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The inverse of a function intersects the function on $y=x$ line.

This is what I was taught. It works fine for $y=x^2, x^3$ ,

Eg $y = x^2$ meet $x= y^2$ at$ (1,1)$ but..

For a function like $ y =-x^3$ It seems to intersect at $ x+y = 0 , $ Why, is the first statement wrong. Also can it so happen , that an inverse of a function meets the function on a point other than on line $ y=±x $??

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  • $\begingroup$ Huh??? What is your question? $\endgroup$ – David G. Stork Jun 5 at 17:02
  • $\begingroup$ Inverse functions are symmetric with respect to the line $y=x$ $\endgroup$ – J. W. Tanner Jun 5 at 17:11
  • $\begingroup$ What’s the problem? That statement doesn’t say that they only intersect along that line. $\endgroup$ – amd Jun 5 at 17:14
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Consider the curve $y=1-x$. It's inverse is $y=1-x$, i.e. it is self inverse. This means it intersects all along its curve, despite only intersecting $y=x$ once.

Now suppose a curve $y=f(x)$ intersects the line $y=x$ at $x_0$. This means that $$y_0=f(x_0)=x_0.$$

Applying $f$ to both sides yields $$ f(y_0)=f(x_0)=x_0, $$ and hence the inverse of the curve intersects the curve at its intersection with $y=x$.

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Inverse functions are symmetric with respect to the line $y=x$.

They don't necessarily contain a point such as $(1,1)$ on that line,

but if they do (e.g., $y=x^2, x^3, ...$), then they intersect there.

In fact, $y=-x^3$ intersects its inverse at $(0,0)$ on that line.

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