Domain and range dominoes

Excerpt: This card sorting activity makes students think carefully about what we mean by the domain and range of a function and how to express them precisely. They will get lots of practise whilst arranging the cards into a loop.

Warm-up

Describe in words, as accurately as possible, the domain and range of the functions illustrated here.

$f(x)=\sqrt{3x}(1-x)$

$f(x)=\sqrt{3x}(1-x)$

$f(x)=\dfrac{1}{x-1}$

$f(x)=\dfrac{1}{x-1}$

Can you write the domains and ranges using mathematical notation?

The function on the left is undefined for any negative value of $x$ because of the square root. It is defined at $x=0$ , so we could describe its domain as any real number greater than or equal to zero. $f(0)=0$ and as $x$ increases $f(x)$ increases up to $\frac{2}{3}$ then decreases. The function’s range is any real number less than or equal to $\frac{2}{3}$ . $\text{domain: }x\in\mathbb{R}\text{, }x\ge0\quad\text{ range: }f(x)\in\mathbb{R}\text{, }f(x)\le\frac{2}{3}$

The symbols $\in\mathbb{R}$ are mathematical notation meaning “is a member of the set of all real numbers”.

The function on the right is defined for all real values of $x$ except for $x=1$ . Its range is all real numbers greater than zero and all real numbers less than zero. Notice that it is impossible to obtain the output value $f(x)=0$ . $\text{domain: }x\in\mathbb{R}\text{, }x\ne1\quad\text{ range: }f(x)\in\mathbb{R}\text{, }f(x)\ne0$

Sometimes, the domain and range might be written in an abbreviated form such as $\text{domain: }x\ne1\quad\text{ range: }f(x)\ne0 \text{ .}$

We refer to real numbers as opposed to imaginary or complex numbers. You may learn about those elsewhere.

Check your answers

Problem

Match up each function with its domain and range. They form a set of domino cards that can be arranged in a sequence.

To turn the sequence into a closed loop, you will need to complete the cards with missing information. A separate sheet containing just those cards is available here.

$f(x)=x^2+1$	domain: $x\neq-1$ range: $f(x)\neq0$

| f(x)=|x| $f(x)=\|x\|$ | domain: $x\in \mathbb{R}$ range: $f(x)\ge1$ | | --- | --- |

$f(x)=\sqrt(3){x}$ $f(x)=\sqrt[3]{x}$	domain: $x\in \mathbb{R}$ range: $0\!<\!f(x)\!\le\!1$

$f(x)=\sqrt{x}$ $f(x)=\sqrt{x}$	domain: $x\in \mathbb{R}$ range: $f(x)\ge-1$

$f(x)=\frac{1}{x}$ $f(x)=\frac{1}{x}$	domain: $x\neq0$ range: $f(x)\ne-1$

$f(x)=x+\frac{1}{x}$ $f(x)=x+\frac{1}{x}$	domain: $x\neq-1$ , $x\neq1$ range: $f(x)\le-1$ or $f(x)>0$

$f(x)=\frac{1}{x^2}$ $f(x)=\frac{1}{x^2}$	domain: $x\in \mathbb{R}$ range: $f(x)\ge0$

$f(x)=\frac{1}{x^2+1}$ $f(x)=\frac{1}{x^2+1}$	domain: $x\ge0$ range: $f(x)\ge0$

$f(x)=\frac{1}{x^2-1}$ $f(x)=\frac{1}{x^2-1}$	domain: $x>0$ range: $f(x)>0$

$f(x)=\frac{1}{1+x}$ $f(x)=\frac{1}{1+x}$	domain: $x\ge0$ range: $f(x)\le\frac{2}{3}$

$f(x)=\frac{1}{\sqrt{x}}$ $f(x)=\frac{1}{\sqrt{x}}$	domain: $x\in \mathbb{R}$ range: $f(x)\in \mathbb{R}$

$f(x)=\sqrt{3x}(1-x)$ $f(x)=\sqrt{3x}(1-x)$	domain: $x\neq0$ range: $f(x)\neq0$

$f(x)=\frac{1}{x}-1$ $f(x)=\frac{1}{x}-1$	domain: range:

$f(x)=\qquad\qquad$ $f(x)=\qquad\qquad$	domain: range:

Things you may have noticed

Unrestricted domains

In this exercise we are using domain to mean all the possible values of $x$ . In other contexts, we may artificially restrict the domain of a function. For instance we might only be interested in the function $f(x)=x^2$ in the domain $0\le x<2$ .

Similar inequalities

With some of the functions you have to be careful to distinguish between, for instance, $x>0$ and $x\ge0$ .

Multiple inequalities

One of the functions has a range that is between $0$ and $1$ , including $1$ but not including $0$ . This can be written neatly as $0<f(x)\le1$ .

Another has a range that includes all numbers $>0$ and all numbers $\le-1$ but not numbers in between. This has to be written as two separate inequalities.

Turning points of $x+\frac{1}{x}$

Finding the range of $f(x)=x+\frac{1}{x}$ depends on knowing the value of the function at its turning points (maximum and minimum) and this is not given on the card. There is more than one way to find these values; here are a few.

Using calculus

We could differentiate the function and thus find when its gradient is zero.

Using an inequality

By substitution we can find that

$f(1)=2$

. Is there any positive

$x$

value for which

$f(x)<2$

? If so,

\begin{align*} x+\frac{1}{x}&<2\\ x^2+1&<2x \text{ .} \end{align*}

Rearranging and factorising leads to an inequality with no real solutions, so $2$ must be the minimum value of $f(x)$ for positive $x$ . The local maximum can be found by symmetry.

When rearranging the inequality, why does it matter that the question specified positive $x$ values?

By adding gradients

Notice that $f(x)$ is the sum of two functions: $x$ and $\frac{1}{x}$ . So the gradient of $f(x)$ is the sum of the gradients of $x$ and $\frac{1}{x}$ (it is possible to prove this, but we aren’t going to do so here!).

$y=x$ has gradient $1$ , so the turning points of $f(x)$ will be where $y=\frac{1}{x}$ has gradient $-1$ . Can you find the $x$ values where this happens?

scan of dominoes arranged in a loop

Here is one way of arranging the cards