Will inverse functions, and functions always meet at the line $y=x$?

Question

If I have a function, the inverse function, by definition will be a reflection of the original function in the line $y=x$, so if I wanted to find the point of intersection, instead of solving it with equating both of the functions to equaling each other, could I assume that the point of intersection, between the two functions, will always be at the point $y=x$?

This would enable me to solve problems much easier, instead of having to solve quartic equations, for example.

Answer 1

Not necessarily. If $f(x)=-x$, this implies that $f^{-1}(x)=-x$, (since $f(f(x))=f(-x)=x$), so the graphs of $f$ and $f^{-1}$ intersect everywhere.

Another example is $g(x)=-\frac{1}{x}$, which is also its own inverse, but doesn't intersect $y=x$ at all.

Answer 2

As already pointed out by sky90 and Marra in the comments, in general a function and its inverse do not need to have an intersection. This can be seen from the example given in the comments. Another example would be $f(x)=\exp(x)$ and its inverse $f^{-1}(x) = \log(x)$, whose graphs never intersect.

Note that apart from the case mentioned by Chappers, where a function is its own inverse and there is an infinite number of intersections, you can also find examples with a finite set of intersections, e.g. $f(x) = -x^3$ and its inverse $f^{-1}(x) = -\sqrt[3]{x}$, whose graphs intersect at the points $(-1,1)$, $(0,0)$, $(1,-1)$, where the first and the latter are clearly not on the line $y=x$.

Answer 3

Building upon the other answers:

Generally, $f$ and $f^{-1}(x)$ intersect at every $x$ and $f(x)$ for which $f(f(x))=x$. Vizualize this as a pair of points mirrored on the line $f(x)=x$. Especially, they intersect at every $x$ for which $f(x)=x$, which are the points precisely on this 'mirror'.

Chappers constructed a function for which this holds for every $x$, and thus meets its inverse everywhere. It is just a line straight through the 'mirror', which stays the same when mirrored.

But it is easy to see that $f(f(x))$ does not necessarily hold at all, for example when $f(x)$ always stays greater, or always stays less than $x$, as in mkausp's examples $exp$ and $log$. They never cross the mirror and so can not contain a pair of mirrored points.

Answer 4

You are only half right. They will always be mirrored about the straight line $ x=y$. Intersection of $x=y$ and the function ( or its inverse) supplies all the real solution points, only if portions of the graph lies below line $ x=y$.

Wherever x occurs rub off and put y.Wherever original y occurs rub off and put x. That procedure gives you the inverse function mirrored about $x=y$ whether or not it cuts $x=y$ at real points.

For example the circle

$$ x^2+ y^2 + 5 x + 3 y + 6 = 0 $$

has roots $(-3,-1) $ in x or y by your method.

So $(-3,-3), (-1,-1) $ are points of intersection without further calculation.

Answer 5

You can prove that if $f$ is a strictly increasing function (of course it is 1-1 and so it has an inverse), then the points of intersection of $C_f$ (the graph of $f$) and $C_{f^{-1}}$ lie on the line $y=x$. Assume a point $A(x,y)$ that is an intersection point of $C_f$ and $C_{f^{-1}}$. Since $x,y\in\mathbb{R}$, one of the following holds: either $x=y$, or $x>y$, or $x<y$. We will just show that the last two lead to a contradiction due to the fact that $f$ is increasing. First of all since $A\in C_f, C_{f^{-1}}$, it holds $y=f(x)=f^{-1}(x)$. Now we take cases:

• Let $x>y$. Then $x>y \Rightarrow x>f^{-1}(x)\Rightarrow f(x) > f(f^{-1}(x)) \Rightarrow y>x $ a contradiction.
• Now let $x<y$. Similarly $x<y \Rightarrow y<x$ a contradiction.

Therefore $y=x$, which means graphically that the points of intersection lie on the line $y=x$. ($f$ increasing $\Rightarrow$ (if $(x,y)\in C_f, C_{f^{-1}}$ then $y=x$)).

However the restriction that $f$ be strictly increasing is absolutely necessary since as others have mentioned it is a quite common case that the curves intersect outside $y=x$. If your problems involve increasing functions you may simply state $f(x) = f^{-1}(x) \Leftrightarrow f(x) = x$ (The converse is trivial: if $f(x)=x$, then $x = f^{-1}(x)$, so $f(x)=f^{-1}(x)$ and it holds for any 1-1 function) and reduce your equation. This technique is useful not only for polynomials but also for functions involving say logarithms and exponentials. Consider $f(x) = \ln(x-1) + x$. Without the above theorem you get to solve a system of equations involving transcendental functions, whereas with the theorem you just prove $f$ is increasing and then you immediately get that the sole point of intersection of $f$ with its inverse is $(2,2)$. This means that although you can't use it with $y = -x^3$ of mkausp, the intersection points of $y=x^3$ are bound to be on $y=x$. Hope this helps.