A function \(f\) is defined by \(f:x\rightarrow \dfrac{1}{x+1}\). Write down in similar form expressions for \(f^{-1}\)

To find the inverse of a function we can write \(y=f(x)\) and rearrange to make \(x\) the subject. This gives

\[\begin{align*} y &= \dfrac{1}{x+1}\\ x+1 &= \dfrac{1}{y}\\ x &= \dfrac{1}{y} - 1\\ &= \dfrac{1-y}{y} \end{align*}\]

Replacing \(y\) with \(x\) in the right hand side, we obtain the inverse function \[f^{-1}:x\rightarrow\frac{1-x}{x}.\]

… and \(ff\).

The composite function of \(f\) with itself can be found by ‘replacing the \(x\) in \(f(x)\) with \(f(x)\)’. This gives \[f\bigl( f(x) \bigr) = \dfrac{1}{ \left( \dfrac{1}{x+1} \right)+1 }.\]

We can now simplify by multiplying top and bottom by \(x+1\), yielding

\[\begin{align*} ff(x) &= \dfrac{1}{ \left( \dfrac{1}{x+1} \right)+1 } \times \dfrac{x+1}{x+1} \\ &= \dfrac{x+1}{1 + (x + 1)} \\ &= \dfrac{x+1}{x+2}. \end{align*}\]

Our solution then reads \[ff:x\rightarrow \frac{x+1}{x+2}.\]

It is required to find the values of \(x\) for which \((i)\) \(f=f^{-1}\), \((ii)\) \(f=ff\).

Show that, in each case, the values of \(x\) are given by the equation \[ x^2+x-1=0.\]

Using the result from above, we set \(f(x) = f^{-1}(x)\), giving

\[\begin{align*} \dfrac{1}{x+1}&=\dfrac{1-x}{x}\\ 1&=\dfrac{(1-x)(1+x)}{x}\\ x&=1-x^2\\ x^2+x-1&=0 \end{align*}\]

as required.

Similarly, setting \(f(x) = ff(x)\) we have

\[\begin{align*} \dfrac{1}{x+1} &= \dfrac{x+1}{x+2}\\ 1&=\dfrac{(x+1)^2}{x+2}\\ x+2 &= (x+1)^2\\ x+2 &=x^2+2x+1\\ x^2+x-1&=0. \end{align*}\]

The question does not require us to solve this equation for \(x\), but we could do so using the quadratic formula. We have

\[\begin{align*} x^2+x-1&=0\\ x &= \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -1}}{2 \times 1} \\ &= \dfrac{-1 \pm \sqrt{5}}{2} . \end{align*}\]