Why $f(g(x)) = x$ and $f(x)=g(x)$ imply $f(x)=x$?

Question

I was watching this video from blackpenredpen where he solves the equation $\sqrt{5-x}=5-x^2$ by writing it in terms of "5". However, there's a comment with an alternate solution using functions that goes as follows:

Set $f(x) = 5 – x^2$ and $g(x) = \sqrt{5–x}$, then $f(g(x))=x$. Since $g(x) = f(x)$, this implies that $f(x) = x$, which is easy to solve because you can rewrite it as $5 – x^2 = x$.

Could someone explain (prove) why $\begin{cases} f(g(x))=x \\ f(x) = g(x)\end{cases} \implies f(x) = x$ ?

I really don't have any idea how to address this. Thanks in advance.

Answer 1

The comment meant to say that the solutions of the equation $f(x)=g(x)=f^{-1}(x)$ would be found when $f(x)=x$.

This is because the inverse of a function is its reflection in the line $y=x$. So, the point that lies on $y=x$ will remain intact. Ponder upon the statement. Maybe the graphs help you:

However, this logic fails if the function is already symmetric about $y=x$. For instance, solution to $f(x)=f^{-1}(x)$ for $f(x)=\frac1x$ is all real numbers.

Answer 2

For a value $x$ to satisfy $f(x) = g(x)$ given that $f(g(x)) = x$, it must be true that $f(f(x)) = x$. Now assume that $f(x) = x$. If this is the case, $f(f(x)) = f(x) = x$. This shows that the value of $t$ satisfying $f(t) = t$ would also satisfy $f(t) = g(t)$ given that $f(g(x)) = x$ for all $x$.

I also have a more intuitive graphical explanation. If $f(g(x)) = x$, then $g(x)$ is the inverse of $f(x)$. What this means graphically is that $g(x)$ is $f(x)$, but flipped over the line $y = x$. Clearly, for $f(x) = g(x)$, they must be on the line $y = x$ since otherwise it would be impossible for $f(x) = g(x)$. This then means that $f(x) = g(x) = x$.

Answer 3

The original comment on the video has a typo. It should be $g(x)=f^{-1}(x)$. However, as discussed in the answers for that comment, the suggested method (which was successfully used here) does not work in this case. The case in which the conclusion $f(x)=x$ is valid (from the correct hypothesis $f(x)=f^{-1}(x)$) is explained here.