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

Introduction to Functions

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.

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

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

Function Notation

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.

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) = 3 + 2 = 5. For x = 0, f(0) = 3 – 2(0) = 3 – 0 = 3. For x = 2, f(2) = 3 – 2(2) = 3 – 4 = –1.

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.

Exercise Section 2.2 (page 195) Question 49

The Domain of a Function

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.

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.

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

Even and Odd Functions

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

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

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

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

Arithmetic Combinations of Functions

Arithmetic Combinations of Functions

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

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

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) = 32 + 4(3) = x2 + 4x = 21.

Composition of Functions

Composition of Functions Figure 2.63

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.

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.

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.

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

Inverse Functions

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

Inverse Functions Figure 2.66

One-to-One Functions

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

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

Finding Inverse Functions Algebraically

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.

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).

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