Lesson 1.4 · Unit 1 · F-BF.3
Two kinds of symmetry, tested algebraically
A graph symmetric about the \( y \)-axis and a graph unchanged by a half-turn about the origin are the two symmetries a function can have. Both have a one-line algebraic test, and most functions have neither.
- A function is even when \( f(-x) = f(x) \) for every \( x \) in the domain.
- Even means symmetric about the \( y \)-axis: the left half is the mirror image of the right.
- A function is odd when \( f(-x) = -f(x) \) for every \( x \) in the domain.
- Odd means symmetric about the origin: rotating the graph \( 180^\circ \) about the origin returns the same graph.
- To test, compute \( f(-x) \) fully and compare it with \( f(x) \) and with \( -f(x) \).
- Most functions are neither, and saying so is a complete answer.
- A power function \( x^n \) is even when \( n \) is even and odd when \( n \) is odd, which is where the names come from.
- The only function that is both is \( f(x) = 0 \).
Where students lose marks: concluding "odd" from the fact that \( f(-x) \ne f(x) \). Failing the even test does not pass the odd test. The odd condition has to be checked separately, and a function can fail both.
The problem. Classify each as even, odd or neither. (a) \( f(x) = x^3 - 4x \). (b) \( g(x) = x^4 - 3x^2 + 1 \). (c) \( h(x) = x^2 + x \). (d) Explain why the only function that is both even and odd is the zero function.
Step one: compute \( f(-x) \) for (a). \( f(-x) = (-x)^3 - 4(-x) \). \( (-x)^3 = -x^3 \), since an odd power keeps the sign. \( -4(-x) = 4x \). So \( f(-x) = -x^3 + 4x \).
Step two: compare for (a). \( f(x) = x^3 - 4x \), so \( -f(x) = -x^3 + 4x \). That is exactly \( f(-x) \). The function is odd. Numerical check: \( f(2) = 8 - 8 = 0 \) and \( f(-2) = -8 + 8 = 0 \). Both zero, consistent. A better check: \( f(1) = 1 - 4 = -3 \) and \( f(-1) = -1 + 4 = 3 \). Opposite signs, as odd requires.
Step three: compute \( g(-x) \) for (b). \( g(-x) = (-x)^4 - 3(-x)^2 + 1 \). \( (-x)^4 = x^4 \) and \( (-x)^2 = x^2 \), since even powers destroy the sign. So \( g(-x) = x^4 - 3x^2 + 1 = g(x) \). The function is even. Check: \( g(2) = 16 - 12 + 1 = 5 \) and \( g(-2) = 16 - 12 + 1 = 5 \). Equal.
Step four: compute \( h(-x) \) for (c). \( h(-x) = (-x)^2 + (-x) = x^2 - x \).
Step five: compare both ways for (c). Against \( h(x) = x^2 + x \): not equal, since the linear terms differ in sign. Not even. Against \( -h(x) = -x^2 - x \): not equal, since the squared terms differ in sign. Not odd. The function is neither. Both comparisons were necessary, and this is the case the warning is about.
Step six: note the pattern. Part (a) had only odd powers of \( x \), and came out odd. Part (b) had only even powers, counting the constant 1 as \( x^0 \), and came out even. Part (c) mixed them, and came out neither. That is a reliable shortcut for polynomials: all-odd exponents means odd, all-even means even, mixed means neither. It does not extend to non-polynomials, so the algebraic test remains the method.
Step seven: set up (d). Suppose \( f \) is both even and odd. Then for every \( x \) in the domain, \( f(-x) = f(x) \) from evenness, and \( f(-x) = -f(x) \) from oddness.
Step eight: finish (d). The two right-hand sides must be equal, since both equal \( f(-x) \): \( f(x) = -f(x) \). Adding \( f(x) \) to both sides gives \( 2f(x) = 0 \), so \( f(x) = 0 \). This holds for every \( x \), so \( f \) is the zero function. Conversely the zero function does satisfy both conditions, since \( 0 = 0 \) and \( 0 = -0 \), so it genuinely is both and is the only such function. The result is worth having because it rules out a whole class of wrong answers: a classification of "both" is never correct unless the function is identically zero.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- Classify \( f(x) = x^2 \).
Show the full solution
Even
- Classify \( f(x) = x^5 \).
Show the full solution
Odd
- Classify \( f(x) = \left| x \right| \).
Show the full solution
\( \left| -x \right| = \left| x \right| \). Even
- Classify \( f(x) = x + 2 \).
Show the full solution
\( f(-x) = -x + 2 \), which is neither \( f(x) \) nor \( -f(x) \). Neither
- What symmetry does an even function's graph have?
Show the full solution
Symmetry about the \( y \)-axis
- Classify \( f(x) = \dfrac{1}{x^3} \).
Show the full solution
\( f(-x) = \dfrac{1}{(-x)^3} = \dfrac{1}{-x^3} = -\dfrac{1}{x^3} = -f(x) \). Odd Check: \( f(2) = \dfrac{1}{8} \) and \( f(-2) = -\dfrac{1}{8} \). Opposite. Correct.
