Functions and Relations
A set is a group of unique numbers. The numbers in a set are known as elements.
To define a set we use braces. For example if we want the set A to contain the numbers 1, 2, 3 and 4, then: \[A = \{1, 2, 3, 4\}\]
If a set C is empty - the null set - then we can represent this with: \[C = \emptyset\]
If a set A contains the number 2, then we can represent this with: \[2 \in A\]
If a set A does not contain the number 5, then we can represent this with: \[5 \notin A\]
If the whole of set A is contained within set B, then A is a subset of B: \[A \subseteq B\]
If the whole of set A is not contained within B, then A is not a subset of B: \[A \nsubseteq B\]
The set of numbers contained in both sets A and B is known as the intersection: \[A \cap B\]
The set of numbers either in set A or set B is known as the union: \[A \cup B\]
If the sets A and B have no numbers in common, we say A and B are disjointed: \[A \cap B = \emptyset\]
Numbers can be categorised into the following sets:
Symbol | Name | Definition | Example |
---|---|---|---|
\[\mathbb{N}\] | Natural numbers | Positive whole numbers not including zero | \[1,2,3,4\ldots\] |
\[\mathbb{Z}\] | Integers | Positive and negative whole numbers including zero | \[\ldots-2,-1,0,1,2\ldots\] |
\[\mathbb{Q}\] | Rational numbers | Any number which can be represented as a fraction | \[\frac{1}{2}, \frac{5}{3}, 8\] |
\[\mathbb{R}\] | Real numbers | All positive, negative numbers and zero | \[0, 2, 3.1415, \pi\] |
If we only want positive real numbers or negative real numbers, we simply add a plus (+) or minus (-) sign next to the symbol. This applies for integers, rational numbers and real numbers. For example: \[\mathbb{R}^+,\,\mathbb{Q}^-\]
Note: All natural numbers are integers, all integers are rational numbers and all rational numbers are real numbers. That is: \[\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}\]
Interval notation is a concise way of writing sets by referencing end points.
Square brackets are used when the end points are inclusive. For example, all real values from a to b, including a and b: [a,b]
Parentheses (round brackets) are used when the end points are not inclusive. For example, all real values from a to b, excluding a and b: (a,b)
We can use both square and round brackets in the same interval. For example, all real values from a to b, including a, but excluding b: [a,b)
If we want to exclude certain numbers from an interval we use a backslash (\). For example, all real values from 2 to 5, including 2, but excluding 5 and excluding 3: [2,5)\{3}
If we want to include all numbers below a certain point or all numbers above a certain point we reference infinity. For example, the intervals of all the numbers below 3 inclusive and all the numbers above 8 exclusive are show below: \[(-\infty, 3] \text{ and } (8, \infty)\] Note: we always use parentheses (round brackets) for the infinity bound.
The domain of a function is all the valid values x can be. If no domain is given in the question, the implied (maximal) domain is the assumed domain. The implied domain is simply all the valid values x can be. However, in methods, often a restrict domain will be provided. A restricted domain is a subset of the implied domain and dictates which values x can be.
The range of a function is all the values y can be for a given domain. Below is a table comprising of how to find the domain and range for all fundamental functions (excluding polynomials) in methods.
Function | (Implied) Domain | Range |
\[y = \frac{a}{nx+b}+c\] | \[\text{All values, but } nx+b\ne0\] | \[\mathbb{R} \setminus \{c\}\] |
\[y = \frac{a}{(nx+b)^2}+c\] \[(a>0)\] | \[\text{All values, but } nx+b\ne0\] | \[(c, \infty)\] |
\[y = \frac{a}{(nx+b)^2}+c\] \[(a<0)\] | \[\text{All values, but } nx+b\ne0\] | \[(-\infty, c)\] |
\[y = a\sqrt{nx+b}+c\] \[(a>0)\] | \[\text{All values for which } nx+b \geqslant 0\] | \[[c, \infty)\] |
\[y = a\sqrt{nx+b}+c\] \[(a<0)\] | \[\text{All values for which } nx+b \geqslant 0\] | \[(-\infty, c]\] |
\[y = a\sqrt[e]{nx+b}+c\] \[(a>0) ,\, e=2k ,\, k \in \mathbb{Z}\] \[(e \text{ is even})\] | \[\text{All values for which } nx+b \geqslant 0\] | \[[c, \infty)\] |
\[y = a\sqrt[e]{nx+b}+c\] \[(a<0) ,\, e=2k ,\,k \in \mathbb{Z}\] \[(e \text{ is even})\] | \[\text{All values for which } nx+b \geqslant 0\] | \[(-\infty, c]\] |
\[y = a\sqrt[e]{nx+b}+c\] \[e=2k+1 , \,k \in \mathbb{Z}\] \[e \text{ is odd}\] | \[\mathbb{R}\] | \[\mathbb{R}\] |
\[(nx+b)^{\frac{p}{q}}+c \\ p \text{ is a positive odd integer}\\ q \text{ is a positive odd integer}\] | \[\mathbb{R}\] | \[\mathbb{R}\] |
