Exercise 2.3 answers: Relations and Functions

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(i) (2,1), (5,1), (8,1), (11,1), (14,1), (17,1)

1. List the first numbers in the pairs: 2, 5, 8, 11, 14, 17 — every one is different.
2. Since no first number repeats, each input goes to exactly one output, so this is a function.
3. The domain is the set of first numbers: {2, 5, 8, 11, 14, 17}.
4. The range is the set of second numbers actually used: {1}.

AnswerYes, it is a function. Domain = {2, 5, 8, 11, 14, 17}, Range = {1}.

(ii) (2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)

1. List the first numbers: 2, 4, 6, 8, 10, 12, 14 — all different.
2. Since every first number appears only once, this is a function.
3. Domain = {2, 4, 6, 8, 10, 12, 14}.
4. Range = the second numbers used = {1, 2, 3, 4, 5, 6, 7}.

AnswerYes, it is a function. Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}.

(iii) (1,3), (1,5), (2,5)

1. The first number 1 appears twice: once paired with 3, once paired with 5.
2. One input (1) is sent to two different outputs (3 and 5), so this is not a function.
3. Even so, we can still list what would be its domain: the first numbers used are {1, 2}.
4. And its range: the second numbers used are {3, 5}.

AnswerNo, it is not a function, because 1 has two different images. Domain = {1, 2}, Range = {3, 5}.

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(i) f(x) = − |x|

1. |x| is defined for every real number, so f(x) = −|x| is defined for every real x too.
2. So the domain is all real numbers, R.
3. |x| is never negative, so −|x| is never positive — it is always 0 or less.
4. As x ranges over all reals, −|x| takes every value from 0 down to −∞.

AnswerDomain = R, Range = (−∞, 0].

(ii) f(x) = √(9−x²) .

1. A square root needs the value inside it to be 0 or more, so we need 9 − x² ≥ 0.
2. That means x² ≤ 9, so x must lie between −3 and 3.
3. So the domain is [−3, 3].
4. On this interval, 9−x² is smallest (0) at x = ±3 and largest (9) at x = 0.
5. So √(9−x²) ranges from 0 to 3.

AnswerDomain = [−3, 3], Range = [0, 3].

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(i) f (0)

1. Substitute x = 0 into f(x) = 2x − 5.
2. f(0) = 2(0) − 5 = −5.

Answer−5

(ii) f (7)

1. Substitute x = 7.
2. f(7) = 2(7) − 5 = 14 − 5 = 9.

Answer9

(iii) f (–3)

1. Substitute x = −3.
2. f(−3) = 2(−3) − 5 = −6 − 5 = −11.

Answer−11

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(i) t (0)

1. Substitute C = 0 into t(C) = 9C/5 + 32.
2. t(0) = 0 + 32 = 32.

Answer32

(ii) t (28)

1. Substitute C = 28.
2. t(28) = 9×28/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4.

Answer82.4 (i.e. 412/5)

(iii) t (–10)

1. Substitute C = −10.
2. t(−10) = 9×(−10)/5 + 32 = −18 + 32 = 14.

Answer14

(iv) The value of C, when t(C) = 212

1. Set 9C/5 + 32 = 212.
2. Subtract 32 from both sides: 9C/5 = 180.
3. Multiply both sides by 5/9: C = 100.

AnswerC = 100

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(i) f (x) = 2 – 3x, x ∈ R, x > 0.

1. As x increases from just above 0, 3x increases from just above 0 with no upper limit.
2. So 2 − 3x decreases from just below 2, going down towards −∞.
3. x = 0 itself is not allowed, so the value 2 (which would need x = 0) is never actually reached.

AnswerRange = (−∞, 2) — every real number less than 2.

(ii) f (x) = x² + 2, x is a real number.

1. x² is never negative, and its smallest value is 0, at x = 0.
2. So the smallest value of x² + 2 is 2.
3. As x moves away from 0 in either direction, x² + 2 grows without any upper bound.

AnswerRange = [2, ∞).

(iii) f (x) = x, x is a real number.

1. This is the identity rule: every real number is the output for exactly one input, itself.

AnswerRange = R, all real numbers.

Watch this explained “Nine squares, five answers”, 5:56 into Each standard function is pinned down by its picture as much as by its rule