1The idea in one paragraph

A function is even if putting in −x gives the same output as x: its graph is unchanged by reflection in the y-axis. A function is odd if putting in −x gives the negative of the output: its graph is unchanged by a half turn about the origin. Most functions are neither, and that is fine. The second idea is the inverse: a function can be reversed only if it is one-to-one, so a function that is not must have its domain restricted until it is, and the choice of domain decides which inverse you get. The third idea joins the two: a function that is its own inverse, f⁻¹ = f, has a graph that is unchanged by reflection in the line y = x.

2Even and odd functions

Even: f(−x) = f(x) for every x in the domain. Odd: f(−x) = −f(x) for every x in the domain.

Both definitions need the domain to be symmetric about 0: if x is in it, −x must be too.

What they look like. Reflecting a graph in the y-axis gives y = f(−x) (2.11). If that is the same graph, f(−x) = f(x): so an even function is symmetric in the y-axis. A half turn about the origin sends (x, y) to (−x, −y), which is a reflection in the y-axis followed by one in the x-axis, giving y = −f(−x). If that is the same graph, f(−x) = −f(x): so an odd function has rotational symmetry of order 2 about the origin. Figure 1 shows one of each.

In Figure 1(a), y = x⁴ − 5x² + 4 takes the same value −2.19 at x = 1.5 and at x = −1.5. In Figure 1(b), y = x³ − 4x takes the value −3 at x = 1 and +3 at x = −1.

Where the names come from. A power xⁿ is even when n is even, since (−x)⁴ = x⁴, and odd when n is odd, since (−x)³ = −x³. So a polynomial with only even powers (a constant counts, as x⁰) is even, and one with only odd powers is odd. x⁴ − 5x² + 4 is even; x³ − 4x is odd; x³ + x² is neither.

Two facts worth having. An odd function that is defined at 0 must pass through the origin, since f(0) = −f(0) forces f(0) = 0. And the only function that is both odd and even is f(x) = 0.

3How to test a function

Work out f(−x), simplify it, and compare it with f(x) and with −f(x).

Worked example 1 · four functions (Paper 1).

(a) f(x) = x² cos 3x. f(−x) = (−x)² cos(−3x) = x² cos 3x = f(x), using cos(−θ) = cos θ. Even.

(b) g(x) = x/(x² + 1). g(−x) = −x/(x² + 1) = −g(x). Odd.

(c) h(x) = x³ + x². h(−x) = −x³ + x², which is neither h(x) nor −h(x). To prove "neither", one number is enough: h(1) = 2 and h(−1) = 0, so h(−1) ≠ h(1) and h(−1) ≠ −h(1). Neither.

(d) k(x) = ln((1 + x)/(1 − x)), for −1 < x < 1. k(−x) = ln((1 − x)/(1 + x)) = −ln((1 + x)/(1 − x)) = −k(x), using ln(1/A) = −ln A. Odd.

For a "show that", the marks are for writing f(−x) out and simplifying it into the required form, then the conclusion. A graph from the GDC suggests the answer but does not prove it. For "neither", a single counterexample with numbers is the proof.

4Periodic functions

The guide includes periodic functions. From the unit circle (3.x), the point at angle −θ is the reflection of the point at angle θ in the horizontal axis, so it has the same cosine and the opposite sine:

cos(−x) = cos x (even) sin(−x) = −sin x (odd) tan(−x) = −tan x (odd)

Figure 2 shows the three graphs. The cosine curve is a mirror image in the y-axis; the sine and tangent curves are unchanged by a half turn about the origin. The reciprocal functions follow: sec x is even, cosec x and cot x are odd.

A sum of periodic functions need not be either: sin x + cos x is neither, since at x = π/4 it is √2 and at x = −π/4 it is 0.

5Sums, products and composites

These rules come straight from the definitions and save a lot of algebra. They can also be asked as proofs.

Products behave like the signs of the powers: x · x = x², so odd × odd is even, just as (−) × (−) = (+). So x sin x is even, x² sin x is odd, sin x tan x is even, and sin(x³) is odd. An even function plus an odd one is usually neither: x² + x.

One use you will meet in calculus (5.x): for an odd function the areas either side of the origin cancel, so the integral from −a to a of an odd function is 0.

