For which $x$-values does a function equal its inverse?

For which $x$-values does a function equal its inverse?

Question

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

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 $x^2+x-1=0.$

Suggestion

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

How can we find the inverse function? What happens if we make $x$ the subject of $y=f(x)$?

… and $ff$.

Would it help to view the composite function $ff(x)$ as $f(f(x))$?

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 $x^2+x-1=0.$

From the above, we should now know $f(x)$, $f^{-1}(x)$ and $ff(x)$. Can we form equations using these?

Solution

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 $ff:x\rightarrow \frac{x+1}{x+2}.$

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

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