2.2 A Pocketful of Functions: Some Basic Examples 61 10. Which of the following functions is continuous at x = 2? (a) f(x)=x +3 (b) f(x)= x+3 x−2 (c) f(x)= x 2 +x−6 x−2 11. Which of the following functions is continuous at x = 1? (a) f(x)=5 (b) f(x)= 5x x (c) f(x)= 5x(x−1) x−1 (d) f(x)= x 2 (x−1) x−1 2.2 A POCKETFUL OF FUNCTIONS: SOME BASIC EXAMPLES Throughout this text we will be looking at functions analytically and graphically. Consider any equation relating two variables. Every point that lies on the graph of the equation satisfies the equation. Conversely, every point whose coordinates satisfy the equation lies on the graph of the equation. The correspondence between pairs of x and y that constitute the coordinates of a point on a graph and that satisfy an equation was a key insight of Rene Descartes (1596–1650) that unified geometry and algebra. 8 As a consequence of this insight, the fields of geometry and algebra became irrevocably intertwined, opening the door to much of modern mathematics, including calculus. We have been discussing functions and their graphs in a very general way, but it is always strategically wise to include a few simple, concrete examples for reference. In this section you will become familiar with a small sampling of functions, functions that you can carry about and pull out of your back pocket at any moment. As we go along in the text this collection will grow; you will learn that these functions belong to larger families of functions sharing some common characteristics, and you’ll also be introduced to a greater variety of families of functions. But for the time being, we will become familiar with a few individual functions. A Few Basic Examples Consider the following function: f(x)=x g(x) = x 2 h(x) =|x| j(x)= 1 x The function f is the identity function; its output is identical to its input. The function g is the squaring function; its output is obtained by squaring its input. The function h is the absolute value function; its output is the magnitude (size) of its input. In other words, if the input is positive (or nonnegative), then the output is identical to the input; if the input is not positive, then change its sign to obtain the 8 We have already discussed this correspondence when looking at the graph of a function. In fact, it holds for the graph of any equation, whether defining a function or not. 62 CHAPTER 2 Characterizing Functions and Introducing Rates of Change output. We can define the absolute value function |x| analytically as follows. |x|= x if x ≥ 0; −x if x<0. The function j is the reciprocal function; the output is the reciprocal of the input. output = 1 input . The graphs of these functions are given below. Familiarize yourself with these functions by doing the exercises that follow. x g g(x) = x 2 f(x) = x h(x) = |x| j(x) = 1 1 x f 1 1 x h 1 1 x j 1 1 – 1 x Figure 2.11 EXERCISE 2.5 Look at each of the functions f(x)=x,g(x) = x 2 , h(x) =|x|,and j(x)= 1 x one by one and answer the following questions. (a) What are the domain and range of the function? (b) Where is the function positive? Negative? (c) Where is the function increasing? Decreasing? (d) Is the function continuous? If it is not continuous everywhere, where is it discontinuous? (e) Is the function 1-to-1? Answers are provided at the end of the section. EXERCISE 2.6 Consider the function j(x)= 1 x . (a) Why is j(x) never equal to zero? (b) Why is zero not in the domain of j ? (c) As x increases without bound what happens to the values of j?Asxdecreases (becomes more and more negative) without bound what happens to the values of j?Howis this information displayed on the graph of j ? How is it displayed on your graphing calculator’s rendition of the graph? (d) Find j(x) for x = 0.01, 0.001, and 0.0001. If x is a positive number, as x gets increasingly close to zero, what happens to the values of j? How is this information displayed on the graph? (e) Find j(x) for x =−0.01, −0.001, and −0.0001. If x is a negative number, as x gets increasingly close to zero, what happens to the values of j? How is this information 2.2 A Pocketful of Functions: Some Basic