Algebra 2 · F.BF.B.3

Even and odd functions

Can you tell from a rule alone, without drawing anything, whether its graph is its own mirror image?

Where you are 10 in Features of a graph

When you reach for it

Symmetry about the vertical axis and the origin describes two mirror images a graph can have: a fold along the y-axis, where each point (a, b) has the partner (-a, b), and a half turn about the origin, where the partner is (-a, -b). A drawing can suggest either one, but it shows only a window, and only as exactly as a pencil line allows.

The rule settles it exactly, for every input at once. Reach for this test when you have a formula and want to know its symmetry: before sketching, so that you plot only the half with x \ge 0 and mirror the rest; when checking a sketch; and later, when symmetry shortens a calculation. The same idea tests equations that are not functions, such as a circle.

The procedure

Even and odd functions

The graph of f is symmetric about the y-axis when every point (a, f(a)) has its partner (-a, f(a)) on the graph too. The point of the graph above -a is (-a, f(-a)), so the partner is there exactly when f(-a) = f(a). Asking this for every a gives the definition. A function f is even when

f(-x) = f(x) \quad \text{for every } x \text{ in its domain.}

The graph is symmetric about the origin when every point (a, f(a)) has the partner (-a, -f(a)), which means f(-a) = -f(a). A function f is odd when

f(-x) = -f(x) \quad \text{for every } x \text{ in its domain.}

Both definitions need a symmetric domain: whenever x is in it, so is -x, or f(-x) has no meaning.

So the test is a calculation. Find f(-x) by putting -x in for every x in the rule, as in evaluating at an expression, and simplify. If the result is f(x), the function is even. If it is -f(x), the opposite of every term, it is odd. Otherwise the function is neither, which is the usual case: unlike integers, functions need not be one or the other.

In reflecting a graph, y = f(-x) was the graph flipped across the y-axis and y = -f(-x) the graph turned a half turn. An even function is one the flip leaves unchanged, and an odd function one the half turn leaves unchanged, since -f(-x) = f(x) says the same as f(-x) = -f(x).

Why the names come from powers

For a whole number n, (-x)^n = (-1)^n x^n, and (-1)^n is 1 when n is even and -1 when n is odd:

\begin{aligned} (-x)^n &= x^n &&\text{for even } n,\\ (-x)^n &= -x^n &&\text{for odd } n. \end{aligned}

So every even power x^2, x^4, x^6, \ldots is an even function and every odd power x, x^3, x^5, \ldots an odd function; a constant c is c\,x^0, an even power. That is where the names come from, and the squaring and cubing functions are the first examples. A polynomial built only from even powers is even, one built only from odd powers is odd, and one with terms of both kinds is neither: replacing x by -x flips the signs of the odd-power terms and keeps the others, so the result matches neither f(x) nor -f(x).

Past polynomials the names keep their meaning but lose the exponents: |x| and \displaystyle \sqrt{9 - x^2} are even, and the reciprocal function \displaystyle \frac{1}{x} is odd.

Showing that it is not

One input shows a function is not even: if f(-a) \ne f(a) for a single a, the rule f(-x) = f(x) fails. Likewise one input with f(-a) \ne -f(a) shows it is not odd. The other direction needs algebra: f(-2) = f(2) and f(-5) = f(5) say nothing about x = 3. To show a function is even or odd, simplify f(-x) for a general x.

Equations that are not functions

An equation in x and y, such as x^2 + y^2 = 25, may fail the vertical line test, so there is no f to evaluate. The same idea works on the equation itself: a point (a, b) is on the graph when it makes the equation true, so each test asks whether the partner point makes it true as well.

ReplacePartner of (a, b)Equation unchanged means symmetry about
y by -y(a, -b)the x-axis
x by -x(-a, b)the y-axis
both(-a, -b)the origin

“Unchanged” means the new equation simplifies to the old one, perhaps after multiplying both sides by -1, so it has exactly the same solutions. When it does not, one point that satisfies the old equation and fails the new one proves it.

A graph symmetric about the x-axis is almost never a function: a point (a, b) with b \ne 0 and its partner (a, -b) stand on the same vertical line. So the x-axis test belongs to equations, not functions.