6Inverses when the domain must be restricted

From 2.5: a function has an inverse only if it is one-to-one, every output coming from exactly one input. A graph that fails the horizontal line test has no inverse on its full domain. The fix is to restrict the domain to a piece on which it is one-to-one, usually cutting at a turning point.

Worked example 2 · a quadratic (Paper 1). Let f(x) = x² − 6x + 5, x ≤ k.

(a) Find the largest value of k for which f⁻¹ exists. Complete the square: f(x) = (x − 3)² − 4. The vertex is at x = 3, and f is decreasing to its left, so f is one-to-one on x ≤ 3. Any k > 3 would include points on both sides of the vertex with equal outputs, such as f(2) = f(4) = −3. So k = 3.

(b) Find f⁻¹(x), stating its domain.

The range of f on x ≤ 3 is f(x) ≥ −4, since the vertex is the lowest point. So the domain of f⁻¹ is x ≥ −4, and its range is f⁻¹(x) ≤ 3.

Figure 3 shows the kept half of the parabola, the discarded half dashed, and the inverse as the reflection of the kept half in y = x. The vertex (3, −4) reflects to the end point (−4, 3). Had the right half, x ≥ 3, been chosen, the inverse would be 3 + √(x + 4): the sign of the root is decided by the domain you kept, and that sign is the mark most often lost.

Worked example 3 · an exponential. f(x) = e²ˣ − 3, x ∈ ℝ. It is increasing, so no restriction is needed, but the range matters.

The range of f is f(x) > −3, because e²ˣ > 0; so the domain of f⁻¹ is x > −3. Figure 4 shows the pair. The horizontal asymptote y = −3 of f reflects in y = x into the vertical asymptote x = −3 of f⁻¹, and the intercept (0, −2) of f reflects into (−2, 0).

Worked example 4 · a harder inverse (Paper 1 or Paper 3). Let f(x) = (eˣ − e⁻ˣ)/2, x ∈ ℝ. Show that f is odd, and find f⁻¹(x).

f(−x) = (e⁻ˣ − eˣ)/2 = −f(x), so f is odd. f is increasing (eˣ increases and e⁻ˣ decreases), so it is one-to-one on ℝ, and its range is ℝ. For the inverse, the trick is to see a quadratic in eʸ:

As a check, f⁻¹ should be odd too, since the reflection of a graph with half-turn symmetry about the origin in y = x keeps that symmetry.

Trigonometric functions are the most famous restricted domains. sin x is not one-to-one on ℝ, but it is on −π/2 ≤ x ≤ π/2, and that restriction defines arcsin; cos x is restricted to 0 ≤ x ≤ π for arccos, and tan x to −π/2 < x < π/2 for arctan. These are the inverse trigonometric functions you use in 3.x and in 2.16.

7Self-inverse functions

A function is self-inverse if f⁻¹ = f. Applying it twice gets you back where you started:

f is self-inverse if and only if (f ∘ f)(x) = x for every x in its domain.

On the graph, f⁻¹ is the reflection of f in y = x. If that reflection is f itself, the graph is symmetric in the line y = x. Its domain and range are the same set.

The simple examples in Figure 5(b): y = 5 − x (any line of gradient −1), y = 4/x (any k/x, which is why 1/x in 2.8 is its own inverse), and y = √(4 − x²) for 0 ≤ x ≤ 2, a quarter circle. And f(x) = x, trivially.

Worked example 5 · proving it (Paper 1). Show that f(x) = (2x + 3)/(x − 2), x ≠ 2, is self-inverse.

So f ∘ f is the identity and f⁻¹ = f. Figure 5(a) shows the graph: the vertical asymptote x = 2 and the horizontal asymptote y = 2 are reflections of each other in y = x, as they must be, and so are the intercepts (−1.5, 0) and (0, −1.5).

The general rule for (ax + b)/(cx + d). Its asymptotes are x = −d/c and y = a/c. For the graph to be symmetric in y = x they must swap, so −d/c = a/c, that is d = −a. Any function (ax + b)/(cx − a) is self-inverse, provided it is not a constant. So (3x + 5)/(x + k) is self-inverse exactly when k = −3.

Watch the domain. For f(x) = √(4 − x²) on 0 ≤ x ≤ 2, f(f(x)) = √(4 − (4 − x²)) = √(x²) = |x|, and |x| = x only because x ≥ 0 there. On −2 ≤ x ≤ 2 the same formula is not self-inverse; it is not even one-to-one.