Sketch the following functions and determine which of them are self-inverse.

In order to sketch the graph of \(y=g(x)\) it is helpful to rewrite it as \[y=\frac{x+1}{x-1}=\frac{x-1+2}{x-1} = 1+\frac{2}{x-1} .\] This reveals that the graph is a transformation of \(y=\dfrac{1}{x}\).

graph of x+1 over x-1

We compose \(g(x)\) with itself: \[ gg(x) = g\left( \frac{x+1}{x-1} \right) = \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} = \frac{x+1+(x-1)}{x+1-(x-1)} = \frac{2x}{2} = x \]

So \(g(x)\) is self-inverse.

Strictly, for \(g\) to be self-inverse, we should check that its range exactly matches its domain since its output is used as its own input in the composition.

What are the domain and range of \(g(x)\)?

By rewriting \(m\) we find that, as for \(g\), its graph is a transformation of \(y=\dfrac{1}{x}\).

Composing, we find \[\begin{align*} mm(x) &= m\left( \frac{x-1}{x+1} \right) = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}\\ &= \frac{x-1-(x+1)}{x-1+(x+1)} = \frac{-2}{2x} = -\frac{1}{x} , \end{align*}\]

so \(m(x)\) is not self-inverse.

What is different about \(g\) that makes it self-inverse while \(m\) is not?

graph of x-1 over x+1

What happens if we compose \(mm(x)\) with itself to make the function \(mmmm(x)\) (which could be written as \(m^4(x)\))?

Of the remaining functions, we find that \(h\), \(k\) and \(p\) are self-inverse.

What do the graphs of the self-inverse functions have in common?

We have sketched the four self-inverse functions on the same set of axes.

sketch of functions g h k and p

We remember that the graphs of any function and its inverse are reflections of one another in the line \(y=x\). So it is no surprise that the graph of each self-inverse function is a reflection of itself in \(y=x\).

Can you match up the four functions with their graphs?

Sketch and find equations for some other self-inverse functions.

By thinking about the reflection property we can see that any straight line parallel to the purple line in the graph above will represent a self-inverse function. Can you write down a generalised form of the equation?

Also, any translation of \(y=\dfrac{1}{x}\) will be self-inverse so long as the origin is transformed to a point on the line \(y=x\). What about stretches, reflections and other transformations?

Can you find any other families of self-inverse functions?