Why is it that when $f$ is an increasing function then the points of intersection of $f$ and $f^{-1}$ lie on the line $y=x$?
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Suppose $(x,y)$ is an intersection point of $f$ and $f^{-1}$. Then $f(x)=y$ and $f^{-1}(x)=y$, thus $f(x)=y$ and $f(y)=x$. If $x < y$ this would mean that $f$ is not increasing on $[x,y]$, whereas $x>y$ would mean that $f$ is not increasing on $[y,x]$. Hence $x=y$, implying that $(x,y)$ lies on the line $y=x$.
user133281
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How exactly does $x<y \implies f \text{ is decreasing on }[x,y]$? All I see is that $f(x)=y$ and $f(y)=x<y$. – D Poole Aug 28 '14 at 18:04
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It means that the function cannot be increasing on $[x,y]$, since in that case we would have $f(y) \geq f(x)$. – user133281 Aug 28 '14 at 18:10
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Functions can be both non increasing and non decreasing. – D Poole Aug 29 '14 at 03:13
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You're absolutely right, which is why I edited my answer. – user133281 Aug 29 '14 at 07:09
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Nevermind, I missed that the question says that $f$ is increasing – D Poole Aug 29 '14 at 16:19