Look at a butterfly's wings or the arch of the Sydney Harbour Bridge. Symmetry is everywhere in nature and design. In mathematics, special functions called even and odd functions capture this symmetry in precise, testable rules.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Consider the parabola $y = x^2$. What happens to the $y$-value when you replace $x$ with $-x$? Now consider $y = x^3$. Does the same thing happen? What kind of symmetry do you think each graph might have?
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Wrong: √(a + b) = √a + √b.
Right: The square root of a sum is not the sum of square roots; √(a+b) cannot be simplified this way.
📚 Core Content
Some functions have special symmetry properties that make them easier to analyse and sketch. There are two main types: even functions and odd functions.
A function is even if:
$$f(-x) = f(x)$$
This means replacing $x$ with $-x$ leaves the output unchanged. Geometrically, the graph of an even function is a mirror image across the $y$-axis.
Examples: $f(x) = x^2$, $f(x) = x^4$, $f(x) = |x|$, $f(x) = \cos(x)$
A function is odd if:
$$f(-x) = -f(x)$$
This means replacing $x$ with $-x$ flips the sign of the output. Geometrically, the graph of an odd function has rotational symmetry of $180^\circ$ about the origin. If you rotate the graph halfway around the point $(0, 0)$, it looks exactly the same.
Examples: $f(x) = x$, $f(x) = x^3$, $f(x) = x^5$, $f(x) = \sin(x)$
🧮 Worked Examples
🧪 Activities
$f(x) = x^6$
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$f(x) = x^3 - x$
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$f(x) = x^2 + x + 1$
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$f(x) = \dfrac{1}{x}$
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1 If the graph is known to be even, describe how you would complete it for $x < 0$. What specific transformation maps each point $(a, b)$ on the right side to a corresponding point on the left side?
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2 If the graph is known to be odd, describe how you would complete it for $x < 0$. What specific transformation maps each point $(a, b)$ on the right side to a corresponding point on the left side?
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3 An even function satisfies $f(2) = 5$. Without doing any further calculation, state the value of $f(-2)$ and explain why.
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Earlier you were asked: Consider $y = x^2$. What happens when you replace $x$ with $-x$? Now consider $y = x^3$. Does the same thing happen? What symmetry might each graph have?
For $y = x^2$, replacing $x$ with $-x$ gives $(-x)^2 = x^2$, so the output stays the same. This is the hallmark of an even function, and the graph has reflection symmetry in the $y$-axis. For $y = x^3$, replacing $x$ with $-x$ gives $(-x)^3 = -x^3$, so the output flips sign. This is the hallmark of an odd function, and the graph has $180^\circ$ rotational symmetry about the origin. Symmetry in functions is not just beautiful — it is a powerful tool that lets us predict behaviour, simplify calculations, and sketch graphs with far less effort.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. Explain the difference between an even function and an odd function, both algebraically and geometrically. Use the functions $f(x) = x^2$ and $g(x) = x^3$ as examples in your explanation. 3 MARKS
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9. Determine whether each function is even, odd, or neither. Show the algebraic test for each. (a) $f(x) = 3x^4 - x^2$ (b) $f(x) = x^5 + 2x^3$ (c) $f(x) = x^2 - 2x + 3$ 4 MARKS
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10. A student claims that if a function contains only odd powers of $x$, it must be an odd function. Evaluate this claim. Is it always true? Provide a proof if it is true, or a counterexample if it is false. 3 MARKS
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A. $f(x) = x^6$ — Even. $f(-x) = (-x)^6 = x^6 = f(x)$.
B. $f(x) = x^3 - x$ — Odd. $f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x)$.
C. $f(x) = x^2 + x + 1$ — Neither. $f(-x) = x^2 - x + 1$, which equals neither $f(x)$ nor $-f(x)$.
D. $f(x) = \frac{1}{x}$ — Odd. $f(-x) = \frac{1}{-x} = -\frac{1}{x} = -f(x)$.
1. Reflect the right-hand side in the $y$-axis. Each point $(a, b)$ maps to $(-a, b)$.
2. Rotate the right-hand side $180^\circ$ about the origin. Each point $(a, b)$ maps to $(-a, -b)$.
3. $f(-2) = 5$. For an even function, $f(-x) = f(x)$, so the outputs at $x = 2$ and $x = -2$ must be equal.
1. B — $f(-x) = f(x)$ defines an even function.
2. B — $f(-x) = -f(x)$ for $f(x) = x^3$.
3. C — $180^\circ$ rotational symmetry about the origin.
4. A — All powers are even; $f(-x) = f(x)$.
5. B — $f(-x) = -f(x)$, so odd.
Q8 (3 marks): An even function satisfies $f(-x) = f(x)$ and has reflection symmetry in the $y$-axis [1]. For example, $f(x) = x^2$ gives $f(-x) = (-x)^2 = x^2 = f(x)$ [0.5]. An odd function satisfies $f(-x) = -f(x)$ and has $180^\circ$ rotational symmetry about the origin [1]. For example, $g(x) = x^3$ gives $g(-x) = (-x)^3 = -x^3 = -g(x)$ [0.5].
Q9 (4 marks):
(a) $f(-x) = 3(-x)^4 - (-x)^2 = 3x^4 - x^2 = f(x)$ → Even (b) $f(-x) = (-x)^5 + 2(-x)^3 = -x^5 - 2x^3 = -(x^5 + 2x^3) = -f(x)$ → Odd (c) $f(-x) = (-x)^2 - 2(-x) + 3 = x^2 + 2x + 3$. Not equal to $f(x)$ or $-f(x)$ → NeitherAward 1 mark each for correct classification with working.
Q10 (3 marks): The student's claim is true [1]. If a function consists only of odd powers of $x$, each term satisfies $(-x)^{\text{odd}} = -x^{\text{odd}}$. When summed, $f(-x) = -f(x)$ [1–2]. For example, $f(x) = x^3 + 2x$ gives $f(-x) = -x^3 - 2x = -f(x)$. Therefore any polynomial with only odd powers is an odd function.
Face the boss using your knowledge of symmetry, odd and even functions. Pool: lessons 1–5.
Tick when you've finished all activities and checked your answers.