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## Algebra (all content)

### Course: Algebra (all content)>Unit 7

Lesson 21: Verifying that functions are inverses (Algebra 2 level)

# 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?
• Yes a function and its inverse are both inverses of each other.
• So basically what Sal is saying at is if f(g(x)) is equal to g(f(x)), they're inverse functions, right?
• Correct. If they are not equal, then the functions are not inverses.
(1 vote)
• 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?
• 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.
• 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))?
• 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).
• 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?
• 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!
• 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?
• 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?
• 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).
• I know that this seems pretty obvious, but ¿Is always a function the inverse of its inverse?
• yes it is, but only if the inverse exists
• 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
• ``sqrt(x-(3/5))``