Let f(x) = x² + x. What is f(−x), and what does this tell you about f?
Af(−x) = x² + x = f(x), so f is even
Bf(−x) = x² − x, which is neither f(x) nor −f(x), so f is neither even nor odd
Cf(−x) = −x² − x = −f(x), so f is odd
Df(−x) = −x² + x, and since the leading term has even degree, f is even
Substituting −x: f(−x) = (−x)² + (−x) = x² − x. Comparing: f(x) = x² + x and −f(x) = −x² − x. Since x² − x equals neither x² + x nor −x² − x, the function is neither even nor odd. This illustrates the key misconception: a polynomial with both even- and odd-degree terms is neither. The polynomial shortcut (even ↔ only even-degree terms; odd ↔ only odd-degree terms) requires *all* terms to have the same parity — mixing them breaks both symmetries.
Question 2 Multiple Choice
A student claims: 'Since f(x) = x³ + 1 has an odd-degree leading term, it must be an odd function.' Which response is correct?
AThe student is correct — the leading term's degree determines the function's parity
BThe student is wrong — f(−x) = −x³ + 1, which is not equal to −f(x) = −x³ − 1, so f is neither even nor odd
CThe student is wrong — f is even because the constant term 1 acts as an even-degree term
DThe student is wrong — functions with a constant term are always even
f(−x) = (−x)³ + 1 = −x³ + 1. For f to be odd, we need f(−x) = −f(x) = −x³ − 1. Since −x³ + 1 ≠ −x³ − 1 (the constant term has the wrong sign), f is not odd — and f(−x) ≠ f(x) either, so it's not even. The student's error is applying the polynomial shortcut to just the leading term. The correct shortcut: a polynomial is odd only if *every* term has odd degree (no constants, no even-degree terms).
Question 3 True / False
Most polynomial function is either even or odd.
TTrue
FFalse
Answer: False
Most polynomial functions are neither even nor odd. A polynomial is even only if it contains exclusively even-degree terms (e.g., 3x⁴ − x² + 5), and odd only if it contains exclusively odd-degree terms (e.g., 2x³ − x). Any polynomial mixing both types — such as x² + x or x³ + 1 — is neither. Since most polynomials mix degree types, most are neither. The even/odd classification must be checked algebraically for each function, and 'neither' is the most common outcome.
Question 4 True / False
If f is an even function, then its graph is symmetric about the y-axis.
TTrue
FFalse
Answer: True
This follows directly from the definition. f being even means f(−x) = f(x) for all x — the function assigns the same output to x and −x. Geometrically, this means the point (x, f(x)) and the point (−x, f(x)) are both on the graph, and these points are reflections of each other across the y-axis. Every point on the right half of the graph has a mirror image at the same height on the left half. Y-axis symmetry is the visual equivalent of the algebraic condition f(−x) = f(x).
Question 5 Short Answer
Why is it insufficient to test whether a function is even or odd by checking just a few specific input values, and what is the correct procedure?
Think about your answer, then reveal below.
Model answer: A function must satisfy f(−x) = f(x) (even) or f(−x) = −f(x) (odd) for every x in its domain — a single counterexample rules out a symmetry, but no finite number of confirmations proves the property holds everywhere. The correct procedure is to substitute −x for x throughout the entire expression and simplify algebraically. If the result is identically equal to f(x), the function is even; if equal to −f(x), it is odd; if neither, it is neither.
For example, checking f(0) is useless for determining parity: f(−0) = f(0) always, so the origin is consistent with both even and odd and provides no information. A more subtle trap: a function might satisfy f(−x) = f(x) at many 'nice' values but fail at others. Only the algebraic substitution confirms the identity holds for all x simultaneously. Additionally, one must verify that the domain is symmetric about 0 — a function defined only for x ≥ 0 cannot be even or odd by convention.