Examples 63 displayed on the graph? How is this displayed on your calculator’s rendition of the graph? (f) Fill in the blanks with >, <,or=as appropriate. i. If x>0then j(x) 0. ii. If x<0then j(x) 0. iii. If |x| > 1 then |j(x)| 1. iv.If0<|x|<1then |j(x)| 1. (g) Translate the statements from part (f) into plain prose. Your translation should be accessible to someone who knows nothing about functions and absolute values. EXERCISE 2.7 Consider the function g(x) = x 2 . (a) Fill in the blanks with >, <,or=as appropriate. i. If |x| 1, then g(x) > |x|. ii. If |x| 1, then g(x) < |x|. (b) Translate the statements from part (a) into plain prose. Your translation should be accessible to someone who knows nothing about functions and absolute values. (c) Consider h(x) =|x|. i. For what values of x is h(x) > g(x)? ii. For what values of x is h(x) < g(x)? REMARKS The graph of f(x)=x given in Figure 2.11 has the same scale on the x- and y-axes and therefore the line cuts the angle between the axes evenly in two. If different scales were used on the axes this would not be true. 9 For x ≥ 0 the functions f(x)=x and h(x) =|x|are identical. The graph of h(x) has a sharp corner at x = 0. No matter how much you magnify the portion of the graph around that point, the corner remains sharp. The graph of the squaring function g(x) = x 2 is called a parabola. Notice that g(−3) = (−3) 2 , not −3 2 . Therefore g(3) = g(−3) and, more generally, g(x) = g(−x). The graph of g is symmetric about the y-axis; the graph is a mirror image on either side of the y-axis. You could fold the graph along the y-axis and the function would exactly match itself on either side of the y-axis. The characteristic g(x) = g(−x) guarantees this symmetry. The graph of j(x)= 1 x has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. If y = 1 x , then y is inversely proportional to x. 9 If you use a graphing calculator to look at the graph of a function you can set the range and domain so that the scales are very different. You should be aware that your choice of domain and range can make the line f(x)=x look almost horizontal or almost vertical. 64 CHAPTER 2 Characterizing Functions and Introducing Rates of Change Definition A is inversely proportional to B if A = k B for some constant k. As B increases A decreases, and vice versa. This type of relationship arises frequently in the world around us. For example, the amount of time required to complete a trip is inversely proportional to the rate at which one travels. Informal Introduction to Asymptotes The function Q(x) has a vertical asymptote at x = a if Q is undefined at x = a and as x gets closer and closer to a the magnitude of Q(x), written |Q(x)|, increases without bound. Because Q is undefined at its vertical asymptotes, the graph of a function will never cross its vertical asymptote. Around a vertical asymptote, the graph will resemble one of the four options shown below. If Q has a one-sided vertical asymptote at x = a then as x approaches a from one side |Q(x)| increases without bound. Figure 2.12 For the function j(x) = 1 x ,asxapproaches 0 from the right, the value j(x) grows without bound. We can write this symbolically: as x → 0 + j(x)→∞. “As x → 0 + ” means “as x approaches zero from the right”; “j(x)→∞” means “j(x) grows without bound.” On the other hand, as x approaches 0 from the left, the value of j(x) gets increasingly negative without bound. We can write this symbolically: as x → 0 − j(x)→−∞. The function Q has a horizontal asymptote at y = c if, as the magnitude of x increases without bound, the value of Q(x) gets closer and closer to c. A horizontal asymptote tells us only about the behavior of the function for |x| very large; therefore, it can be crossed. A one- sided horizontal asymptote tells us what happens to Q as x either increases or decreases 2.2 A Pocketful of Functions: Some Basic Examples 65 without bound. It can be crossed. As |x| increases without bound, j(x)= 1 x approaches zero. We can express this symbolically by writing as x →∞ j(x)→0 and as x →−∞ j(x)→0. Even and Odd Symmetry A function may be even, odd, or neither. Definition A