\[(nx+b)^{\frac{p}{q}}+c \\ p \text{ is a positive odd integer}\\ q \text{ is a positive even integer}\] | \[\text{All values for which } nx+b\geqslant 0\] | \[[c,\infty)\] |
\[(nx+b)^{\frac{p}{q}}+c \\ p \text{ is a positive even integer}\\ q \text{ is a positive odd integer}\] | \[\text{All values, but } nx+b \neq 0\] | \[[c,\infty)\] |
\[r^2 = (x-h)^2+(y-k)^2\] | \[[h-r, h+r]\] | \[[k-r, k+r]\] |
\[y = A\cdot e^{nx+b}+c\] \[A>0\] | \[\mathbb{R}\] | \[(c, \infty)\] |
\[y = A\cdot e^{nx+b}+c\] \[A<0\] | \[\mathbb{R}\] | \[(-\infty, c)\] |
\[y = A\log{(nx+b)}+c\] | \[\text{All values for which } nx+b > 0\] | \[\mathbb{R}\] |
\[y = A\sin(nx+b)+c\] | \[\mathbb{R}\] | \[[c-\lvert A \rvert, c+\lvert A \rvert]\] |
\[y = A\cos(nx+b)+c\] | \[\mathbb{R}\] | \[[c-\lvert A \rvert, c+\lvert A \rvert]\] |
\[y = A\tan(nx+b)+c\] | \[\text{All values, but } \cos(nx+b) \neq 0\] | \[\mathbb{R}\] |
The standard function looks like: \[f: [0,5] \rightarrow \mathbb{R} , f(x) = x + 1\] Now let's dissect each part.
f | The name of the function. |
[0,5] | The domain of the function |
\[\mathbb{R}\] | The codomain - you do not need to know what this is. For methods simply put ℝ |
f(x) | The name of the function with the variable in brackets |
x + 1 | The expression for the function |
There are two types of functions and two other types of relations.
One-to-one function | Exactly one x value corresponds to exactly one y value |
Many-to-one function | There is at least one case where many x values correspond to a single y value |
One-to-many relation | There is at least one case where exactly one x value corresponds to many y values |
Many-to-many relation | There is at least one case where many x values correspond to many y values |
The vertical line test determines if a certain graph represents that of a function. If a vertical line can be drawn anywhere on a graph and it only ever cuts through the graph at at most one point, then the graph is that of a function.
The horizontal line test determines if the graph of a function is one-to-one. If a horizontal line can be drawn anywhere on a graph of a function (test with vertical line test first) and it only ever cuts through the graph at at most one point, then the graph is that of a one-to-one function. If it cuts through the graph at more than one point, then the graph is that of a many-to-one function.
A piecewise function has different equations for different subsets of the domain of the function. To graph them, first mark in any endpoints which you can and then follow the steps provided in the graphing section for the different functions in other chapters. An example of the formatting for a piecewise function is given below: \[ f(x) = \begin{cases} \sin(x)+2 &x < 0 \\ (x-5)^2 & 0 \leqslant x \leqslant 5 \\ \log_e{x} & 5 < x \\ \end{cases}\\ \]
This is hardly ever tested! In fact, we have no recollection of there being a VCAA question in the last 10 years which requires this knowledge. An even function is one where f(x) = f(-x). That is, it is symmetrical about the y-axis. An odd function is one where f(-x) = -f(x). That is, the graph remains unchanged when rotated 180 degrees about the origin.
Some notation: \[f(x) + g(x) = (f + g)(x)\] \[f(x) - g(x) = (f-g)(x)\] \[f(x) \cdot g(x) = (fg)(x)\]
When adding or multiplying two functions together the domain of the function is simply the intersection of the individual functions. That is: \[\text{dom}(fg) = \text{dom}(f) \cap \text{dom}(g)\] \[\text{dom}(f + g) = \text{dom}(f) \cap \text{dom}(g)\] \[\text{dom}(f - g) = \text{dom}(f) \cap \text{dom}(g)\]
Here we look at how to graph a function (f + g)(x) when presented with f(x) and g(x). You are not expected to produce a perfect graph. There are a few key points which we detail below which you must get correct. For all the other points in the graph, something of the correct shape roughly in the right spot will be awarded full marks.
1 | Look at the points where f(x) = 0, (f+g)(x) will equal the value of g(x) at those x-values. Mark those points in. |
2 | Look at the points where g(x) = 0, (f+g)(x) will equal the value of f(x) at those x-values. Mark those points in. |
3 | Look for values for x where f(x) + g(x) ≈ 0. Mark those points in. |
4 | Connect the dots up appropriately, attempting to roughly add the y-coordinates of the two graphs as you go. |
Use ‘common sense’ checks on your graph. For example, if f(x) > 0 and g(x) > 0, then the y-coordinate of (f+g)(x) will be the largest for those x-values. The inverse is also true.
