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# Secondary Magazine - Issue 129: Sixth Sense

Created on 17 December 2015 by ncetm_administrator
Updated on 02 February 2016 by ncetm_administrator

# Sixth Sense

Students of mathematics toy with the concept of functions from quite an early age – how many of us were challenged at some stage to “think of a number, double it, add seven, double that, add 2, quarter that, subtract the number you first thought of, and hey presto …” – though the formal notation and language of functions often catches out even some of the best at A2. Can your top-module-scoring student explain, coherently and concisely and correctly, the difference between the co-domain and the range of a function?

Using inverse and composite functions, including formal notation, certainly crops up in the AQA Level 2 Further Mathematics Qualification, which sits alongside GCSE mathematics, and other boards have historically included some of this content at GCSE Higher tier and in Additional Maths. Under the new structure, all higher tier GCSE students will be expected to interpret reverse processes and combined processes using the formal notation of inverses and composition.

Students’ lack of conceptual understanding of functional notation is unmissably apparent when they are considering transformations of graphs: very few students, even those with A* grades embarking on Further Mathematics in Year 12, have an explanation that relates $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )$ and $\inline \dpi{80} \fn_jvn \small y=f\left ( x+2 \right )$ other than “it’s +2 in the brackets so it’s 2 to the left. Or the right. I can never remember.” Even if the student on the spot can remember whether it’s up, down, left, right, in, out or shake it all about, the Sixth Sense we’re nurturing in AS and A2 mathematicians is the ability to justify rather than the ability to parrot-recall!

The hurdle, it seems, is in moving from informal, wordy instruction sets to formal functional notation and understanding. To tackle this, you might need to go right back to the Year 2 carpet … Even very young learners are quite able to cope with

A number machine takes an input, doubles it and then subtracts 3 to get an output. What’s the output if the input is 4?

since all they need is an understanding of the words and some core arithmetic skills.

Even before being introduced to algebra, many of them make a reasonable attempt at the harder question

If the output is 9, what was the input?

by reversing the process step-by-step.

Composite questions are similarly straightforward when discussed in words:

Machine A triples an input and adds 5. Machine B squares inputs. What happens if we connect the machines together, so that the output from A goes straight into B, and input 2 into this combined machine? What happens if we connect them the other way round?

Judicious use of “input-output diagrams” where processes are chained together is clearly helpful here too.

So, what goes wrong when we formalise? These natural processes which one might describe as “easy-ish” suddenly become “ferociously hard” even for the highest attaining students. Indeed, by the time they meet

$\inline \dpi{80} \fn_jvn \small f\left ( x \right )=3x+5$

they will have spent some time studying algebra, and so their first instinct might be that this is some kind of equation and ask “can we subtract 5 from both sides?” This confusion is added to by questions such as

Given $\inline \dpi{80} \fn_jvn \small f\left ( x \right )=3x+5$, solve $\inline \dpi{80} \fn_jvn \small f\left ( x \right )=14$

Should we use a different notation for functions? It could well be less confusing to explain and interpret

$\inline \dpi{80} \fn_jvn \small f:x\rightarrow 3x+5$

at GCSE and, in the sixth form,

$\inline \dpi{80} \fn_jvn \small f:\mathbb{R}\rightarrow \mathbb{R}$

$\inline \dpi{80} \fn_jvn \small x\rightarrow 3x+5$

(Ah-ha, there’s the co-domain!)

At GCSE and A-level, rightly or wrongly, students need to learn to read $\inline \dpi{80} \fn_jvn \small f\left ( x \right )=3x+5$ as a definition and $\inline \dpi{80} \fn_jvn \small f\left ( x \right )=14$ as an equation (unless, of course, it is a definition of a constant function – students’ lack of confidence is completely forgivable!).

A natural approach at level 3 is via the language and notation of sequences. For whatever reason, something about the statement

Find the first four terms of the sequence that is generated using $\inline \dpi{80} \fn_jvn \small u_{n}=3n+5$

causes less (but certainly not no!) confusion. So, once students are fluent with this notation, why not try

Find the first four terms of the sequence that is generated using $\inline \dpi{80} \fn_jvn \small f\left ( n \right )=n^{2}-3$

to encourage fluency with functional notation?

After some practice, they will happily step away from sequences and respond to more abstract questions, such as

If $\inline \dpi{80} \fn_jvn \small f\left ( x \right )=3x-1$, what is $\inline \dpi{80} \fn_jvn \small f\left ( 1.3 \right )$? $\inline \dpi{80} \fn_jvn \small f\left ( 1.3 \right )$? $\inline \dpi{80} \fn_jvn \small \dpi{80} \fn_jvn \small f\left ( \frac{7}{2} \right )$?

and then

what is $\inline \dpi{80} \fn_jvn \small \dpi{80} \fn_jvn \small f\left ( k\right )$? $\inline \dpi{80} \fn_jvn \small f\left ( 3x\right )$? $\inline \dpi{80} \fn_jvn \small f\left ( x^{2}-1\right )$?