Sums and products

For functions f and g with the same symmetric domain:

Sum f + gProduct fg
both eveneveneven
both oddoddeven
one even, one oddneither, unless one is 0odd

The product column behaves like exponents: x^2 \cdot x^3 = x^5, and an even exponent plus an odd one is odd. Each entry takes one line. For two odd functions,

\begin{aligned} (fg)(-x) &= f(-x)\,g(-x)\\ &= \big({-f(x)}\big)\big({-g(x)}\big)\\ &= f(x)\,g(x), \end{aligned}

so the product is even. A quotient \displaystyle \frac{f}{g} follows the product column, since \displaystyle \frac{1}{g} has the same type as g. Example 7 runs the sum row backwards, splitting a function into an even part and an odd part.

Worked examples

Example 1

Three polynomials

Decide whether each function is even, odd or neither.

(a) f(x) = x^6 - 5x^2 \qquad (b) g(x) = 2x^5 - x \qquad (c) h(x) = x^2 + x

Solution

(a) Put -x in for each x. Both powers are even, so each term keeps its sign:

f(-x) = (-x)^6 - 5(-x)^2 = x^6 - 5x^2 = f(x).

f is even.

(b) Both powers are odd, so each term changes sign:

g(-x) = 2(-x)^5 - (-x) = -2x^5 + x.

That is not g(x), but the opposite of every term of g is -g(x) = -2x^5 + x, the same expression. So g(-x) = -g(x) and g is odd.

(c) h(-x) = (-x)^2 + (-x) = x^2 - x, which is neither h(x) = x^2 + x nor -h(x) = -x^2 - x. To be sure they really differ, try x = 1: h(1) = 2 and h(-1) = 0. Since h(-1) \ne h(1), h is not even, and since h(-1) \ne -h(1) = -2, it is not odd. h is neither, as a mix of an even and an odd power must be.

Example 2

A fraction, and a domain that is not symmetric

Decide whether each function is even, odd or neither.

(a) \displaystyle r(x) = \frac{x}{x^2 + 1} \qquad (b) \displaystyle s(x) = \frac{x^2 + 1}{x - 1}

Solution

(a) The denominator is never 0, so the domain is every real number. Replacing x by -x everywhere,

r(-x) = \frac{-x}{(-x)^2 + 1} = -\frac{x}{x^2 + 1} = -r(x).

r is odd: an odd function divided by an even one, as the table predicts.

(b) Look at the domain first. s is undefined at x = 1, but \displaystyle s(-1) = \frac{2}{-2} = -1 exists. The input -1 has no partner 1, so s is neither. No algebra is needed: a domain that is not symmetric rules out both.

Example 3

Roots, absolute values and a shifted parabola

Decide whether each function is even, odd or neither.

(a) \displaystyle f(x) = \sqrt{9 - x^2} \qquad (b) g(x) = x|x| \qquad (c) h(x) = (x - 1)^2

Solution

(a) The domain is -3 \le x \le 3, symmetric about 0. Since (-x)^2 = x^2,

f(-x) = \sqrt{9 - (-x)^2} = \sqrt{9 - x^2} = f(x),

so f is even.

(b) A number and its opposite have the same absolute value, |-x| = |x|. So

g(-x) = (-x)\,|-x| = -x|x| = -g(x),

and g is odd: an odd factor x times an even factor |x|.

(c) Multiply out first: h(x) = x^2 - 2x + 1, so h(-x) = x^2 + 2x + 1. Only the x term changed sign, so this is neither h(x) nor -h(x). With x = 1: h(1) = 0 and h(-1) = 4, neither equal nor opposite. h is neither. Its graph is symmetric, but about the line x = 1; the test only asks about the axes and the origin.

Example 4

An equation that is not a function

Test the graph of y^2 = x + 3 for symmetry about the x-axis, the y-axis and the origin.

Solution

The x-axis: replace y by -y. Since (-y)^2 = y^2, the result is y^2 = x + 3 again, so the graph is symmetric about the x-axis. The point (1, 2) is on it, since 2^2 = 1 + 3, so its partner (1, -2) is too.

