Higher Mathematics · Classes 9-10 · Chapter 1

Function, Domain and Range.

Know the radius of a circle and you know its circumference: one value, never two. That “one and only one” is the whole idea of a function. After that, one question is left: which x can you put in, and which x breaks it?

1. What a function is

The radius r and circumference p of a circle are linked by p = 2πr (Example 20). Every possible r fixes one and only one p, so the variable p is a function of the variable r: p = f(r), where f(r) = 2πr.

Definition 5: let X and Y be nonempty sets. If, under some rule, every element of X is associated with one and only one element of Y, that rule is a function from X to Y. Functions are written with symbols such as f, g, F, G.

Definition 6: a function f from X to Y is written f : X → Y. X is the domain of f and Y is its codomain.

Definition 7: if under f : X → Y the element x ∈ X is associated with y ∈ Y, then y is the image of x and x is the pre-image of y. This is written y = f(x).

Definition 8: the elements of Y that are images of elements of X form the range of f: range f = {f(x) : x ∈ X}. The range is always a subset of the codomain Y.

Example 21: for f : x → 2x + 1, x ∈ Z, dom f = Z. Range f = {y : y = 2x + 1, x ∈ Z}, the set of all odd integers.

2. Ordered pairs and arrow diagrams

There is one test for a function: how many arrows leave each x of the domain. Pick a relation and tap the x values.

A set of ordered pairs describes a function only when different ordered pairs have different first entries. Example 22: F = {(0, 0), (1, 1), (−1, 1), (2, 4), (−2, 4), (3, 9), (−3, 9)} is a function. dom F = {0, 1, −1, 2, −2, 3, −3}, range F = {0, 1, 4, 9}, and F(x) = x². Now sort six relations yourself.

In an arrow diagram, from every point of the domain one and only one arrow must start and end at one point of the codomain. In the first diagram of Example 26, 2 → 4, 2 → 5 and 3 → 4, 3 → 5, so it is not a function. Two arrows arriving at the same codomain point still give a function, and some codomain points may receive no arrow at all. Count the arrows leaving, not the arrows arriving.

3. Images, pre-images and the range

With a finite domain, find the range by substituting every x. Example 27: f(x) = 2x² + 1 on X = {1, 2, 3} gives f(1) = 3, f(2) = 9, f(3) = 19, so the range is {3, 9, 19}. Activity: f(x) = 4x + 2 on D = {−1, 3, 5} gives range {−2, 14, 22}.

Example 28: for f(x) = mx + c the images of 2 and 4 are 7 and −1. So 2m + c = 7 and 4m + c = −1, which give m = −4, c = 15. The image of 5 is f(5) = −5. For the pre-image of 3, set −4x + 15 = 3, which gives x = 3.

Substituting and working backwards both run on the same function. For F(x) = 2x − 1: F(−2) = −5, F(0) = −1, F(2) = 3, F((a + 1)/2) = a, F(x + 1) = 2x + 1, F(1/2) = 0. Backwards, F(x) = 5 gives x = 3, and F(x) = y gives x = (y + 1)/2.

The same method works on an arrow diagram. For f : x → px + q with 8 → −3 and 6 → 21: 8p + q = −3 and 6p + q = 21. Subtracting gives 2p = −24, so p = −12.

Now run it yourself. Pick a function and give x for the image, or give y for the pre-image. Where an input cannot go in, the reason is written out.

4. Finding the domain

A function F is fixed once its domain and the unique image F(x) of every x in it are fixed. When the domain is not stated, it is taken to be the set of all x for which F(x) is defined. Wherever the function is defined, that is the domain.

Under a square root the expression must be ≥ 0. Example 29: for F(x) = √(1 − x), 1 − x ≥ 0 gives x ≤ 1, so dom F = {x ∈ R : x ≤ 1}. F(−3) = 2, F(0) = 1, F(1/2) = 1/√2, F(1) = 0. But F(2) is undefined, because 2 ∉ dom F. In the same way √(x − 1) needs x ≥ 1 and √(1 − 2x) needs x ≤ 1/2. An absolute value is never negative: for real x, |x − 1| takes every value 0 or more, that is [0, ∞).

In a fraction, remove every x that makes the denominator zero. For f(x) = (2x − 1)/(2x + 3), 2x + 3 = 0 happens at x = −3/2, so dom f = R \ {−3/2}. 1/(x − 2) needs x ≠ 2, and 3/(2x − 1) needs x ≠ 1/2.

When the two rules meet: a square root in the denominator needs the expression inside to be > 0, because zero would make the denominator zero. 2/√(2 − x) needs 2 − x > 0, so x < 2. 1/√(5 − 2x) needs x < 5/2.

Now take a function and find its domain one step at a time.

5. Substituting into roots and fractions

For F(x) = √(x − 1): F(10) = 3, F(5) = 2, F(1) = 0. Backwards, F(x) = 2 gives x − 1 = 4, so x = 5. F(x) = 5 gives x = 26. And F(x) = y (y ≥ 0) gives x = y² + 1.

The book's sample question follows from that rule. For F(x) = −√(x − 4), F(a² + 4) = −√(a²) = −|a|. And F(x) = −2 gives √(x − 4) = 2, so x = 8. Likewise for √(x − 1), F(a² + 1) = |a|.

Substituting into fractions takes patience, but there is one rule. For f(x) = (2x − 1)/(2x + 3), f(−1/3) = (−5/3)/(7/3) = −5/7. For F(x) = 2x/(x − 1), F(−3) = −6/−4 = 3/2. For f(x) = (x − 3)/(2x + 1), f(−2) = −5/−3 = 5/3.

6. Check yourself

Three questions in the exam's style. Start with the first.

Work the second one out yourself.

In the last one, watch the condition on the domain.

The whole note on seven cards.

Source: Higher Mathematics, Classes 9-10 (NCTB), Chapter 1 “Set and Function”, the “Function, Domain and Range” part.