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2.02 Function notation

Lesson

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)=302d30d2, 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

Consider the equation $x-4y=8$x4y=8.

Assume that $y$y is a function of $x$x.

  1. Rewrite the equation using function notation $f\left(x\right)$f(x).

  2. Find the value of $f\left(12\right)$f(12).

Question 2

Consider the function $f\left(x\right)=2x^3+3x^2-4$f(x)=2x3+3x24.

  1. Evaluate $f\left(0\right)$f(0).

  2. Evaluate $f\left(\frac{1}{4}\right)$f(14).

Question 3

If $Z(y)=y^2+12y+32$Z(y)=y2+12y+32, find $y$y when $Z(y)=-3$Z(y)=3.

  1. Write both solutions on the same line separated by a comma.

Question 4

Consider the function $g\left(x\right)=ax^3-3x+5$g(x)=ax33x+5.

  1. Determine $g\left(k\right)$g(k).

  2. Form an expression for $g\left(-k\right)$g(k).

  3. Is $g\left(k\right)=g\left(-k\right)$g(k)=g(k)?

    Yes

    A

    No

    B
  4. Is $g\left(k\right)=-g\left(-k\right)$g(k)=g(k)?

    Yes

    A

    No

    B

 

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 $(f-g)(x)$(fg)(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

Given the following values:

$f\left(2\right)=4$f(2)=4, $f\left(7\right)=14$f(7)=14, $f\left(9\right)=18$f(9)=18, $f\left(8\right)=16$f(8)=16

$g\left(2\right)=8$g(2)=8, $g\left(7\right)=28$g(7)=28, $g\left(9\right)=36$g(9)=36, $g\left(8\right)=32$g(8)=32

  1. Find $\left(f+g\right)$(f+g)$\left(2\right)$(2)

question 6

If $f(x)=3x-5$f(x)=3x5 and $g(x)=5x+7$g(x)=5x+7, find each of the following:

  1. $(f+g)(x)$(f+g)(x)

  2. $(f+g)$(f+g)$\left(4\right)$(4)

  3. $(f-g)(x)$(fg)(x)

  4. $(f-g)$(fg)$\left(10\right)$(10)

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 $(g\circ f)(x)$(gf)(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

Consider the functions $f\left(x\right)=-2x-3$f(x)=2x3 and $g\left(x\right)=-2x-6$g(x)=2x6.

  1. Find $f\left(7\right)$f(7).

  2. Hence, or otherwise, evaluate $g\left(f\left(7\right)\right)$g(f(7)).

  3. Now find $g\left(7\right)$g(7).

  4. Hence, evaluate $f\left(g\left(7\right)\right)$f(g(7)).

  5. Is it true that $f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)$f(g(x))=g(f(x)) for all $x$x?

    Yes

    A

    No

    B

Question 8

Consider $f\left(x\right)=x^2+3$f(x)=x2+3 and $g\left(x\right)=4x-9$g(x)=4x9.

  1. Define $f\left(2x\right)$f(2x).

  2. Show that $f\left(2x\right)=g\left(f\left(x\right)\right)$f(2x)=g(f(x))

Outcomes

1.1.24

use function notation, determine domain and range, recognise independent and dependent variables

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