The y-axis: replace x by -x, giving y^2 = -x + 3, a different equation. One point proves it: the partner of (1, 2) is (-1, 2), and 2^2 = 4 while -1 + 3 = 2.

The origin: replace both, giving (-y)^2 = -x + 3, again y^2 = -x + 3. The partner of (1, 2) is (-1, -2), and (-2)^2 = 4 \ne 2.

x y −3 3 6 −2 2
(1, 2)
(1, -2)
y^2 = x + 3
Figure 1Example 4. Replacing y by -y leaves y^2 = x + 3 unchanged, so each point (a, b) has its partner (a, -b).

The graph is symmetric about the x-axis only. It is a parabola opening to the right from (-3, 0), and it fails the vertical line test, as a graph with this symmetry must.

Example 5

A circle, and all three tests at once

Test the graph of x^2 + y^2 = 25 for symmetry about the x-axis, the y-axis and the origin.

Solution

Each variable appears only squared, and (-x)^2 = x^2, (-y)^2 = y^2. So replacing y by -y, or x by -x, or both, gives back x^2 + y^2 = 25 each time. The graph is symmetric about the x-axis, the y-axis and the origin.

The point (3, 4) is on it, since 9 + 16 = 25. The three symmetries give (3, -4), (-3, 4) and (-3, -4) with no further arithmetic.

x y −5 5 −5 5
(3, 4)
(-3, 4)
(3, -4)
(-3, -4)
x^2 + y^2 = 25
Figure 2Example 5. One point found by hand gives three more, one for each symmetry.

So the points with x \ge 0 and y \ge 0 decide the whole graph, the circle of radius 5 about the origin (the equation of a circle).

Example 6

Products without multiplying out

Decide whether each is even, odd or neither without multiplying out. Check (b) by expanding.

(a) x^3(x^4 + 1) \qquad (b) (x^5 - 2x)(x^3 + x) \qquad (c) (x^2 - 3)^3

Solution

(a) x^3 is odd and x^4 + 1 is even, so the product is odd.

(b) Each factor has only odd powers, so both are odd, and odd times odd is even. Expanding gives x^8 + x^6 - 2x^4 - 2x^2, only even powers, which agrees.

(c) (x^2 - 3)^3 is three copies of the even function x^2 - 3 multiplied together, and even times even is even, so it is even.

Example 7

Making a tuning fork outline symmetric

Thorsten draws the U-shaped bend of a tuning fork, with the y-axis along the center line of the handle and lengths in centimeters:

\begin{aligned} f(x) &= \frac{x^4}{8} + \frac{x^3}{10} + x^2,\\ &\qquad -2 \le x \le 2. \end{aligned}

The prongs rise from the ends x = -2 and x = 2, and they are meant to match. Is the outline symmetric about the center line? If not, replace it with an even function by averaging it with its own mirror image.

Solution

The domain is symmetric. Replacing x by -x changes the sign of the cube term only:

f(-x) = \frac{x^4}{8} - \frac{x^3}{10} + x^2,

which is not f(x). The ends show it: f(2) = 2 + 0.8 + 4 = 6.8 and f(-2) = 2 - 0.8 + 4 = 5.2, so the right end sits 1.6 cm higher.

Any function on a symmetric domain is an even function plus an odd one:

E(x) = \frac{f(x) + f(-x)}{2}, \qquad O(x) = \frac{f(x) - f(-x)}{2}.

Adding them gives f(x). Replacing x by -x swaps f(x) and f(-x), which leaves E alone and turns O into -O, so E is even and O is odd. Here

E(x) = \frac{x^4}{8} + x^2, \qquad O(x) = \frac{x^3}{10}.

All of the lopsidedness is in O. The even part E is f averaged with its mirror image, and its ends are E(2) = E(-2) = 6, halfway between 6.8 and 5.2. Thorsten’s symmetric outline is \displaystyle y = \frac{x^4}{8} + x^2.

Boundaries

The domain must be symmetric. \displaystyle \sqrt{x} and \displaystyle \frac{1}{x - 2} are neither, because some input has no opposite in the domain.