A good deal of time spent developing procedural fluency with functional notation is needed before we try anything too ambitious, but there are plenty of opportunities for conceptual variation as we go along:

A function $\dpi{80} \fn_jvn \small f$ takes an input, squares it and subtracts 7.

(a) Complete the following table:

 Input Output $\dpi{80} \fn_jvn \small 3$ $\dpi{80} \fn_jvn \small -12$ $\dpi{80} \fn_jvn \small 74$ $\dpi{80} \fn_jvn \small 93$ $\inline \dpi{80} \fn_jvn \small p$ $\inline \dpi{80} \fn_jvn \small q^{2}-7$ $\inline \dpi{80} \fn_jvn \small 2x$ $\inline \dpi{80} \fn_jvn \small y$

(b) Define the function in the form $\dpi{80} \fn_jvn \small f\left ( x \right )=$...

(c) Which row in your table explains how to “undo” the effect of $\dpi{80} \fn_jvn \small f$?

(d) Define this “undo” or “inverse” process in the form $\inline \dpi{80} \fn_jvn \small g\left ( x \right )=$...

and clearly this is a great opportunity to practise mental arithmetic and algebraic skills – the definition at the start of the question could be considerably more demanding. Once your students are happy with the interpretation of "find $\inline \dpi{80} \fn_jvn \small f\left ( 3 \right )$" as requiring them to find the output when $\inline g$ is used as an input for $\dpi{80} \fn_jvn \small f$, it is not too giant a leap to suggest that "find $\inline \dpi{80} \fn_jvn \small f\left ( g\left ( \right 2) \right )$" requires them to find the output when $\inline \dpi{80} \fn_jvn \small g\left ( \right 2)$ is used as an input for $\dpi{80} \fn_jvn \small f$, which in turn needs them to find the output when 2 is used as the input for $\inline \dpi{80} \fn_jvn \small g$. It is worth spending time using lots of brackets before explaining that mathematicians often ditch the outer pair so that $\inline \dpi{80} \fn_jvn \small fg\left ( 2 \right )$ is commonly accepted to mean exactly the same thing as $\inline \dpi{80} \fn_jvn \small f\left ( g\left ( \right 2) \right )$.

Working with composite functions is an excellent example of a scenario where specialising before generalising helps enormously: the leap from finding $\inline \dpi{80} \fn_jvn \small fg\left ( 2 \right )$ to finding $\inline \dpi{80} \fn_jvn \small fg\left ( x \right )$ is a large one and should not be attempted prematurely. Ultimately, though, we are aiming to get our students to see $\dpi{80} \fn_jvn \small x$ here as a shorthand for “input” rather than as anything special, or unknown in the “solving equations” sense, so that when they see

Given $\inline \dpi{80} \fn_jvn \small f\left ( x \right )=2x^{2}$ and $\inline \dpi{80} \fn_jvn \small g\left ( x \right )=4x-1$ find $\inline \dpi{80} \fn_jvn \small fg\left ( x \right )$ and $\inline \dpi{80} \fn_jvn \small gf\left ( x \right )$

they can suggest that $\inline \dpi{80} \fn_jvn \small fg\left ( x \right )$ takes an input, multiplies it by $\inline \dpi{80} \fn_jvn \small 4$, subtracts $\inline \dpi{80} \fn_jvn \small 1$ and then takes this whole output, squares it and finally doubles the result. That is,

$\inline \dpi{80} \fn_jvn \small fg\left ( x \right )=f\left ( 4x-1 \right )=2\left ( 4x-1 \right )^{2}$

and similarly that

$\inline \dpi{80} \fn_jvn \small gf\left ( x \right )=g\left ( 2x^{2} \right )=4\left ( 2x^{2} \right )-1=8x^2{}-1$

All of this discussion of inputs and outputs, with x representing the input, may at some point remind learners of how and why we draw graphs in mathematics: often we have a relationship, typically between an input $\dpi{80} \fn_jvn \small x$ and an output $\dpi{80} \fn_jvn \small y$, and we plot and join up points that meet the conditions specified by the relationship – the points whose coordinates satisfy the relationship.

