Question 1: Is the graph shown below that of a function?
Solution to Question 1:

Vertical line test: A vertcal line at x = 0 for example cuts the graph at two points. The graph is not that
of a function.
Question 2: Does the equation
y2 + x = 1
represents a function y in terms of x?
Solution to Question 2:


Solve the above equation for y

For one value of x we have two values of y and this is not a function.
y = + SQRT(1 - x) or y = - SQRT(1 - x)
Question 3: Function f is defined by
f(x) = - 2 x 2 + 6 x - 3
find f(- 2).
Solution to Question 3:

Substitute x by -2 in the formula of the function and calculate f(-2) as follows

f(-2) = - 2 (-2) 2 + 6 (-2) - 3
f(-2) = -23
Question 4: Function h is defined by
h(x) = 3 x 2 - 7 x - 5
find h(x - 2).
Solution to Question 4:

Substitute x by x - 2 in the formula of function h
h(x - 2) = 3 (x - 2) 2 - 7 (x - 2) - 5

Expand and group like terms
h(x - 2) = 3 ( x 2 - 4 x + 4 ) - 7 x + 14 - 5
= 3 x 2 - 19 x + 7
Question 5: Functions f and g are defined by
f(x) = - 7 x - 5 and g(x) = 10 x - 12
find (f + g)(x)
Solution to Question 5:

(f + g)(x) is defined as follows

(f + g)(x) = f(x) + g(x) = (- 7 x - 5) + (10 x - 12)

Group like terms to obtain
(f + g)(x) = 3 x - 17
Question 6: Functions f and g are defined by
f(x) = 1/x + 3x and g(x) = -1/x + 6x - 4
find (f + g)(x) and its domain.
Solution to Question 6:

(f + g)(x) is defined as follows
(f + g)(x) = f(x) + g(x)
= (1/x + 3x) + (-1/x + 6x - 4)

Group like terms to obtain
(f + g)(x) = 9 x - 4

The domain of function f + g is given by the intersection of the domains of f and g
Domain of f + g is given by the interval (-infinity , 0) U (0 , + infinity)
Question 7: Functions f and g are defined by
f(x) = x 2 -2 x + 1 and g(x) = (x - 1)(x + 3)
find (f / g)(x) and its domain.
Solution to Question 7:

(f / g)(x) is defined as follows
(f / g)(x) = f(x) / g(x) = (x 2 -2 x + 1) / [ (x - 1)(x + 3) ]

Factor the numerator of f / g and simplify
(f / g)(x) = f(x) / g(x) = (x - 1) 2 / [ (x - 1)(x + 3) ]
= (x - 1) / (x + 3) , x not equal to 1

The domain of f / g is the intersections of the domain of f and g excluding all values of x that make the
numerator equal to zero. The domain of f / g is given by
(-infinity , -3) U (-3 , 1) U (1 , + infinity)
Question 8: Find the domain of the real valued function h defined by
h(x) = SQRT ( x - 2)
Solution to Question 8:

For function h to be real valued, the expression under the square root must be positive or equal to 0.
Hence the condition
x - 2 >= 0

Solve the above inequality to obtain the domain in inequality form
x >= 2

and interval form
[2 , + infinity)
Question 9: Find the domain of
g(x) = SQRT ( - x 2 + 9) + 1 / (x - 1)
Solution to Question 9:

For a value of the variable x to be in the domain of function g given above, two conditions must be
satisfied: The expression under the square root must not be negative
- x 2 + 9 >= 0

and the denomirator of 1 / (x - 1) must not be zero
x not equal to 1
or in interval form
(-infinity , 1) U (1 , + infinity)

The solution to the inequality - x 2 + 9 >= 0 is given by the interval
[-3 , 3]

Since x must satisfy both conditions, the domain of g is the intersection of the sets
(-infinity , 1) U (1 , + infinity) and [-3 , 3]
[-3 , 1) U (1 , +3]
Question 10: Find the range of
f(x) = | x - 2 | + 3
Solution to Question 10:

| x - 2 | is an absolute value and is either positive or equal to zero as x takes real values, hence
| x - 2 | >= 0

Add 3 to both sides of the above inequality to obtain
| x - 2 | + 3 >= 3

The expression on the left side of the above inequality is equal to f(x), hence
f(x) >= 3

The above inequality gives the range as the interval
[3 , + infinity)
Question 11: Find the range of
f(x) = -x 2 - 10
Solution to Question 11:

-x 2 is either negative or equal to zero as x takes real values, hence
-x 2 <= 0

Add -10 to both sides of the above inequality to obtain
-x 2 - 10 <= -10

The expression on the left side is equal to f(x), hence
f(x) <= -10

The above inequality gives the range of f as the interval
(-infinity , -10]
Question 12: Find the range of
h(x) = x 2 - 4 x + 9
Solution to Question 12:

h(x) is a quadratic function, so let us first write it in vertex form using completing the square
h(x) = x 2 - 4 x + 9
= x2 - 4 x + 4 - 4 + 9
= (x - 2) 2 + 5

(x - 2) 2 is either positive or equal to zero as x takes real values, hence
(x - 2) 2 >= 0

Add 5 to both sides of the above inequality to obtain
(x - 2) 2 + 5 >= 5

The above inequality gives the range of h as the interval
[5 , + infinity)
Question 13: Functions g and h are given by
g(x) = SQRT(x - 1) and h(x) = x 2 + 1
Find the composite function (g o h)(x).
Solution to Question 13:

The definition of the absolute value gives
(g o h)(x) = g(h(x))
= g(x 2 + 1)
= g(x 2 + 1)
= SQRT(x 2 + 1 - 1)
=|x|

So
(g o h)(x) = | x |
Question 14: How is the graph of f(x - 2) compared to the graph of f(x)?
Solution to Question 14:

The graph of f(x - 2) is that of f(x) shifted 2 units to the right.
Question 15: How is the graph of h(x + 2) - 2 compared to the graph of h(x)?
Solution to Question 15:

The graph of h(x + 2) - 2 is that of h(x) shifted 2 units to the left and 2 unit downward.
Question 16: Express the perimeter P of a square as a function of its area.
Solution to Question 16:

A square shape with side x has perimeter P given by
P=4x

and an area A given by
A = x2

Solve the equation P = 4 x for x
x=P/4

and substitute into the formula for A to obtain
A = (P / 4) 2
= P 2 / 16

For any square shape the area A may be expressed as a function the perimeter P as follows
A = P 2 / 16
Exercises:
1. Evaluate f(3) given that f(x) = | x - 6 | + x 2 - 1
2. Find f(x + h) - f(x) given that f(x) = a x + b
3. Find the domain of f(x) = SQRT(-x 2 - x + 2)
4. Find the range of g(x) = - SQRT(- x + 2) - 6
5. Find (f o g)(x) given that f(x) = SQRT(x) and g(x) = x 2 - 2x + 1
6. How do you obtain the graph of - f(x - 2) + 5 from the graph of f(x)?
Answers to Above Exercises:
1. f(3) = 11
2. f(x + h) - f(x) = a h
3. [-2 , 1]
4. (- infinity , - 6]
5. (f o g)(x) = | x - 1 |
6. Shift the graph of f 2 units to the right then reflect it on the x axis, then shift it upward 5 units.