Determine if f(x) = x³ - x is even odd or neither

Question

Determine whether $f(x) = x^3 - x$ is an even function, odd function, or neither.

Solution — Step by Step

Even function: $f(-x) = f(x)$ for all $x$ in the domain. Graph is symmetric about the y-axis.
Odd function: $f(-x) = -f(x)$ for all $x$ in the domain. Graph has 180° rotational symmetry about the origin.
Neither: If neither condition holds.

The test is purely algebraic: substitute $-x$ for $x$ and simplify.

f(-x) = (-x)^3 - (-x)

= -x^3 + x

= -(x^3 - x)

= -f(x)

We found: $f(-x) = -f(x)$ .

This matches the definition of an odd function.

We can verify it does NOT equal $f(x) = x^3 - x$ (which would make it even). For example, $f(1) = 1 - 1 = 0$ , $f(-1) = -1 + 1 = 0$ . This case doesn’t help — let’s try $f(2) = 8 - 2 = 6$ and $f(-2) = -8 + 2 = -6 = -f(2)$ ✓.

Since $f(-x) = -f(x)$ for all $x$ :

\boxed{f(x) = x^3 - x \text{ is an ODD function.}}

Its graph is symmetric about the origin — if you rotate it 180° around the origin, it looks the same.

Why This Works

Each term in $f(x) = x^3 - x$ is an odd power of $x$ : $x^3$ has odd power 3, and $x$ has odd power 1. A sum of odd-power terms is always an odd function, because $(-x)^{\text{odd}} = -(x^{\text{odd}})$ .

This gives a quick pattern check: if every term in a polynomial has an odd power of $x$ , the function is odd. If every term has an even power (including constants, since $c = cx^0$ ), the function is even.

Quick shortcut for polynomials:

All even powers → even function
All odd powers → odd function
Mix of even and odd powers → neither

$f(x) = x^4 + x^2$ is even. $f(x) = x^3 + x$ is odd. $f(x) = x^3 + x^2$ is neither.

Alternative Method

Check graphically: plot $f(x) = x^3 - x$ . Find the roots: $x^3 - x = x(x^2 - 1) = x(x-1)(x+1) = 0$ → $x = -1, 0, 1$ .

The graph passes through $(-1, 0)$ , $(0, 0)$ , and $(1, 0)$ , and extends from negative infinity to positive infinity. If you rotate it 180° about the origin, it lands back on itself — confirming it’s odd.

Common Mistake

Students often confuse “odd function” with “odd number” or assume that any function with an $x^3$ term is odd. The full function must satisfy $f(-x) = -f(x)$ for ALL $x$ — not just at one point. For example, $f(x) = x^3 + 1$ has an $x^3$ term but is neither even nor odd because $f(-x) = -x^3 + 1 \neq -f(x) = -x^3 - 1$ . Always substitute and compare — don’t guess from visual inspection of individual terms.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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