Building blocks

## Solution

Put these functions in pairs – each one with its inverse. One function is missing so you’ll need to fill in the blank card.

Three of the first set of functions are quadratics in completed square form and four involve square roots. We noticed that functions $a$ and $g$ are simpler than the others so we looked at those first.

$a(x)=x^2-2$,
$x\ge0$

$g(x)=\sqrt{x+2}$,
$x\ge-2$

To find the inverse of $a(x)$, we rearrange the equation $y=a(x)$ to make $x$ the subject. $y=x^2-2 \quad\implies\quad x^2=y+2 \quad\implies\quad x=\pm\sqrt{y+2}$ To resolve the $\pm$ ambiguity, note that the definition of $a(x)$ specifies the domain $x\ge0$. So we must require the positive square root. The function $a(x)$ is defined for all positive values of $x$ so $y$ can take any value greater than or equal to $-2$.

So now changing the input from $y$ to $x$ we have the inverse function, $a^{-1}(x)=\sqrt{x+2}$ defined on the domain $x\ge-2$. This is the same as $g(x)$ and we can say that $g(x)$ is the inverse of $a(x)$.

For more about the process of finding inverses, see Making inverse functions.

What would the inverse function be if instead we had $a(x)$ defined only for $x\le0$?

Alternatively, we could verify that $a$ and $g$ are inverses by evaluating their composition. $ag(x) = a(g(x)) = (g(x))^2-2 = (x+2)-2 = x$ This almost confirms it, but we have to think a bit harder to make sure the domains are correct. For what values of $x$ is $ag(x)$ defined?

For them to be a pair of inverse functions we also require that $ga(x)=x$ over a suitable domain.

What are the ranges of $a(x)$ and $g(x)$? How are the domains and ranges of the two functions related?

Having had to think about the domains, we next noticed that $b(x)$ and $c(x)$ differ only in their domains…

$b(x)=(x-2)^2+1$,
$x\ge2$

$h(x)=2+\sqrt{x-1}$,
$x\ge1$

$c(x)=(x-2)^2+1$,
$x\le2$

$e(x)=2-\sqrt{x-1}$,
$x\ge1$

Rearranging $y=b(x)$ we have $x=2\pm\sqrt{y-1}$ but $x\ge2$ so we need the positive square root here, and as $x$ increases from $2$, $y$ increases from $1$. Hence we have $b^{-1}(x)=2+\sqrt{x-1}, \ x\ge1$ which matches function $h(x)$.

Similarly, rearranging $y=c(x)$ $x=2\pm\sqrt{y-1}$ but for $c(x)$ we have $x\le2$ so \begin{align*} c^{-1}(x)&=2-\sqrt{x-1}, \ x\ge1\\ &= e(x). \end{align*}

You might have expected a difference only in the domains to result in the inverses differing only in their ranges. But instead we have different functions.

$d(x)=1+\sqrt{x-2}$,
$x\ge2$

$f(x)={}$
$\$

Rearranging $y=d(x)$ we have $x=(y-1)^2+2$ which means that $d^{-1}(x)=(x-1)^2+2.$

To find the domain of $d^{-1}(x)$, consider that as $x$ increases from $2$, $d(x)$ increases from $1$. So the inverse function will be defined only for $x\ge1$. Hence $d(x)$ is paired with $f(x)=(x-1)^2+2, \ x\ge1.$

The second set are all rational functions. The only one that looks different is $r(x)$ but it is hard to predict how any of them behave without doing some work, so we started with $p(x)$.

$p(x)=\dfrac{x}{1+x}$, $x\ne-1$

$t(x)=\dfrac{x}{1-x}$, $x\ne1$

To find the inverse of $p(x)$ we want to rearrange its equation to make $x$ the subject. Here are two different ways of doing that.
\begin{align*} y &= \frac{x}{1+x} \\ y+yx &= x \\ yx-x &= -y \\ x(y-1) &= -y \\ x &= \frac{-y}{y-1} \end{align*}
\begin{align} y=\frac{x}{1+x} &= \frac{1+x-1}{1+x} = 1-\frac{1}{1+x} \label{eq:p-hyperbola}\\ \implies\quad \frac{1}{1+x} &= 1-y \notag\\ \implies\quad x &= \frac{1}{1-y}-1 \label{eq:t-hyperbola}\\ &= \frac{y}{1-y} \notag \end{align}

And so we have the inverse function, $p^{-1}(x) = \frac{x}{1-x} = t(x).$

We get to the same result, but you might find the second method and equations $\eqref{eq:p-hyperbola}$ and $\eqref{eq:t-hyperbola}$ more informative when it comes to sketching the graphs.

In each case the domain is specified to exclude the value of $x$ where the fraction is undefined.

$q(x)=\dfrac{-x}{1+x}$, $x\ne-1$

$s(x)=\dfrac{-x}{1+x}$, $x\ne-1$

Rewriting and rearranging as above, we have \begin{align*} y= q(x) &= \frac{-(1+x)+1}{1+x} = -1+\frac{1}{1+x} \\ \implies\quad x &= \frac{1}{1+y}-1 \end{align*}

and so $q^{-1}(x) = \frac{1}{1+x}-1 = \frac{-x}{1+x} = q(x) = s(x).$

The functions $q$ and $q^{-1}$ are identical which means that this is a self-inverse function.

What does this imply about the composition $qq(x)$ or $q(q(x))$?

$r(x)=\dfrac{x+2}{x+1}$, $x\ne-1$

$u(x)=\vphantom{\dfrac{x+1}{x+1}}$

As before we have \begin{align*} r(x) = \frac{x+1+1}{x+1} &= 1+\frac{1}{x+1} \\ \implies\quad r^{-1}(x) = u(x) &= \frac{1}{x-1}-1 \quad\text{for }x\ne1 \end{align*}

### Things you might have noticed

• In each case, the graphs of a function and its inverse are mirror images of one another in the line $y=x$.

• The graphs of self-inverse functions are symmetrical about the line $y=x$.

• The domain of a one-to-one function $f$ is the same as the range of its inverse $f^{-1}$ and the range of $f$ is the same as the domain of $f^{-1}$.

• In these examples, we have chosen a domain for each function that is as big as possible whilst allowing the function to have an inverse.
• For the quadratic functions, we chose one half or the other so that the function was one-to-one rather than many-to-one.
• For the rational functions, we chose domains that excluded only the problematic values where the functions are undefined (corresponding to vertical asymptotes).
• You could define functions with the same equations on different domains. They could still be inverted so long as they were one-to-one.