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

Solution to Question 2:

• Solve the above equation for y
  
  y ²= 1 - x
  
  y = + SQRT(1 - x) or y = - SQRT(1 - x)
  
  
• For one value of x we have two values of y and this is not a function.

Question 3: Function f is defined by

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

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

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

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

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

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

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

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

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

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
  
  = x ² - 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

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 = 4 x
  
  
• and an area A given by
  
  A = x ²
  
  
• 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.