“For every x” is the whole condition. Agreement at a few inputs proves nothing; one disagreement disproves it.

Even and odd are neither the only choices nor opposites. Most functions are neither, and f(x) = 0 is both, the only function that is (Exercise 22).

The test sees only the axes and the origin. A graph symmetric about another line or point, such as y = (x - 1)^2 about x = 1, passes none of these tests. A shift left or right moves the mirror off the axis, and a shift up or down moves an odd function’s center off the origin, which is why combining a shift with other moves usually destroys evenness or oddness.

For functions, symmetry about the x-axis is ruled out unless every point is on the x-axis. Test it on equations.

Where this goes next

The powers that named these functions return in Degree, end behavior and the shape of a polynomial graph, where an even or odd leading power decides how the two ends of a polynomial graph behave. The half-turn symmetry of an odd function is the point symmetry of Rotational and point symmetry. In trigonometry, Unit circle symmetry and even-odd identities shows that cosine is even and sine is odd, the even-odd identities. In calculus, the properties of definite integrals include the fact that over an interval centered at 0 an odd function’s integral is 0, the two sides canceling, and an even function’s is twice the integral over the right half.

Exercises

Polynomials

Decide whether each function is even, odd or neither.

  1. f(x) = x^4 + 7x^2 - 3
  2. g(x) = 4x^3 - 9x
  3. h(x) = x^3 + 1
  4. q(x) = -x^7 + 3x^5 - x^3
  5. r(x) = (x + 2)^2

Beyond polynomials

Decide whether each function is even, odd or neither.

  1. \displaystyle f(x) = \frac{x^2}{x^4 + 1}
  2. \displaystyle g(x) = \frac{x^3}{x^2 - 9}
  3. \displaystyle h(x) = \frac{x}{x - 2}
  4. \displaystyle k(x) = \sqrt{x^2 + 4}
  5. \displaystyle m(x) = x\sqrt{x^2 + 1}

Equations

Test the graph of each equation for symmetry about the x-axis, the y-axis and the origin.

  1. y = x^2 - 4
  2. x = y^4 - 3
  3. xy = 4
  4. 4x^2 + 9y^2 = 36
  5. y^3 = x
  6. x^2 + y^2 - 6y = 0

Sums, products and parts

  1. Without multiplying out, decide whether each is even, odd or neither: (a) x^2(x^3 - x), (b) (x^3 + x)^2, (c) \displaystyle \frac{x^5 - x}{x^2 + 3}.
  2. f is odd and g is even, both defined for every real number. Decide whether each is even, odd or neither: (a) f(x)^3, (b) g(x) + 5, (c) f(x) + 5.
  3. Write p(x) = x^3 + 2x^2 - 5x + 6 as the sum of an even function and an odd function.
  4. Vlasta models the bend of another tuning fork as \displaystyle v(x) = \frac{x^4}{10} - \frac{x^3}{5} + \frac{x^2}{4} for -2 \le x \le 2, in centimeters. (a) Is the outline symmetric about the y-axis? (b) Find the heights v(2) and v(-2) of the two ends. (c) Find the even part of v and the height of its ends.

