For an invertible function, is the graph of its inverse always the mirror image of the function's graph in the line $y=x$? And, if yes, then why is it so?
$\endgroup$4 Answers
$\begingroup$Ans to your first question is Yes.
It is so, because if $f(x_1)=y_1$, then $f^{−1}(y_1)=x_1$. It's easy to prove that the line segment joining $(x_1,y_1)$ and $(y_1,x_1)$ has slope −1 and is bisected by the line y=x.
$\endgroup$ $\begingroup$If you mean the graph of the functions, then yes, the graph of the inverse $g:B\to A$ of a function $f:A\to B$, where $A,B\subseteq\Bbb R$, is always obtained from the graph of $f$ by reflection in the diagonal with equation $x=y$, which reflection is the map $(x,y)\mapsto(y,x)$ that swaps coordinates.
The reason is that by definition for each $y\in B$ the value $g(y)$ equals the value $x\in A$ for which $f(x)=y$ (such $x$ should exist and be unique, otherwise no inverse of $f$ exists in the first place). In other words $g(y)=x$ holds if and only if $f(x)=y$ holds. Now the graph of $f$ is the set of points $\{\,(x,f(x))\mid x\in A\,\}$, which one can write somewhat more cumbersome as $\{\,(x,y)\mid x\in A,y\in B,y=f(x)\,\}$. But now we can apply the equivalence mentioned to give $\{\,(x,y)\mid x\in A,y\in B,x=g(y)\,\}$ and finally simplify again to obtain $\{\,(g(y),y)\mid y\in B\,\}$. Finally applying the reflection in $x=y$ gives the set $\{\,(y,g(y))\mid y\in B\,\}$, which is the graph of$~g$.
$\endgroup$ $\begingroup$Yes, because they share lots of things in common like the range of the inverse function $f^{-1}$ which equals the domain of the function $f$ and vice versa. So they should for sure become symmetric; because they're basically 100% similar but in reverse just like you in a mirror.
$\endgroup$ $\begingroup$Yes, I used to find the same thing confusing, this is the best way I found to think about it: the first step is to believe that swapping x and y is the same as reflecting wrt $y=x$ (this should not be difficult).
Then, if $g$ is the inverse of $f:A\to B$, it means that $x = g(y)$. In other words, I imagine that I can write a formula $g(y_0)$ that gives me the $x$ corresponding to a given $y_0$ for every possible $y_0$. If I consider the points satisfying my original equation $y=f(x)$ and the new $x=g(y)$, these are the same curves in the plane.
However, if I want to graph the function corresponding to "formula" $g$, I will do as I always do, and plot $y = g(x)$. This, compared to the original graph $x=g(y)$, corresponds to switching x and y, i.e. the reflection of the original graph.
Note that I didn't discuss details on domain and codomain, the above cosiderations assumes they are "good enough".
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