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I've been working through a textbook and noticed that if a function intersects with its inverse function it's always on the line y = x. Why is this?

Anonymous
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2 Answers2

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Well, the statement is not true. Take any bijective function for which $f(1) = 2$ and $f(2)=1$ - and just make the graph not be symmetric in the diagonal ($y=x$). Then the graph of $f$ and that of $f^{-1}$ may intersect on the diagonal, but they will certainly also meet at $(1,2)$ and at $(2, 1)$. (Sorry for the poor drawing...) By the way, the condition "not symmetric in the diagonal" is not really necessary; at the extreme, the function $f(x)=-x$ is its own inverse, so it intersects itself at all the points on it. To avoid such brain twisters, just assume the graphs only intersect in a few points.

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If $f(5) = 22$ then $f^{-1}(22) = 5$. $\qquad\longleftarrow$This is the main idea, without which this will be incomprehensible.

Similarly:

If $f(9) = 9$ then $f^{-1}(9) = 9$. And the point $(9,9)$ is on the line $y=x$, since $9=9$.

Michael Hardy
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  • It's not even clear - is this meant to be a proof of the statement the OP proposed? Fortunately it doesn't seem like you are claiming that... –  Aug 28 '16 at 01:07
  • @mathguy : Obviously it's not supposed to be a proof. It's an explanation of the central idea: the roles of $x$ and $y$, in this case $5$ and $22$, get interchanged. But if $x$ is the same as $y$ (in this case $9$ is the same as $9$) then you have a point on the line $y=x$, playing the same role in the statement $f(9)=9$ (involving $f$, not $f^{-1}$ and in the statement $f^{-1}(9) = 0$ (involving $f^{-1}$, not $f$) so it's on the graph of both $f$ and $f^{-1}$. $\qquad$ – Michael Hardy Aug 28 '16 at 01:35
  • But then it is also not related to the OP's question, which was "why the intersection points are always on the line $y=x$." If you had no intention to answer, then this should have been a comment, not an answer. –  Aug 28 '16 at 01:57
  • @mathguy : It very directly addresses that very question. $\qquad$ – Michael Hardy Aug 28 '16 at 02:10
  • "But if $x$ is the same as $y$" does not address that question. The question was in the opposite direction: if the point is on both graphs, do $x$ and $y$ have to be the same. –  Aug 28 '16 at 02:13
  • @mathguy : The answer I wrote should make it perfectly clear what the answer to that is. $\qquad$ – Michael Hardy Aug 28 '16 at 15:38