This is likely going to be the hardest concept to understand in the first semester of your studies. Below we have tried to explain the intuition and have presented the formulas necessary to solve questions.
A composite function is where we have a ‘function within a function’. It is mathematically represented as: \[f(g(x)) = f \circ g(x) \]
What this means is that we replace all the \(x\) values in \(f(x)\) with whatever \(g(x)\) is. For example: \[ \begin{aligned} f(x) &= x^2 + 2x + 1 \\ g(x) &= \sin(x) \\ f(g(x)) &= \sin^2(x) + 2\sin(x) + 1 \end{aligned} \]
For a composite function \(f(g(x)), \, ran(g) \subseteq dom(f)\). That is, the range of g(x) must be in the domain of f(x). Thus, the domain of the function f(g(x)) is also determined by g(x).
\[ \text{Example 1.1: Let } f: \mathbb{R} \rightarrow \mathbb{R} , \, f(x) = x^2 - 1 \text{ and } g: (-\infty, 3] \rightarrow \mathbb{R}, \, g(x) = \sqrt{x-3} \\ \text{i) Show that } g(f(x)) \text{ is not defined.} \\ \text{ii) Find the largest possible domain for } f(x) \text{ such that } g(f(x)) \\ \text{ is defined and hence state the domain for } g(f(x)) \\ \text{ }\\ \begin{aligned} \text{i) } \text{ran}(f) &= [-1, \infty) \\ \text{dom}(g) &= (-\infty, 3] \\ \text{ran}(f) &\nsubseteq \text{dom}(g) \\ \end{aligned}\\ \text{Thus, } g(f(x)) \text{ is not defined}. \\ \text{ }\\ \begin{aligned} \text{ii. We require: } \text{ran}(f) &\subseteq \text{dom}(g) \\ \text{ran}(f) &= [-1, \infty) \\ \text{dom}(g) &= (-\infty, 3] \\ \text{Restrict, } \text{ran}(f) &= [-1, 3] \\ \text{ }\\ \text{Let }f(x) &= 3 \\ x^2 - 1 &= 3 \\ x^2 &= 4 \\ x &= \pm 2 \\ \text{ }\\ \text{So, } \text{dom}(f) &= [-2,2] \\ \text{And, } \text{dom}(g \circ f) &= [-2,2] \\ \end{aligned} \]
Inverse functions are calculated by interchanging x and y and solving for y of a 1-1 function. Thus, the domain of the inverse function is the range of the original function and the range of the inverse function is the domain of the original function. \[\text{dom}(f^{-1}) = \text{ran}(f)\] \[\text{ran}(f^{-1})=\text{dom}(f)\] An inverse function is represented with: \[y=f^{-1}(x)\]
When asked to find the intersection of a function and its inverse simply equate: \[f^{-1}(x) = f(x)\] The intersection points do not necessarily lie on the line y = x. An example of this is: \[f(x)=-x^3 \text{ and } f^{-1}(x)=-\sqrt[3]{x}\]
When graphing inverse functions, the inverse function is simply a reflection in the line y = x of the original function.
\[ \text{Example 1.2: For the function }\,f : [2,3] \rightarrow \mathbb{R}, \,f(x) = 2x + 3\text{, find the inverse function of }f(x) \\ \text{ and its domain and range.}\\ \text{ }\\ \text{Let } f(x)=y \text{ and for the inverse, interchange x and y}\\ \begin{aligned} x&=2y+3\\ 2y&=x-3\\ y&=\frac{x-3}{2}\\ f^{-1}(x)&=\frac{x-3}{2}\\ \end{aligned}\\ \text{dom}(f) = [2,3] = \text{ran}(f^{-1})\\ \text{ran}(f) = [f(2), f(3)] = [7,9] = \text{dom}(f^{-1})\\ \text{ }\\ \] \[ \text{Example 1.3: For the function }\,g : [1,\infty) \rightarrow \mathbb{R}, \,g(x) = (x+3)(x-1) \\ \text{, find the inverse function of }g(x)\text{ and its domain and range.}\\ \text{ } \\ \text{Let } g(x)=y \text{ and for the inverse, interchange x and y}\\ \begin{aligned} x&=(y+3)(y-1)\\ x&=y^2+2y-3\\ x&=y^2+2y+1-1+3\\ x&=(y+1)^2-4\\ y+1&=\pm\sqrt{x+4}\\ y&=\pm\sqrt{x+4}-1\\ \because \text{dom}(g^{-1})&=\text{ran}(g) = [0,\infty) &\\ \therefore g^{-1}(x)&=\sqrt{x+4}-1 \end{aligned}\\ \text{dom}(g) = [1, \infty) = \text{ran}(f^{-1})\\ \text{ran}(g) = [g(1), \infty) = [0, \infty) = \text{dom}(f^{-1})\\ \]