# Inverse of a function intersecting at y =X line

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 $$y=±x$$??

• Huh??? What is your question? – David G. Stork Jun 5 at 17:02
• Inverse functions are symmetric with respect to the line $y=x$ – J. W. Tanner Jun 5 at 17:11
• What’s the problem? That statement doesn’t say that they only intersect along that line. – amd Jun 5 at 17:14

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 $$y=x$$.

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.