function f is even if f(−x) = f(x)for all x in the domain of f . This means that the height of the graph of f is the same at x and at −x. The graph of an even function is a mirror image about the y-axis. A function is odd if f(−x) =−f(x)for all x in the domain of f . An odd function is said to be symmetric about the origin. For further description see the answers to Exercise 2.8. x f a -a x f a -a x f x f (-a, f(-a)) (a, f(a)) even f(x) = f(-x) even odd odd f(x) = -f(-x) (-a, f(-a)) (a, f(a)) Figure 2.13 EXERCISE 2.8 In this exercise we return to the four functions we introduced at the beginning of this section: f(x)=x, g(x) = x 2 , h(x) =|x|, and j(x) = 1 x (a) Which of functions f , g, h, and j are even? (b) Which of functions f , g, h, and j are odd? (c) Describe in words the characteristic of the graph of an odd function. (The graph of an odd function is not a mirror about the x-axis.) Answers to Exercise 2.8 are provided at the end of the section. Working with Absolute Values To look at the absolute value of a quantity is to look only at its magnitude, to make it nonnegative. 66 CHAPTER 2 Characterizing Functions and Introducing Rates of Change Analytic Principle for Working with Absolute Values To deal with an absolute value, remove it by breaking the problem up into two cases. Case (1): If the expression inside the absolute value is nonnegative, then simply remove the absolute value signs. Case (2): If the expression inside the absolute value is negative, then change its sign and remove the absolute values. It follows that: |x − 3|= x−3 for x ≥ 3, −(x − 3) for x ≤ 3. Geometric Principle for Working with Absolute Values Equivalent to the preceding analytic definition is the “geometric” definition: The absolute value of x is the distance between x and 0 on the number line. More generally, this geometric definition tells us that |x − a| is the distance between x and a on the number line. It follows that |x − 3| is the distance between x and 3, while |x + 2| (which equals |x − (−2)|)isthe distance between x and −2. We can use this geometric interpretation to solve equations and inequalities involving absolute values. For example, we can interpret the inequality |x − 100|≤0.1 as “x differs from 100 by no more than 0.1,” or x ∈ [99.9, 100.1]. ◆ EXAMPLE 2.4 Solve: (a) |x − 1|=2 (b) |x + 1|≤2 (c) |2x + 3| > 1 SOLUTIONS (a) Geometric Approach: |x − 1|=2 (the distance between x and 1) = 2 –2 –101234 22 Figure 2.14 x =−1orx=3 2.2 A Pocketful of Functions: Some Basic Examples 67 Analytic Approach: Case (1): x − 1 ≥ 0 Case (2): x − 1 < 0 x − 1 = 2 −(x − 1) = 2 x = 3 − x + 1 = 2 x =−1 Check that when x = 3, x − 1 ≥ 0 and when x =−1, x − 1 < 0. Alternatively, check the answers in the original problem. (b) Geometric Approach: |x + 1|≤2 |x−(−1)|≤2 (the distance between x and −1) ≤ 2 – 4 –3 –2 –10 1 2 22 3 Figure 2.15 x is between −3 and 1, including both endpoints x ∈ [−3, 1] Analytic Approach: Solve the corresponding equation: |x + 1|=2. Case (1): x + 1 ≥ 0 Case (2): x + 1 < 0 x + 1 = 2 −(x + 1) = 2 x = 1 − x − 1 = 2 x =−3 These two solutions partition the number line into three intervals. To determine whether each of these intervals contains acceptable values of x, substitute one number from each of these intervals into the original equation. This technique gives us −3 ≤ x ≤ 1. (c) Geometric Approach: |2x + 3| > 1 |2x − (−3)| > 1 (the distance between 2x and −3) > 1 We’ll focus on 2x and solve for x later. 68 CHAPTER 2 Characterizing Functions and Introducing Rates of Change 11 –4–3–2–101–6–5 Figure 2.16 2x<−4or2x>−2 x<−2orx>−1 Analytic Approach: Solve the corresponding equation |2x + 3|=1. Case (1): 2x + 3 ≥ 0 Case (2): 2x + 3 < 0 2x + 3 = 1 −(2x + 3) = 1 2x =−22x+3=−1 x=−12x=−4 x=−2 These two solutions once again divide the number line into three intervals. Check to see if a number inside each of these intervals is a solution to determine whether all numbers within the interval are solutions. Here we discover x<−2orx>−1. ◆ EXERCISE 2.9 Solve for x. Do these problems in two ways, first using the analytic definition of the absolute value and then using the geometric definition