- Classify \( f(x) = x^4 + x^2 - 6 \).
Show the full solution
\( f(-x) = x^4 + x^2 - 6 = f(x) \), since both powers are even and the constant is unchanged. Even The shortcut confirms it: every exponent present is even, counting \( -6 \) as \( -6x^0 \). Check: \( f(1) = 1 + 1 - 6 = -4 \) and \( f(-1) = -4 \). Equal.
- Classify \( f(x) = \dfrac{x^2}{x^3 - x} \), and state its domain.
Show the full solution
Domain first. The denominator factors as \( x(x^2 - 1) = x(x-1)(x+1) \), zero at \( 0 \), \( 1 \) and \( -1 \). Domain: all reals except those three. Note the domain is symmetric about zero, which it must be for either classification to be possible. Test. \( f(-x) = \dfrac{(-x)^2}{(-x)^3 - (-x)} = \dfrac{x^2}{-x^3 + x} = \dfrac{x^2}{-(x^3 - x)} = -\dfrac{x^2}{x^3 - x} = -f(x) \). Odd Check at \( x = 2 \): \( f(2) = \dfrac{4}{8 - 2} = \dfrac{2}{3} \), and \( f(-2) = \dfrac{4}{-8 + 2} = -\dfrac{2}{3} \). Opposite. Correct. Note on the shortcut. The all-even or all-odd exponent rule applies to polynomials, not to quotients. Here an even numerator over an odd denominator produced an odd function, which the rule would not have predicted.
- Explain why a function whose domain is \( x \ge 0 \) cannot be even or odd, unless its
domain is just \( \{0\} \).
Show the full solution
Both tests require evaluating \( f(-x) \). For the test to mean anything, \( -x \) must be in the domain whenever \( x \) is. A domain of \( x \ge 0 \) fails that at once: if \( x = 2 \) is in the domain then \( -2 \) must be too, and it is not. So \( f(-2) \) does not exist and the equation \( f(-2) = f(2) \) cannot hold or fail; it is not a statement about anything. The general requirement. A function can be even or odd only if its domain is symmetric about zero. That is a precondition, checked before any algebra. Why \( \{0\} \) is the exception. The set \( \{0\} \) is symmetric about zero, since \( -0 = 0 \). A function defined only there is even if \( f(0) \) is anything, and odd only if \( f(0) = 0 \), since oddness would demand \( f(0) = -f(0) \). A worked case. \( f(x) = \sqrt{x} \) has domain \( x \ge 0 \) and is therefore neither, for this reason rather than for any algebraic one. By contrast \( f(x) = \sqrt[3]{x} \) has all reals as its domain and is genuinely odd. The practical order. Check the domain for symmetry first. If it is not symmetric, the answer is "neither" and no computation is needed. The tests require \( -x \) to be in the domain whenever \( x \) is, and a domain of \( x \ge 0 \) is not symmetric about zero
- Prove that the product of two odd functions is even, and that the product of an even and
an odd function is odd.
Show the full solution
Set up. Let \( f \) and \( g \) be functions on a domain symmetric about zero, and let \( p(x) = f(x)g(x) \). First claim: odd times odd is even. Suppose \( f \) and \( g \) are both odd, so \( f(-x) = -f(x) \) and \( g(-x) = -g(x) \). Then \[ p(-x) = f(-x)\,g(-x) = \big(-f(x)\big)\big(-g(x)\big) = f(x)g(x) = p(x) \] The two negatives multiply to a positive, so \( p \) is even. Check with an example. \( f(x) = x \) and \( g(x) = x^3 \) are both odd, and their product is \( x^4 \), which is even. Confirmed. Second claim: even times odd is odd. Suppose \( f \) is even and \( g \) is odd, so \( f(-x) = f(x) \) and \( g(-x) = -g(x) \). Then \[ p(-x) = f(-x)\,g(-x) = f(x)\big(-g(x)\big) = -f(x)g(x) = -p(x) \] One negative survives, so \( p \) is odd. Check with an example. \( f(x) = x^2 \) is even and \( g(x) = x \) is odd, and their product is \( x^3 \), which is odd. Confirmed. The third case, for completeness. Even times even is even, by the same argument with no negatives at all. So the three rules together behave exactly like multiplying signs, or like adding exponents in \( x^m \cdot x^n = x^{m+n} \) and asking whether the sum is even or odd. Why the analogy is not a coincidence. The names come from the power functions, where evenness and oddness are literally the parity of the exponent. Since multiplying powers adds exponents, and parity adds the way these rules multiply, the general result had to come out matching. A caution. The corresponding statements for sums are weaker: even plus even is even and odd plus odd is odd, but even plus odd is usually neither. \( x^2 + x \) from the worked example is exactly that case. Proved by substituting the definitions and tracking the signs