Unit 2 - Functions
Required Prior Knowledge
Questions
If \(f\left(x\right)=2x^{2}-3x\), determine:
a) \(f\left(-x\right) \)
b) \(-f\left(x\right)\)
If \(g\left(x\right)=x^{3}+x\), determine:
c) \(g\left(-x\right)\)
d) \(-g\left(x\right)\)
Solutions
Get Ready
Questions
Draw the graph of a function that has the \(y\)-axis as a line of symmetry.
Why is it not possible to draw the graph of a function that has the \(x\)-axis as a line of symmetry?
Draw the graph of a function that has rotational symmetry about the point \(\left(0,0\right)\).
Solutions
Notes
An even function is one that is symmetrical about the \(y\)-axis.
More formally, a function is said to be even if and only if$$f\left(-x\right)=f\left(x\right)\quad\forall x\text{ in the domain}$$
For example, consider the function \(f\left(x\right)=x^{2}\).
We can see in the graph of this function below that it is symmetrical about the \(y\)-axis.
Notice that \(f\left(-2\right)=\left(-2\right)^{2}=4=\left(2\right)^{2}=f\left(2\right)\).
That is when you input \(-2\) you get the same output as when you input \(+2\).
We can show this is true for all values of \(x\):$$f\left(-x\right)=\left(-x\right)^{2}=x^{2}=\left(x\right)^{2}=f\left(x\right)$$
An odd function is one that has rotational symmetry about the origin.
More formally, a function is said to be odd if and only if$$f\left(-x\right)=-f\left(x\right)\quad\forall x\text{ in the domain}$$
For example, consider the function \(f\left(x\right)=x^{3}\).
We can see in the graph of this function below that it has rotational symmetry about the origin.
Notice that \(f\left(-2\right)=\left(-2\right)^{3}=-8=-\left(2\right)^{3}=-f\left(2\right)\).
That is when you input \(-2\) you get the negative of the output when you input \(+2\).
We can show this is true for all values of \(x\):$$f\left(-x\right)=\left(-x\right)^{3}=-x^{3}=-\left(x\right)^{3}=-f\left(x\right)$$
Use the applet below to look at functions of the form \(y=x^{n}\) for different values of \(n\in\mathbb{Z}\).
For different values of \(n\), can you determine if the function will be even or odd?
Why do you think we call them even and odd functions?
Examples and Your Turns
Example
Determine algebraically whether the following functions are odd, even or neither.
Confirm your answers graphically.
(a) \(f\left(x\right)=\frac{1}{x}\)
(b) \(g\left(x\right)=x^{4}+x^{2}\)
(c) \(h\left(x\right)=x^{2}-2x+3\)
Your Turn
Determine algebraically whether the following function is odd, even or neither.
Confirm your answers graphically.$$f\left(x\right)=3-2x^{2}$$
Your Turn
Determine algebraically whether the following function is odd, even or neither.
Confirm your answers graphically.$$g\left(x\right)=3x^{4}-2x^{2}+5$$
Your Turn
Determine algebraically whether the following function is odd, even or neither.
Confirm your answers graphically.$$h\left(x\right)=\frac{x^{2}}{x^{4}+1}$$
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$$\begin{align}h\left(-x\right)&=\frac{\left(-x\right)^{2}}{\left(-x\right)^{4}+1}\\&=\frac{x^{2}}{x^{4}+1}\\&=h\left(x\right)\end{align}$$Therefore, \(h\left(x\right)\) is even.
Your Turn
Determine algebraically whether the following function is odd, even or neither.
Confirm your answers graphically.$$g\left(x\right)=\begin{cases} x-6, & \text{if } -6\le x<0 \\6-x, & \text{if } 0\le x<6 \end{cases}$$
Your Turn
Determine algebraically whether the following function is odd, even or neither.
Confirm your answers graphically.$$h\left(x\right)=\begin{cases} \left(x+6\right)^{2}, & \text{if } -6\le x<0 \\-\left(x-6\right)^{2}, & \text{if } 0\le x<6 \end{cases}$$
Your Turn
Suppose \(f\left(x\right)=\left(2x+3\right)\left(x+a\right)\), where \(a\in\mathbb{R}\), is an even function. Find the value of \(a\).
Your Turn
Suppose \(g\left(x\right)=\left(x+1\right)\left(\frac{1}{x}+b\right)\), where \(b\in\mathbb{R}\), is an even function. Find the value of \(b\).
Your Turn
Consider a function \(h\left(x\right)\) such that for all real numbers \(x\), the following equation holds:$$h\left(x\right)+2h\left(-x\right)=3x^{3}-5x$$Determine whether the function \(h\left(x\right)\) is even, odd or neither. Justify your answer clearly.
Your Turn
Suppose \(f\left(x\right)\) is a function such that \(f\left(x\right)+f\left(-x\right)=0\) for all \(x\) in its domain.
(a) Determine if \(f\left(x\right)\) is even, odd or neither.
Suppose \(g\left(x\right)\) is a function such that \(g\left(x\right)-g\left(-x\right)=0\) for all \(x\) in its domain.
(b) Determine if \(g\left(x\right)\) is even, odd or neither.
Let \(h\left(x\right)=f\left(x\right)g\left(x\right)\).
(c) Determine if \(h\left(x\right)\) is even, odd or neither.
Your Turn
Let \(f\left(x\right)\) be an arbitrary function defined for all real numbers.
