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### Course: Precalculus>Unit 1

Lesson 5: Verifying inverse functions by composition

# 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 not inverses.

## 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)?
• That's a really good question. I think if one function equals the inverse of the other function then it must work vice versa.
• can you have one function = x but not the other?
• 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.
• 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?
• They have to both equal x to be inverse.
• 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?
• As long as both functions give the same answer, yes.
• 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)?
(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
f(g(x)) = 1/3x^4-18x^2+252

I hope that helps!
• Hi
If we work out that f(g(x)) = 0, does that automatically mean that g(f(x)) = 0?
• 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!
• 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. ?
• 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