(x)) doesn't equal x? both sides have to be x so it should automatically be clear they aren't inverses(1 vote)
- I agree. If f(g(x)) is not equal to x, then f is not an inverse to g and no need to go on. I think Sal was just illustrating a point.(3 votes)

## Video transcript

- [Voiceover] Let's say that
f of x is equal to two x minus three, and g of x, g of x is equal to 1/2 x plus three. What I wanna do in this video is evaluate what f of g of x is, and then I wanna evaluate
what g of f of x is. So first, I wanna evaluate f of g of x, and then I'm gonna evaluate
the other way around. I'm gonna evaluate g of f of x. But let's evaluate f of g of x first. And I, like always, encourage
you to pause the video and see if you can work through it. This is going to be equal to, f of g of x is going to be equal to, wherever we see the x in
our definition for f of x, the input now is g of x, so
we'd replace it with the g of x. It's gonna be two times g of x. Two times g of x minus three. And this is going to
be equal to two times, well, g of x is all of that business, two times 1/2 x plus three, and then we have the minus three. And now we can distribute this two, two times 1/2 x is just
going to be equal to x. Two times three is going to be six. So x plus six minus three. This is going to equal x plus three. X plus three, all right, interesting. That's f of g of x. Now let's think about what
g of f of x is going to be. So g of, our input, instead of being,
instead of calling our input x, we're gonna call our input f of x. So g of f of x is going to be equal to 1/2 times our input, which
in this case is f of x. 1/2 time f of x plus three. You can view the x up
here as the placeholder for whatever our input happens to be. And now our input is going to be f of x. And so, this is going to
be equal to 1/2 times, what is f of x? It is two x minus three. So, two times x minus three, and we have a plus three. And now we can distribute the 1/2. 1/2 times two x is going to be x. 1/2 times negative three is negative 3/2s. And then we have a plus three. So let's see, three is
the same thing as 6/2s. So 6/2s minus 3/2s is going to be 3/2s. So this is going to be
equal to x plus 3/2s. So notice, we definitely
got different things for f of g of x and g of f of x. And we also didn't do a round trip. We didn't go back to x. So we know that these are
not inverses of each other. In fact, we just have to
do either this or that to know that they're not
inverses of each other. These are not inverses. So we write it this way. F of x does not equal the inverse of g of x. And g of x does not equal the inverse of f of x. In order for them to be inverses, if you have an x value right over here, and if you apply g to it,
if you input it into g, and then that takes you to g of x, so that takes you to g
of x right over here, so that's the function g, and then you apply f to it, you would have to get
back to the same place. So g inverse would get us back to the same place. And clearly, we did not
get back to the same place. We didn't get back to x, we
got back to x plus three. Same thing over here. We see that we did not get, we did not go get back to x, we got to x plus 3/2s. So they're definitely not
inverses of each other.