1. Functions and Their Graphs RAFIZAH KECHIL, UiTM PULAU PINANG
(Source: Cengage Learning)

2. Introduction to Functions

3. What You Should Learn
Determine whether relations between two variables are functions.
Use function notation and evaluate functions.
Find the domains and ranges of functions.

4. Introduction to Functions
Set A is the domain.
Inputs: 1, 2, 3, 4, 5, 6
Set B contains the range.
Outputs: 9, 10, 12, 13, 15
Figure 2.20

5. Introduction to Functions
Set A is the domain.
Inputs: 1, 2, 3, 4, 5, 6
Set B contains the range.
Outputs: 9, 10, 12, 13, 15

6. Function Notation

7. Function Notation Input Output Equation x f (x) f (x) = 1 – x2
x f (x) f (x) = x2 – 4x + 7
t f (t) f (t) = t2 – 4t + 7
s g (s) g (s) = s2 – 4s + 7
f is the name of the function, f (x) is the value of the function at x.

8. Function Notation The function given by f (x) = 3 – 2x
has function values denoted by f (–1), f (0), f (2), and so on.
For x = –1, f(–1) = 3 – 2(–1) = = 5.
For x = 0, f(0) = 3 – 2(0) = 3 – 0 = 3.
For x = 2, f(2) = 3 – 2(2) = 3 – 4 = –1.

9. Example 4 – A Piecewise-Defined Function
Evaluate the function when x = –1, 0, and 1.
Solution:
Because x = –1 is less than 0, use f (x) = x2 + 1 to obtain
f(–1) = (–1)2 + 1
= 2.
For x = 0, use f (x) = x – 1 to obtain
f(0) = (0) – 1
= –1.

10. Exercise
Section 2.2 (page 195) Question 49

11. The Domain of a Function

12. The Domain of a Function
The domain is the set of all real numbers for which the expression is defined.
has a domain that consists of all real x other than x = ±2.
These two values are excluded from the domain because division by zero is undefined.
Domain excludes x-values that result in division by zero.

13. The Domain of a Function
The domain of a function excludes values that would cause division by zero or that would result in the even root of a negative number:
is defined only for x  0. So, its implied domain is the interval
[0, ).
Domain excludes x-values that result
in even roots of negative numbers.

14. EXERCISES
Section 2.3 (page 207) Question : 10,14

15. Even and Odd Functions

16. Example 8 – Even and Odd Functions
a. The function g(x) = x3 – x is odd because g(–x) = –g(x), as follows.
g(–x) = (–x)3 – (–x)
= –x3 + x
= –(x3 – x)
= – g(x)
Substitute –x for x.
Simplify.
Distributive Property
Test for odd function

17. Example 8 – Even and Odd Functions
cont’d
b. The function h(x) = x2 + 1 is even because h(–x) = h(x),
as follows.
h(–x) = (–x)2 + 1
= x2 + 1
= h(x)
Substitute –x for x.
Simplify.
Test for even function

18. Example 8 – Even and Odd Functions
cont’d
The graphs and symmetry of these two functions are shown in Figure 2.37.
(a) Symmetric to origin: Odd Function
(b) Symmetric to y-axis: Even Function
Figure 2.37

19. What You Should Learn Add, subtract, multiply, and divide functions.
Find the composition of one function with another function.

20. Arithmetic Combinations of Functions

21. Arithmetic Combinations of Functions

22. Arithmetic Combinations of Functions
f (x) + g(x) = (2x – 3) + (x2 – 1)
= x2 + 2x – 4
Sum

23. Arithmetic Combinations of Functions
f (x) – g(x) = (2x – 3) – (x2 – 1)
= –x2 + 2x – 2
f (x)g(x) = (2x – 3)(x2 – 1)
= 2x3 – 3x2 – 2x + 3
Difference
Product
Quotient

24. Example 1 – Finding the Sum of Two Functions
Given f (x) = 2x + 1 and g(x) = x2 + 2x – 1 find (f + g)(x). Then evaluate the sum when x = 3.
Solution:
(f + g)(x) = f (x) + g(x) = (2x + 1) + (x2 + 2x – 1)
When x = 3, the value of this sum is
(f + g)(3) = (3)
= x2 + 4x
= 21.

25. Composition of Functions

26. Composition of Functions
Figure 2.63

27. Example 5 – Composition of Functions
Given f (x) = x + 2 and g(x) = 4 – x2, find the following.
a. (f  g)(x) b. (g  f )(x) c. (g  f )(–2)
Solution:
a. The composition of f with g is as follows.
(f  g)(x) = f(g(x))
= f(4 – x2)
= (4 – x2) + 2
= –x2 + 6
Definition of f  g
Definition of g (x)
Definition of f (x)
Simplify.

28. Example 5 – Solution b. The composition of g with f is as follows.
cont’d
b. The composition of g with f is as follows.
(g  f )(x) = g(f (x))
= g(x + 2)
= 4 – (x + 2)2
= 4 – (x2 + 4x + 4)
= –x2 – 4x
Note that, in this case, (f  g)(x)  (g  f)(x).
Definition of g  f
Definition of f(x)
Definition of g(x)
Expand.
Simplify.

29. Example 5 – Solution
cont’d
c. Using the result of part (b), you can write the following.
(g  f)(–2) = –(–2)2 – 4(–2)
= – 4 + 8
= 4
Substitute.
Simplify.
Simplify.

30. EXERCISES
Section 2.6 (Page 235) Question 6,10,14,18,38,46

31. Inverse Functions

32. What You Should Learn
Verify that two functions are inverse functions of each other.
Find inverse functions algebraically.

33. Inverse Functions
Figure 2.66

34. One-to-One Functions

35. Example 5(a) – Applying the Horizontal Line Test
The graph of the function given by f (x) = x3 – 1 is shown in Figure 2.70.
Because no horizontal line intersects the graph of f at more than one point, you can conclude that f is a one-to-one function and does have an inverse function.
Figure 2.70

36. Example 5(b) – Applying the Horizontal Line Test
cont’d
The graph of the function given by f (x) = x2 – 1 is shown in Figure 2.71.
Because it is possible to find a horizontal line that intersects the graph of f at more than one point, you can conclude that f is not a one-to-one function and does not have an inverse function.
Figure 2.71

37. Finding Inverse Functions Algebraically

38. Example 6 – Finding an Inverse Function Algebraically
Find the inverse function of .
Solution:
Write original function.
Replace f (x) by y.
Interchange x and y.
Multiply each side by 2.

39. Example 6 – Solution
cont’d
Note that both f and f –1 have domains and ranges that consist of the entire set of real numbers. Check that
f (f –1 (x)) = x and f –1 (f (x)) = x.
Isolate the y-term.
Solve for y.
Replace y by f –1 (x).

40. EXERCISES
Section 2.7 (Page 245) Question 8,20,26,54,72