Inverse Functions: Evaluating a Question on Compositions

[DrewD]

How would you answer the following question?

If $(f\circ g)(x)=x$, then $g$ is the inverse function of $f$. True/False?

This is on a test that I gave and I now think it is a bad question, but I'm tired and I want to hear some other people's impressions. The wording of the question is exactly what I wrote above.

 

[symbolipoint]

It is a good question, very fair. Independent variable is x. g is a function of x, and f is a function of x. Think what the hypothesis says. Put x into g and then put g into f; and the result is x, the number that you started with. Function f reversed what function g did. INVERSE!

If you give a function a number x, and the result becomes x, then obviously the function gives you exactly what you gave it. That is what a function composed with its inverse does.

 

[micromass]

It's a very good question.

 

[DrewD]

The reason I have an issue is the example $f(x)=x^2$ and $g(x)=\sqrt x$. For all $x$ in the domain of $(f\circ g)(x)$ the statement is true because the composition is only defined when $x\geq0$. However, $f$ is not the inverse of $g$. I had this on my test for a few years for the reasons stated, but I am not convinced that the implication hold in this direction. That is, I think if $f$ and $g$ are inverses, the equation holds, but if the equation holds, $f$ and $g$ are only inverses given certain qualifications about the domains.

 

[Mark44]

The reason I have an issue is the example $f(x)=x^2$ and $g(x)=\sqrt x$. For all $x$ in the domain of $(f\circ g)(x)$ the statement is true because the composition is only defined when $x\geq0$. However, $f$ is not the inverse of $g$. I had this on my test for a few years for the reasons stated, but I am not convinced that the implication hold in this direction. That is, I think if $f$ and $g$ are inverses, the equation holds, but if the equation holds, $f$ and $g$ are only inverses given certain qualifications about the domains.

That was also the counterexample I thought of, as well. If f and g are inverses, the composition should be commutative, but with these two functions, $(g \circ f)(x) \neq x$.

 

[symbolipoint]

The reason I have an issue is the example $f(x)=x^2$ and $g(x)=\sqrt x$. For all $x$ in the domain of $(f\circ g)(x)$ the statement is true because the composition is only defined when $x\geq0$. However, $f$ is not the inverse of $g$. I had this on my test for a few years for the reasons stated, but I am not convinced that the implication hold in this direction. That is, I think if $f$ and $g$ are inverses, the equation holds, but if the equation holds, $f$ and $g$ are only inverses given certain qualifications about the domains.

That was also the counterexample I thought of, as well. If f and g are inverses, the composition should be commutative, but with these two functions, $(g \circ f)(x) \neq x$.

Good catch! We must distinguish between inverse relations and inverse functions.