# 1.2 Function Notation (Full Solutions) ```1.2 Function Notation (Full Solutions)
Pg. 22 #1 - 3, 6, 7, 8, 23
Question 1
2
x  11
5
2
f (4)  (4)  11
5
3
 12
5
2 2 2
f ( )  ( )  11
3 5 3
11
 10
15
a) f ( x) 
2
f (5)  (5)  11
5
9
b) f ( x)  3 x 2  2 x  1
f (5)  3( 5) 2  2(5)  1
 66
2
2
2
f (  )  3(  ) 2  2(  )  1
3
3
3
4 4
  1
3 3
1
f (5)  2(5  4) 2
2
2
2
f (  )  2(   4) 2
3
3
f ( 4)  3( 4) 2  2( 4)  1
 57
c) f ( x)  2( x  4) 2
f ( 4)  2( 4  4) 2
 128
2
 10 
 2 
 3
200
2

or 22
9
9
d) f ( x)  6
f (4)  6
e) f ( x) 
f (4) 
2
f ( )  6
3
f (5)  6
1
x
1
4
f) f ( x )  x  5
f ( 4)  4  5
3
1
5
1

5
f (5) 
2
1
f ( ) 
3  2
 
 3
3

2
f (5)  5  5
0
2
2
f ( )    5
3
3

13
3
Question 2
All graphs (unless indicated) are from the standard viewing window of the TI83+/84 calculator.
a) f ( x)  5 x  4
f (0)  5(0)  4
4
b) k ( x)  4 x
k (0)  4(0)
0
c) p ( x )  4
p ( 0 )  4
d) g ( x)  11x 2  3 x  1
g (0)  11(0) 2  3(0)  1
 1
e) f ( x)  (3 x  3)(2 x  2)
f (0)  (3(0)  3)(2(0)  2)
 6
f) The window for this graph has been set as
shown.
2
h( x )   (5  4 x)( x  7)
3
2
h(0)   (5  4(0))(0  7)
3
2
  (35)
3
70
1

or 23
3
3
Question 3
a) given (3, –12)
a (3)  12
The defining equation is f ( x)  4 x.
a  4
b) given (5, 15)
a(5)  15
a3
c) given (1,
The defining equation is f ( x)  3x.
2
)
3
2
3
2
a
3
a (1) 
The defining equation is f ( x) 
2
x.
3
d) given (−3, 3)
a(3)  3
a  1
The defining equation is f ( x)   x.
Question 6
a) The ordered pairs in question 5 part a) form a function. For each value of the independent variable,
there is a single corresponding value for the dependent variable.
b) The ordered pairs in question 5 part b) form a function. For each value of the independent variable,
there is a single corresponding value for the dependent variable.
c) The ordered pairs in question 5 part c) do not form a function. The choice of –5 for the independent
variable corresponds to two different values for the dependent variable.
d) The ordered pairs in question 5 part d) form a function. For each value of the independent variable,
there is a single corresponding value for the dependent variable.
Question 7
a) The mapping diagram defines the set {(1, 1), (2, 4), (3, 9), (4, 16)}
b) The mapping diagram defines the set {(–5, 11), (–4, 6), (–2, –4), (0, –14), (2, –24)}.
c) The mapping diagram defines the set ((–4, 6), (1, 6), (3, 6), (5, 6)}.
Question 8
a–c) For all of the mapping diagrams, the arrows clearly show that each member of the domain maps to a
single member of the range. This means that all of the relations shown are functions.
Question 13
a)
f ( x)  x
0
1
2
3
x
0
1
4
9
y
7
6
5
4
3
2
1
x
1
2
3
4
5
6
7
8
9
10
b) The relation is a function. Each member of the domain corresponds to a single member of the range.
c) The relation can be identified as a function immediately from its equation because there is a single
operation to be applied to each member of the domain. The pattern in the table of values can also be
recognized as a function as no member of the domain is repeated to generate a different value for the
dependent variable.
Question 14
Both the domain and range must be restricted to positive values. The student may also choose to place an
upper restriction on the depth of the river and the corresponding speed.
b)
Question 15
a) At the time of purchase, n = 0.
23 000
 1000
n 1
23 000
V ( 0) 
 1000
0 1
 24 000
The car is worth \$24 000 when first purchased.
V ( n) 
b) i)
23 000
 1000
10  1
 3090.91
The car is worth about \$3091 after 10 years.
V (10) 
ii)
23 000
 1000
12  1
 2769.23
The car is worth about \$2769 after 12 years.
V (12) 
c) Replace the dependent variable with the required value and solve for n.
23 000
V ( n) 
 1000
n 1
23 000
2000 
 1000
n 1
23 000
1000 
n 1
1000n  1000  23 000
1000n  22 000
n  22
It would take 22 years for the car to depreciate to a value of \$2000.
d) V(n) is a function. For any chosen number of years, the defining equation gives a single value for the
worth of the car.
Question 21
a) Internet research suggests that the fastest street-legal cars have a top velocity of about 400 km/h. Using
this value, the range is v  ,8  v  400 .
Use the relation to find the distance which corresponds to the fastest speed.
v(d )  12.6 d  8
400  12.6 d  8
392
 d
12.6
2
 280 
d 

 9 
d  968
The formula shows that a skid mark of 968 m would correspond to a top speed of 400 km/h. Using this as
a maximum expected value, the domain is d  ,0  d  968 .
b)
sk id le ngth (m )
400
300
200
100
ve locity (m /s)
200
400
600
800
c) The relation is a function. For any choice of skid length, there is a single corresponding initial velocity
for the car.
Question 23
Double the first given equation.
f ( x)  2 g ( x)  12 x 2  3 x  8  2 f ( x)  4 g ( x)  24 x 2  6 x  16
Subtract the second equation from the result.
2 f ( x)  4 g ( x)  24 x 2  6 x  16
2 f ( x)  3 g ( x)  18 x 2  6 x  13
g ( x)  6 x 2  3
Substitute into the first given equation.
f ( x)  2(6 x 2  3)  12 x 2  3x  8
f ( x)  12 x 2  3x  8  12 x 2  6
f ( x)  3 x  2
Substitute the requested values into f and g.
f (2)  g (3)  3(2)  2  6(3) 2  3
 8  57
 65
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