Reasoning

  1. Explain why f(-2) = f(2) and f(-5) = f(5) do not show that f is even, while f(-1) \ne f(1) does show that f is not even. Give a function with the first two properties that is not even.
  2. Prove that a function defined for every real number that is both even and odd must be f(x) = 0.
  3. Prove that a graph with any two of the three symmetries (about the x-axis, the y-axis, the origin) has the third.
Answers
  1. Even. Every power is even, so f(-x) = x^4 + 7x^2 - 3 = f(x).
  2. Odd. g(-x) = -4x^3 + 9x = -(4x^3 - 9x) = -g(x).
  3. Neither. h(-x) = -x^3 + 1. With x = 1: h(1) = 2 and h(-1) = 0, not equal and not opposite.
  4. Odd. Every power is odd, so q(-x) = x^7 - 3x^5 + x^3 = -q(x).
  5. Neither. r(x) = x^2 + 4x + 4 and r(-x) = x^2 - 4x + 4. With x = 1: r(1) = 9 and r(-1) = 1, not equal and not opposite.
  6. Even. \displaystyle f(-x) = \frac{(-x)^2}{(-x)^4 + 1} = \frac{x^2}{x^4 + 1} = f(x).
  7. Odd. The domain, x \ne \pm 3, is symmetric, and \displaystyle g(-x) = \frac{-x^3}{x^2 - 9} = -g(x).
  8. Neither. h is undefined at 2 but \displaystyle h(-2) = \frac{-2}{-4} = \frac{1}{2}, so the domain is not symmetric.
  9. Even. \displaystyle k(-x) = \sqrt{(-x)^2 + 4} = \sqrt{x^2 + 4} = k(x).
  10. Odd. \displaystyle m(-x) = -x\sqrt{x^2 + 1} = -m(x): an odd factor times an even one.
  11. The y-axis only. x \to -x gives the same equation. y \to -y gives -y = x^2 - 4, and so does replacing both. (0, -4) is on the graph, but (0, 4), its partner under both of those tests, is not.
  12. The x-axis only. y \to -y gives x = y^4 - 3 again. x \to -x, alone or with y, gives -x = y^4 - 3: (-3, 0) is on the graph but (3, 0) is not.
  13. The origin only. (-x)(-y) = xy, so replacing both gives xy = 4. Replacing one gives xy = -4: (1, 4) is on the graph, (1, -4) and (-1, 4) are not.
  14. All three. Both variables appear only squared, so every replacement gives 4x^2 + 9y^2 = 36 again.
  15. The origin only. Both gives -y^3 = -x, which is y^3 = x. One alone gives -y^3 = x or y^3 = -x: (1, 1) is on the graph, (1, -1) and (-1, 1) are not.
  16. The y-axis only. x \to -x gives the same equation. y \to -y, alone or with x, gives x^2 + y^2 + 6y = 0: (0, 6) is on the graph but (0, -6) is not. It is the circle of radius 3 about (0, 3).
  17. (a) Odd: even times odd; it is x^5 - x^3. (b) Even: odd times odd; it is x^6 + 2x^4 + x^2. (c) Odd: odd divided by even.
  18. (a) Odd: f(-x)^3 = \big({-f(x)}\big)^3 = -f(x)^3. (b) Even: g(-x) + 5 = g(x) + 5. (c) Neither, unless f is the zero function: f(-x) + 5 = -f(x) + 5 equals f(x) + 5 only where f(x) = 0, and never equals -f(x) - 5. If f is 0, then f(x) + 5 is the constant 5, which is even.
  19. p(x) = (2x^2 + 6) + (x^3 - 5x). The even part \displaystyle \frac{p(x) + p(-x)}{2} keeps the even powers and the constant, and the odd part keeps the odd powers.
  20. (a) No: \displaystyle v(-x) = \frac{x^4}{10} + \frac{x^3}{5} + \frac{x^2}{4} \ne v(x). (b) v(2) = 1.6 - 1.6 + 1 = 1 cm and v(-2) = 1.6 + 1.6 + 1 = 4.2 cm. (c) The even part is \displaystyle \frac{x^4}{10} + \frac{x^2}{4}, and its ends are both 2.6 cm, the average of 1 and 4.2.
  21. Even means f(-x) = f(x) for every x: two inputs say nothing about the others, while one failure breaks “every”. For example, f(x) = x^2 + x(x^2 - 4)(x^2 - 25) has f(\pm 2) = 4 and f(\pm 5) = 25, but f(1) = 73 and f(-1) = -71.
  22. Even and odd give f(-x) = f(x) and f(-x) = -f(x), so f(x) = -f(x), 2f(x) = 0 and f(x) = 0 for every x.
  23. Take (a, b) on the graph. With both axes: (a, -b) is on it, then its y-axis partner (-a, -b), so the origin. With the x-axis and the origin: (a, -b), then its origin partner (-a, b), so the y-axis. With the y-axis and the origin: (-a, b), then (a, -b), so the x-axis.
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