Year 10 Extended · Unit 3 Functions, quadratics and transformations · Lesson 1 of 7

What can go in, and what can come out

A function accepts some inputs and rejects others, and its outputs may be limited too. Domain and range say exactly which, and notation such as f(t + 1) says what happens to an input before the rule is applied.

FunctionsGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: expand (x + 5)².
    Answerx² + 10x + 25. The middle term is 2 × 5 × x, and the last is 5² = 25.
  2. From Year 9: if f(x) = 2x + 3, work out f(4).
    Answer11. Put 4 in place of x: 2 × 4 + 3 = 11.
  3. From Year 9: what is the lowest value of (x − 6)² + 2?
    Answer2. A square is never negative, so the least value is 0 + 2 = 2, at x = 6.
Words used in this lesson
domain
the set of inputs a function accepts. For example: x ≥ 7 for √(x − 7).
range
the set of outputs a function can give. For example: y ≥ 0 for a square root.
natural domain
all real numbers except those that break the rule. For example: all x except x = 6 for 1 ÷ (x − 6).
function notation
f(x) names the output of the rule f at the input x. For example: f(x) = 4x − 1.
restricted domain
a domain limited on purpose, not by the rule. For example: x² for 0 ≤ x ≤ 3.
refused input
an input that gives no real output. For example: x = 6 in 3 ÷ (x − 6).

The spreadsheet that shows an error

A stallholder in a Lima market keeps her accounts in a spreadsheet. One column works out 1 ÷ (x − 3) for each row number x. Most rows show a number. One row shows an error, and the row after it shows a number again.

Predict first: which row shows the error, and is there a number this column can never show, however many rows there are? Then use the figure to test your prediction on four different rules.

1
0
all real numbersdomain
y ≥ −3range
1value at this input

f(x) = x² − 4x + 1: at x = 0 the value is 1. Domain: all real numbers. Range: y ≥ −3.

Show me the steps
  1. Keep function 1 and slide x to 6. The value is 36 − 24 + 1 = 13. Every input gives a value, so the domain is all real numbers; the lowest point of the graph is y = −3, so the range stops there.
  2. Choose function 2, 1 ÷ (x − 3), and slide x to 4: the value is 1. At x = 2 it is −1. At x = 3 the machine fails, because x − 3 is 0 and dividing by 0 has no value. The domain is every number except 3.
  3. Choose function 3, √(x − 2). At x = 6 the value is 2. At x = 1 the number under the root is −1, which has no real square root. The domain starts at x = 2, and the outputs start at 0.
  4. Choose function 4, √(16 − x²). At x = 0 the value is 4, the greatest it takes. At x = 5 the number under the root is −9, so the input is refused. The domain is −4 ≤ x ≤ 4 and the range is 0 ≤ y ≤ 4.

Domain is the set of inputs a rule accepts; range is the set of outputs it can give. The natural domain is every real number except the ones that break the rule: a value that makes a denominator 0, or that makes the number under a square root negative.

Function notation with an expression. f(a + 1) means: put the whole of (a + 1) in place of x, brackets and all. To find a range, find the lowest or highest value with the vertex, and on a restricted domain also check the end points, because the greatest value can be at an end.

A range read from a screen can be wrong, because a screen shows only a window. Reason it out, then check it with a graph.

f(x + 7) is not f(x) + 7. If f(x) = x², then f(x + 7) = (x + 7)² = x² + 14x + 49, but f(x) + 7 = x² + 7. The first changes the input before squaring; the second changes the output after.

Worked examples

1. f(x) = 5x − x². Simplify f(m − 1).

  1. Put (m − 1) in place of every x: 5(m − 1) − (m − 1)².
  2. Expand: 5m − 5 − (m² − 2m + 1) = 5m − 5 − m² + 2m − 1.
  3. Collect: −m² + 7m − 6.

2. Find the natural domain of g(x) = √(x − 3) ÷ (x − 8).

  1. The square root needs x − 3 ≥ 0, so x ≥ 3.
  2. The denominator must not be 0, so x ≠ 8.
  3. Both rules apply: x ≥ 3 and x ≠ 8.

3. h(x) = −x² + 6x + 1. Find the range, then the range for 0 ≤ x ≤ 5.

  1. The vertex is at x = −6 ÷ (2 × (−1)) = 3, and h(3) = −9 + 18 + 1 = 10. The parabola opens downward, so for all real x the range is y ≤ 10.
  2. On 0 ≤ x ≤ 5 the vertex x = 3 is inside, so 10 is still the greatest. At the ends: h(0) = 1 and h(5) = −25 + 30 + 1 = 6. The least is 1.
  3. The range is 1 ≤ y ≤ 10.

Your go. The first step is given. Find the natural domain of p(x) = 1 ÷ (x² − 9).

  1. The denominator is 0 when x² − 9 = 0, which is when x = 3 or x = −3.
  2. Now say which inputs are allowed.
Check the answer

Every real number except those two: x ≠ 3 and x ≠ −3. The graph has a break at each of them.

With a GDC (optional)

A GDC can confirm a domain and a range. Graph y = √(x − 6), then open the table: an input below 6 shows an error or no value, which is the domain telling you the input is refused. Use the minimum or maximum tool to find the turning point that sets the end of a range.

A graph window shows only part of the picture, so a range read from the screen can be wrong: check it with the vertex and the end points. Everything here can be done by hand.

Check before you go on

1. What is the natural domain of g(x) = √(2x + 10)?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. g(x) = 5 − 2x². Work out g(−3).

3. k(x) = 12 ÷ (x + 4) is not defined for exactly one value of x. What is it?

Level 2 · Building

4. f(x) = x² − 3x + 4. When you simplify f(t + 1) you get t² + pt + q. Find p.

5. f(x) = x² − 6x + 2 is defined only for −1 ≤ x ≤ 5. What is the greatest value of f?

Level 3 · Stretch

6. How many whole numbers are in the natural domain of s(x) = √(20 − x²)?

7. g(x) = 15 ÷ (x² + 3). What is the greatest value that g takes?

Something to think about

Function 3 is √(x − 2) and function 4 is √(16 − x²). Both have a square root, yet one has a domain that never ends and the other has two ends. What makes the difference? Then try to invent a rule that has x ≥ 2 and x ≠ 5 in its domain.

One way to think about it

For √(x − 2) the number under the root grows without limit as x grows, so only the left end is blocked. For √(16 − x²) the number under the root is positive only between −4 and 4, because x² gets large in both directions, so both ends are blocked. A rule with x ≥ 2 and x ≠ 5 is √(x − 2) ÷ (x − 5): the root asks for x ≥ 2 and the denominator refuses 5.

Exit ticket

8. Which function has the natural domain x > 5?