Functions

Question 1

3. Describe each of the following sets using interval notation.
   a. {x : − 4 ≤ x < 2} {y : − 1 < y <
   √3
   b. 3 }
   c. {x : x > 3} d. {x : x ≤ −3}
   e. R f. R \ {0}

Answer 1

a {x∶ − 4 ≤ x < 2}
= [−4, 2)
b {y∶ − 1 < y <√3}
(−1,√3)
c {x∶x > 3}
(3, ∞)
d {x∶x ≤ −3}
(−∞, −3]
e R or (−∞, ∞)
f (−∞, 0) ∪ (0, ∞)

Question 2

. WE5 For each of the following, state the domain and range and whether the relation is a function or not.
a. {(4, 4), (3, 0), (2, 3), (0, −1)} {(x. y) : y = 4 − x
2 b. }

Answer 2

a {(4, 4), (3, 0), (2, 3), (0, −1)}
Domain is {0, 2, 3, 4} Range is
{−1, 0, 3, 4} This is a function
since each x co-ordinate is used
exactly once.
b {(x, y)∶y = 4 − x
2
}
This is the parabola y = 4 − x
2
, with
maximum turning point at (0, 4).
The maximal domain is R and range
is (−∞, 4]. It is a function since
a vertical line would cut its graph
exactly once.

Question 3

a. If f(x) = 3x + 1, determine
.i. f(0) ii. f(2) iii. f(−2) iv. f(5)

b. If g(x) = √
x + 4 , determine
i. g(0) ii. g(−3) iii. g(5) iv. g(−4)

Answer 3

f (x) = 3x + 1
i f (0) = 3 × 0 + 1 = 0 + 1 = 1
ii f (2) = 3 × 2 + 1 = 6 + 1 = 7
iii f (−2) = 3 × −2 + 1 = −6 + 1 = −5
iv f (5) = 3 × 5 + 1 = 15 + 1 = 16

g (x) =√x + 4
i g (0) =√0 + 4 =√4 = 2
ii g(−3) =√−3 + 4 =√1 = 1
iii g (5) = √5 + 4 =√9 = 3
iv g(−4) = √−4 + 4 =√0 = 0

Question 4

The following two equations represent water being added to a water tank over
15 hours, where w is the water in litres and t is the time in hours.
Equation 1: w = 25t, 0 ≤ t ≤ 5
Equation 2: w = 30t − 25, 5 ≤ t ≤ 15
a. Identify the dependent variable.
b. Determine how many litres of water are in the tank after 5 hours.
c. i. At what rate is the water being added to the tank after 5 hours?
ii. For how long is the water added to the tank at this rate?
d. Sketch a piece-wise graph to represent the water in the tank at any time, t, over the 15-hour period

Answer 4

a The volume of water depends on time, so water is the
dependent variable

b t = 5, can use either equation.
Equation 1 ∶w = 25t, 0 ≤ t ≤ 5
w = 25 (5)
w = 125
Equation 2 ∶w = 30t − 25, 5 ≤ t ≤ 15
w = 30 (5) − 25
w = 125
w = 125 L

c i after t = 5, water levels follow equation 2:
w = 30 t − 25, 5 ≤ t ≤ 15
rate = gradient = 30
rate = 30 L/h
ii looking at the domain 5 ≤ t ≤ 15∶ 15 − 5 = 10 hours

d Point of intersection occurs at t = 5 and w = 125