Where will the function intersect its inverse I am stuck on the following question. This came in my exam today.
It gave the following graph :

The question asked what would be a reasonable point where the graph would intersect with its inverse. Is it $(2,3), (-2,3),(3,3),(3,-3),(3,-4)$?
I only knew that in order to take inverse of a function, we nee to consider the reflection of the curve about the line $y=x$.
But I had no idea how to take the reflection of this given curve.
Also I could answer this question. I want to learn how to do these type of questions.
Can someone kindly help me to solve this?
 A: Claim: For each strictly increasing function $f$ and each $x$ in its domain, $$f(x)=f^{-1}(x)\iff f(x)=x.$$
Proof

*

*RHS $\implies \big(x=f^{-1}(x)\,$ and $\,f(x)=x\big)\implies$ LHS.

*For the foward direction, let's assume for the sake of contradiction that $f(x)=f^{-1}(x)$ and $f(x)\ne x$ are both true for some strictly increasing $f$ and some $x$ in its domain. The former implies that $f(f(x))=x,$ while the latter implies that $f(x)<x\,$ or $\,f(x)>x,$ which, since $f$ is increasing, respectively implies that $f(f(x))<f(x)\,$ or $\,f(f(x))>f(x),$ which, by the previous inequality pair, respectively implies that $f(f(x))<x\,$ or $\,f(f(x))>x,$ that is, that $f(f(x))\ne x.$ We have derived a contradiction, so the forward direction must be true.

P.S. Note that in general, $$f(x)=f^{-1}(x) \kern.6em\not\kern-.6em\implies f(x)=\pm x.$$ For example, let $g(x)=-\dfrac1x\,$ and $\,h(x)=1-x.$ Then $y=g(x)$ and $y=g^{-1}(x)$ have infinitely many intersections, as do $y=h(x)$ and $y=h^{-1}(x);$ yet $y=\pm x$ has finitely many intersections with $y=g(x)$ and $y=h(x).$

From the above-proven theorem, since the given function is increasing and intersects the line $y=x$ exactly once, it must intersect with its inverse exactly once at some $(x,x).$ So, from the given options, $(3,3)$ is a reasonable point where the graph intersects with its inverse.
A: If $f(x) = f^{-1}(x)$ then $f(f^{-1})(x) = f(f(x)) = x $.
It seems that there is a value $a$ such that $f(a) =a $ and such that if $x>a$ we have $f(x)>x$ and such that if $x<a$ we have $f(x)<x$. Because the line $y=x$ intersects the curve only once.
It follows that if $x<a$ we have $f(f(x)) < x$ and if $x>a$ we have $f(f(x)) > x$.
Hence the only feasible point is $(a,a)$ which looking at the options could only be $(3,3)$.