of the absolute value. (a) |3 − x|=2 (b) |2x + 1|=4 ◆ Answers to Exercise 2.9 are provided at the end of the section. Answers to Selected Exercises Answers to Exercise 2.5 The linear function f(x)=x (a) The domain of f is all real numbers. Likewise, the range of f is all real numbers. (b) f is positive for x positive, negative for x negative. (c) The graph of f is a straight line. The graph is always increasing. (d) f is a continuous function. (e) f is 1-to-1. g(x) = x 2 (a) The domain of g is all real numbers. The range of g is all nonnegative real numbers. (b) g is positive for all x except x = 0. g(0) = 0. (c) g is decreasing on (−∞, 0] and increasing on [0, ∞). (d) g is a continuous function. 2.2 A Pocketful of Functions: Some Basic Examples 69 (e) g is not 1-to-1. By restricting the domain to nonnegative real numbers the function can be made 1-to-1. The absolute value function h(x) =|x| (a) The domain of h is all real numbers. The range of h is all nonnegative real numbers. (b) h is positive for all x except x = 0. h(0) = 0. (c) h is decreasing on (−∞, 0] and increasing on [0, ∞). (d) h is a continuous function. (e) h is not 1-to-1. By restricting the domain to nonnegative real numbers the function can be made 1-to-1. j(x)= 1 x (a) The domain of j is all nonzero real numbers. The range of j is all nonzero real numbers. (b) j is positive for x positive, negative for x negative. (c) j is decreasing on (−∞,0)and decreasing on (0, ∞). (d) j is undefined and discontinuous at x = 0. (e) j is 1-to-1. Answers to Exercise 2.8 (a) g and h are even functions. (b) f and j are odd functions. (c) The graph of an odd function is said to have symmetry about the origin. What does this mean? Graph the function for all nonnegative x in the domain. The other half of the graph can be obtained by reflecting this portion of the graph across the y-axis and then across the x-axis. (This is equivalent to rotating the selected portion of the graph 180 ◦ around the origin.) Answers to Exercise 2.9 (a) i. Analytic Approach: Remove the absolute value signs by breaking the problem into two cases—one in which the expression inside the absolute value is positive and the other in which the expression inside the absolute value sign is not positive. Case (1): The expression inside the absolute value is positive. Because 3 − x>0, we replace |3 − x| by 3 − x. 3 − x = 2 −x =−1 x=1 Case (2): The expression inside the absolute value is not positive. Because 3 − x ≤ 0, we replace |3 − x| by −(3 − x) =−3+x. −3+x=2 x=5 Check that these answers satisfy the original equation. ii. Geometric Approach: |3 − x|=2 (the distance between 3 and x) = 2 70 CHAPTER 2 Characterizing Functions and Introducing Rates of Change 01 23456 x = 1 or x = 5 Figure 2.17 (b) i. Analytic Approach: Remove the absolute value signs by breaking the problem down into two cases. Case (1): The expression inside the absolute value is positive. Because 2x + 1 > 0, we replace |2x + 1| by 2x + 1. 2x + 1 = 4 2x = 3 x = 3 2 Case (2): The expression inside the absolute value is not positive. Because 2x + 1 ≤ 0, we replace |2x + 1| by −(2x + 1) =−2x−1. −2x − 1 = 4 −2x = 5 x =− 5 2 Check that both of these answers satisfy the original equation. ii. Geometric Approach: |2x + 1|=4 |2x−(−1)|=4 (the distance between 2x and − 1) = 4 01 234–6 –5 – 4 –2 –1 –3 44 Figure 2.18 2x =−5or2x=3 x=− 5 2 or x = 3 2 PROBLEMS FOR SECTION 2.2 1. This problem applies the definitions of proportionality (also referred to as direct pro- portionality), and inverse proportionality. . (−∞,0 )and decreasing on (0, ∞). (d) j is undefined and discontinuous at x = 0. (e) j is 1 -to- 1. Answers to Exercise 2.8 (a) g and h are even functions. (b) f and j are odd functions. (c) The graph of. Characterizing Functions and Introducing Rates of Change Definition A is inversely proportional to B if A = k B for some constant k. As B increases A decreases, and vice versa. This type of relationship. return to the four functions we introduced at the beginning of this section: f(x)=x, g(x) = x 2 , h(x) =|x|, and j(x) = 1 x (a) Which of functions f , g, h, and j are even? (b) Which of functions