4.02 Function notation

Remember that a function is a special type of relation, where each input only has one output. When dealing with functions, function notation is used to highlight the independent and dependent variables, for ease of notation with substitution, and to express functions in a standardised way that is easier to read and understand.

When writing in function notation, instead of writing "$y=$y=", write "$f(x)=$f(x)=". This can be interpreted as "$f$f is a function of the variable $x$x" and read as "$f$f of $x$x". Common letters to use for general functions are lower case $f$f, $g$g and $h$h. However, any letter can be used and often, variables are chosen that have meaning in context such as $t$t for time or $d$d for distance.

Exploration

Instead of $y=2x+1$y=2x+1 we could write $f(x)=2x+1$f(x)=2x+1. Here $f$f is a function of the variable $x$x which follows the rule double $x$x and add $1$1.

Some examples of other functions written in function notation are:

$P(t)=200\times0.8^t$P(t)=200×0.8t, where population is a function of time.

$H(d)=30-2d-30d^2$H(d)=30−2d−30d2, where height is a function of distance.

In these examples, the letter in the bracket on the left is the input, and the right-hand side gives the rule for the output.

Function notation also allows for shorthand for substitution. If $y=3x+2$y=3x+2 and this is written in function notation as $f(x)=3x+2$f(x)=3x+2, then the question 'what is the value of $y$y when $x$x is $5$5?' can be asked simply as 'what is $f(5)$f(5)?' or 'Evaluate $f(5)$f(5).'

 

Worked example

If $A(x)=x^2+1$A(x)=x2+1 and $Q(x)=x^2+9x$Q(x)=x2+9x, find:

A) $A(5)$A(5)

Think: This means we need to substitute $5$5 in for $x$x in the $A(x)$A(x) equation.

Do: 

$A(5)$A(5)	$=$=	$5^2+1$52+1
	$=$=	$26$26

 

B) $Q(6)$Q(6)

Think: This means we need to substitute $6$6 in for $x$x in the $Q(x)$Q(x) equation.

Do:

$Q(6)$Q(6)	$=$=	$6^2+9\times6$62+9×6
	$=$=	$36+54$36+54
	$=$=	$90$90

 

Practice questions

Question 1

Question 2

 

Algebra of functions

Functions can be added, subtracted, multiplied or divided. If $f$f and $g$g are functions, we can compute $f(x)+g(x)$f(x)+g(x) by adding the function values at $x$x for each function. The short-hand notation for function addition is $(f+g)(x).$(f+g)(x). In a similar way, we form $f(x)-g(x)$f(x)−g(x) or $(f-g)(x)$(f−g)(x) for subtraction, $f(x)\times g(x)$f(x)×g(x) or $(fg)(x)$(fg)(x) for multiplication and $f(x)\div g(x)$f(x)÷​g(x) or $(f/g)(x)$(f/g)(x) for function division.

 

Practice questions

Question 3

question 4

Composition of functions

The idea behind the composition of functions is best explained with an example.

Consider the function given by $f\left(x\right)=2x+1$f(x)=2x+1. Here, the function takes values of $x$x for the input and maps them to values, $y=2x+1$y=2x+1, in the output.

However, suppose that this is only the first part of a two-stage treatment of $x$x where these function values are taken and mapped using another function, say $g\left(x\right)=x^2$g(x)=x2. This means that the $y$y values given by $\left(2x+1\right)$(2x+1) become the squared values $\left(2x+1\right)^2$(2x+1)2. The diagram below captures this idea.

The function values $f\left(x\right)$f(x) have become the input for $g\left(x\right)$g(x). The complete two-stage process can be described by the expression $g\left(f\left(x\right)\right)$g(f(x)), sometimes written $(g\circ f)(x)$(g∘f)(x) and referred to as the "$g$g of $f$f  of $x$x".

Algebraically, this can be written: $g\left(f\left(x\right)\right)=g\left(2x+1\right)=\left(2x+1\right)^2$g(f(x))=g(2x+1)=(2x+1)2. 

Note that if the order of the two-stage processing was reversed, this would develop (in this instance) a different composite function. 

Thus $f\left(g\left(x\right)\right)=f\left(x^2\right)=2\left(x^2\right)+1=2x^2+1$f(g(x))=f(x2)=2(x2)+1=2x2+1. This would be "$f$fof $g$g of $x$x".

 

 

Practice questions

Question 5

Question 6