2.02 Function notation

We have seen that functions are special types of relations. When dealing with functions we often use function notation to emphasise that we are dealing with a special case, to highlight the independent and dependent variable and for ease of notation with substitution and calculus.

When we are writing in function notation, instead of writing "$y=$y=", we 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 we can use variables that have meaning in context. 

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.

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

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

These are all examples of just a different way to write '$y=$y= …' notation as a function. The letter in the bracket on the left is the input and the right hand side gives us a rule for the output.

Function notation also allows for shorthand for substitution. If $y=3x+2$y=3x+2 and we write this 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

 

(c) $A(p)$A(p)

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

Do: 

$A(p)$A(p)

$=$=

$p^2+1$p2+1

 

Practice questions

Question 1

Question 2

Question 3

Question 4

 

Algebra of functions

Functions can be added, subtracted, multiplied or divided. If $f$f and $g$g are functions, we 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 5

question 6

Composition of functions

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

Suppose we think about the function given by $f\left(x\right)=2x+1$f(x)=2x+1. We understand that the function takes values of $x$x for the input and maps them to values, say $y=2x+1$y=2x+1, in the output.

Suppose however that this is only the first part of a two-stage treatment of $x$x. Suppose we now take these function values and map them 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 the idea.

The function values $f\left(x\right)$f(x) have become the input for $g\left(x\right)$g(x). Thus we could describe the complete two-stage process by the expression $g\left(f\left(x\right)\right)$g(f(x)), sometimes written $(g\circ f)(x)$(g∘f)(x) and spoken of as "$g$g of $f$f  of $x$x".

Algebraically, we can write $g\left(f\left(x\right)\right)=g\left(2x+1\right)=\left(2x+1\right)^2$g(f(x))=g(2x+1)=(2x+1)2. 

Note that if we reversed the order of the two-stage processing, we would, in this instance, develop 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 "$f$f of $g$g of $x$x".

 

Practice questions

Question 7

Question 8