(a) Show that the function \(E\left(x\right)=\frac{f\left(x\right)+f\left(-x\right)}{2}\) is an even function.
(b) Show that the function \(O\left(x\right)=\frac{f\left(x\right)-f\left(-x\right)}{2}\) is an odd function.
(c) Hence, prove that any function \(f\left(x\right)\) can be expressed as the sum of an even function and an odd function.
(d) Find the even and odd components for the function \(f\left(x\right)=e^{x}\).
Your Turn
Prove that the sum of two odd functions is an odd function.
Investigation - Part 1
Investigate whether the other sums of pairs of odd and even functions are odd or even.
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Let \(h\left(x\right)=f\left(x\right)+g\left(x\right)\) where \(f\left(x\right)\) is an even function and \(g\left(x\right)\) is an even function.
Thus$$f\left(-x\right)=f\left(x\right)$$and $$g\left(-x\right)=g\left(x\right)$$
From this we can see that$$\begin{align}h\left(-x\right)&=f\left(-x\right)+g\left(-x\right)\\ & =f\left(x\right)+g\left(x\right)\\ & =h\left(x\right)\end{align}$$Thus \(h\left(x\right)\) is an even function.
So the sum of two even functions is an even function.
-
Let \(h\left(x\right)=f\left(x\right)+g\left(x\right)\) where \(f\left(x\right)\) is an even function and \(g\left(x\right)\) is an odd function.
Thus$$f\left(-x\right)=f\left(x\right)$$and $$g\left(-x\right)=g\left(-x\right)$$
From this we can see that$$\begin{align}h\left(-x\right)&=f\left(-x\right)+g\left(-x\right)\\ & =f\left(x\right)-g\left(x\right)\\ & \ne h\left(x\right)\\ & \ne -h\left(x\right)\end{align}$$Thus \(h\left(x\right)\) is not an even function nor an odd function.
So the sum of an even function and odd function is neither odd nor even.
Investigation - Part 2
Investigate whether the products of pairs of odd and even functions are odd or even.
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Let \(h\left(x\right)=f\left(x\right)+g\left(x\right)\) where \(f\left(x\right)\) is an even function and \(g\left(x\right)\) is an even function.
Thus$$f\left(-x\right)=f\left(x\right)$$and $$g\left(-x\right)=g\left(x\right)$$
From this we can see that$$\begin{align}h\left(-x\right)&=f\left(-x\right)g\left(-x\right)\\ & =f\left(x\right)g\left(x\right)\\ & =h\left(x\right)\end{align}$$Thus \(h\left(x\right)\) is an even function.
So the product of two even functions is an even function.
-
Let \(h\left(x\right)=f\left(x\right)+g\left(x\right)\) where \(f\left(x\right)\) is an odd function and \(g\left(x\right)\) is an odd function.
Thus$$f\left(-x\right)=-f\left(x\right)$$and $$g\left(-x\right)=-g\left(x\right)$$
From this we can see that$$\begin{align}h\left(-x\right)&=f\left(-x\right)g\left(-x\right)\\ & =\left(-f\left(x\right)\right)\left(-g\left(x\right)\right)\\ & =f\left(x\right)g\left(x\right)\\ & =h\left(x\right)\end{align}$$Thus \(h\left(x\right)\) is an even function.
So the product of two odd functions is an even function.
-
Let \(h\left(x\right)=f\left(x\right)+g\left(x\right)\) where \(f\left(x\right)\) is an even function and \(g\left(x\right)\) is an odd function.
Thus$$f\left(-x\right)=f\left(x\right)$$and $$g\left(-x\right)=g\left(-x\right)$$
From this we can see that$$\begin{align}h\left(-x\right)&=f\left(-x\right)g\left(-x\right)\\ & =f\left(x\right)\left(-g\left(x\right)\right)\\ & =-f\left(x\right)g\left(x\right)\\ & = -h\left(x\right)\end{align}$$Thus \(h\left(x\right)\) is an odd function.
So the product of an even function and odd function is is odd.
Your Turn
Complete the graph for negative values of \(x\) so that it is even.
You can reveal the answer by clicking the empty circle next to “Complete Even”
Your Turn
Complete the graph for negative values of \(x\) so that it is odd.
You can reveal the answer by clicking the empty circle next to “Complete Odd”
Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.
Taking it Deeper
Conceptual Questions to Consider
Can a function be both even and odd? If so, give an example. If not, explain why not.
If a function is neither even nor odd, can it have any symmetry. Provide an example.
Explain why the sum of even powers of \(x\) always produces an even function.
Common Mistakes / Misconceptions
Can a function be both even and odd? If so, give an example. If not, explain why not.
If a function is neither even nor odd, can it have any symmetry. Provide an example.
Explain why the sum of even powers of \(x\) always produces an even function.
Connecting This to Other Skills
We have used the ideas of graphing functions (2.4) to sketch graphs and visually see even and odd functions.
We saw how we could use piecewise functions (2.10) to define even and odd functions.
Reflecting across the \(y\)-axis is going to be an important part of graph transformations (2.18).
We will see another way that even and odd functions can interact in Composite Functions (2.12).
When we look at the graphs of trigonometric functions (3.7), we will see that they satisfy the properties of odd an even functions.
When studying Maclaurin Series (8.2), we will see how the sum of even polynomial terms can be used to represent even functions, and similarly for odd functions.
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?