Consequently the instruction

draw a graph of $\dpi{80} \fn_jvn \small y=f\left ( x \right )$ where $\dpi{80} \fn_jvn \small f\left ( x \right )=7x-1$

merely says “for any input value, the output value is calculated by multiplying by $\inline \dpi{80} \fn_jvn \small 7$ and subtracting $\inline \dpi{80} \fn_jvn \small 1$” and hence we find ourselves plotting points that satisfy the relationship $\inline \dpi{80} \fn_jvn \small y-coordinate =7\times x-coordinate-1$.

The language of inputs and outputs is very useful when looking at transformations of graphs: many A* students end up learning the “rules” here because they find the justification hard. But the justification is well worth it, and follows naturally from the sequence of ideas we’ve set out above.

Again, specialising first helps: suppose we have a graph of $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )$ in front of us, and we can see that the point $\inline \dpi{80} \fn_jvn \small \left ( 2,7 \right )$ is on the graph. This means that when we input $\inline \dpi{80} \fn_jvn \small 2$ into $\dpi{80} \fn_jvn \small f$ the output is $\inline \dpi{80} \fn_jvn \small 7$.

What is the output if we input 2 into $\inline \dpi{80} \fn_jvn \small 3f\left ( x \right )$?

Hopefully we get the answer “21” in unison. This means that if we drew the graph of $\inline \dpi{80} \fn_jvn \small y=3f\left ( x \right )$ we'd have a point at $\inline \dpi{80} \fn_jvn \small \left ( 2,21 \right )$ – which would be the image of the original point at $\inline \dpi{80} \fn_jvn \small \left ( 2,7 \right )$ after some transformation.

What is the output if we input 2 into $\inline \dpi{80} \fn_jvn \small f\left ( 3x \right )$?

This is a harder question, but hopefully we’ve been going carefully enough to realise that this doesn’t really make sense – is $\inline \dpi{80} \fn_jvn \small 2$ the whole input and thus the same as $\inline \dpi{80} \fn_jvn \small 3x$, or is this a badly-worded question that is actually asking us to find $\inline \dpi{80} \fn_jvn \small f\left ( 6 \right )$? Without a definition of $\dpi{80} \fn_jvn \small f$, we can't do this!

In fact the only thing we can say with any confidence is

In order to get an output of $\inline \dpi{80} \fn_jvn \small 7$ from $\inline \dpi{80} \fn_jvn \small f\left ( 3x \right )$, we would need to input $\inline \dpi{80} \fn_jvn \small x=\frac{2}{3}$ so that we’d be calculating $\inline \dpi{80} \fn_jvn \small f\left ( 3\times \frac{2}{3} \right )=f\left ( 2 \right )$ which we know is equal to $\inline \dpi{80} \fn_jvn \small 7$.

This means that if we drew the graph of $\inline \dpi{80} \fn_jvn \small y=f\left ( 3x \right )$ we’d have a point at $\inline \dpi{80} \fn_jvn \small \left ( \frac{2}{3},7 \right )$ - which would be the image of the original point at $\inline \dpi{80} \fn_jvn \small \left ( 2,7 \right )$ after some transformation. After more of this “pointwise thinking”, we’re now in a position to conclude that

1. If the point $\inline \dpi{80} \fn_jvn \small \left ( a,b \right )$ lies on the curve representing $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )$, then we can be sure that the curve given by $\inline \dpi{80} \fn_jvn \small y=3f\left ( x \right )$ will pass through the point $\inline \dpi{80} \fn_jvn \small \left ( a,3b \right )$. That is, the outputs triple, and the geometric effect is that the graph of $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )$ is stretched by a scale factor of three parallel to the $\inline \dpi{80} \fn_jvn \small y$-axis to become the graph of $\inline \dpi{80} \fn_jvn \small y=3f\left ( x \right )$
2. If the point $\inline \dpi{80} \fn_jvn \small \left ( a,b \right )$ lies on the curve representing $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )$, then we can be sure that the curve given by $\inline \dpi{80} \fn_jvn \small y=f\left ( 3x \right )$ will pass through the point $\inline \dpi{80} \fn_jvn \small \left ( \frac{a}{3},b \right )$. That is, the inputs divide by 3, and the geometric effect is that the graph of $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )$ is stretched by a scale factor of one third parallel to the $\inline \dpi{80} \fn_jvn \small y$-axis to become the graph of $\inline \dpi{80} \fn_jvn \small y=f\left ( 3x \right )$.

Similar thinking explains the transformations between $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )$ and $\inline \dpi{80} \fn_jvn \small y=f\left ( x \right )-5$ and $\inline \dpi{80} \fn_jvn \small y=f\left ( x-5 \right )$: it takes time to embed, but it is of so much more value here than the alternative "learn these rules" approach.

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