Putting it together

This lesson pulls the chapter together through the kinds of question that come up.

A linear function from two values. If f(x) = mx + c with f(2) = 7 and f(5) = 16, subtract: 3m = 9, so m = 3, and then c = 7 – 6 = 1. So f(x) = 3x + 1.

Domain of a rational function: every real number except the zeros of the denominator. For (x + 1)/(x² – 9), x² – 9 = 0 at x = ±3, so the domain is R – {–3, 3}.

Range by solving for x. For f(x) = (2x + 1)/(x – 3), put y = f(x) and rearrange: y(x – 3) = 2x + 1, so x(y – 2) = 3y + 1 and x = (3y + 1)/(y – 2). This gives an x for every y except y = 2. So the range is R – {2}.

Is a piecewise rule a function? Where two pieces overlap, they must agree. g(x) = x + 1 for x ≤ 3 and g(x) = 2x – 2 for x ≥ 3 both cover x = 3, giving 4 and 4: they agree, so g is a function. If they gave different values, x = 3 would have two images, and it would not be.

Not a function? Find one element with two images, or one with none. On {1, 2, 3}, the relation {(1, 1), (2, 3), (1, 2)} fails because of 1.

A property true for all pairs needs a proof with letters, not examples. On Q, let R = {(a, b) : a – b is an integer}.

• (a, a) ∈ R for every a, because a – a = 0 is an integer.
• If (a, b) ∈ R, then b – a = –(a – b) is also an integer, so (b, a) ∈ R.
• If (a, b) and (b, c) are in R, then a – c = (a – b) + (b – c) is a sum of integers, so (a, c) ∈ R.

A property that fails needs just one counter-example. On N, take R = {(a, b) : a = b + 1}. Then (3, 2) ∈ R but (2, 3) ∉ R, so (a, b) ∈ R does not always give (b, a) ∈ R.

Checklist for any function question: check every element of the domain has an image, check every overlap of pieces agrees, keep only pre-images that lie in the domain, and never let a denominator be zero.