1920 Solutions - 1 PRECALCULUS REVIEW 1.1 Real Numbers Functions and Graphs Preliminary Questions 1 Give an example of numbers a and b such that a <

# 1920 Solutions - 1 PRECALCULUS REVIEW 1.1 Real Numbers...

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This preview shows page 1 out of 2543 pages. Unformatted text preview: 1 PRECALCULUS REVIEW 1.1 Real Numbers, Functions, and Graphs Preliminary Questions 1. Give an example of numbers a and b such that a < b and |a| > |b|. solution Take a = −3 and b = 1. Then a < b but |a| = 3 > 1 = |b|. 2. Which numbers satisfy |a| = a? Which satisfy |a| = −a? What about |−a| = a? solution The numbers a ≥ 0 satisfy |a| = a and | − a| = a. The numbers a ≤ 0 satisfy |a| = −a. 3. Give an example of numbers a and b such that |a + b| < |a| + |b|. solution Take a = −3 and b = 1. Then |a + b| = | − 3 + 1| = | − 2| = 2, but |a| + |b| = | − 3| + |1| = 3 + 1 = 4. Thus, |a + b| < |a| + |b|. 4. What are the coordinates of the point lying at the intersection of the lines x = 9 and y = −4? solution The point (9, −4) lies at the intersection of the lines x = 9 and y = −4. 5. In which quadrant do the following points lie? (a) (1, 4) (b) (−3, 2) (c) (4, −3) (d) (−4, −1) solution (a) Because both the x- and y-coordinates of the point (1, 4) are positive, the point (1, 4) lies in the first quadrant. (b) Because the x-coordinate of the point (−3, 2) is negative but the y-coordinate is positive, the point (−3, 2) lies in the second quadrant. (c) Because the x-coordinate of the point (4, −3) is positive but the y-coordinate is negative, the point (4, −3) lies in the fourth quadrant. (d) Because both the x- and y-coordinates of the point (−4, −1) are negative, the point (−4, −1) lies in the third quadrant. 6. What is the radius of the circle with equation (x − 9)2 + (y − 9)2 = 9? solution The circle with equation (x − 9)2 + (y − 9)2 = 9 has radius 3. 7. The equation f (x) = 5 has a solution if (choose one): (a) 5 belongs to the domain of f . (b) 5 belongs to the range of f . solution The correct response is (b): the equation f (x) = 5 has a solution if 5 belongs to the range of f . 8. What kind of symmetry does the graph have if f (−x) = −f (x)? solution If f (−x) = −f (x), then the graph of f is symmetric with respect to the origin. Exercises 1. Use a calculator to find a rational number r such that |r − π 2 | < 10−4 . solution r must satisfy π 2 − 10−4 < r < π 2 + 10−4 , or 9.869504 < r < 9.869705. r = 9.8696 = 12337 1250 would be one such number. 1 March 30, 2011 2 CHAPTER 1 PRECALCULUS REVIEW 2. Which of (a)–(f) are true for a = −3 and b = 2? (a) a < b (b) |a| < |b| (c) ab > 0 1 1 < (f) a b (e) −4a < −4b (d) 3a < 3b solution (a) True. (c) False, (−3)(2) = −6 < 0. (e) False, (−4)(−3) = 12 > −8 = (−4)(2). (b) False, |a| = 3 > 2 = |b|. (d) True. (f) True. In Exercises 3–8, express the interval in terms of an inequality involving absolute value. 3. [−2, 2] solution |x| ≤ 2 4. (−4, 4) solution |x| < 4 5. (0, 4) solution The midpoint of the interval is c = (0 + 4)/2 = 2, and the radius is r = (4 − 0)/2 = 2; therefore, (0, 4) can be expressed as |x − 2| < 2. 6. [−4, 0] solution The midpoint of the interval is c = (−4 + 0)/2 = −2, and the radius is r = (0 − (−4))/2 = 2; therefore, the interval [−4, 0] can be expressed as |x + 2| ≤ 2. 7. [1, 5] solution The midpoint of the interval is c = (1 + 5)/2 = 3, and the radius is r = (5 − 1)/2 = 2; therefore, the interval [1, 5] can be expressed as |x − 3| ≤ 2. 8. (−2, 8) solution The midpoint of the interval is c = (8 − 2)/2 = 3, and the radius is r = (8 − (−2))/2 = 5; therefore, the interval (−2, 8) can be expressed as |x − 3| < 5 In Exercises 9–12, write the inequality in the form a < x < b. 9. |x| < 8 solution −8 < x < 8 10. |x − 12| < 8 solution −8 < x − 12 < 8 so 4 < x < 20 11. |2x + 1| < 5 solution −5 < 2x + 1 < 5 so −6 < 2x < 4 and −3 < x < 2 12. |3x − 4| < 2 solution −2 < 3x − 4 < 2 so 2 < 3x < 6 and 32 < x < 2 In Exercises 13–18, express the set of numbers x satisfying the given condition as an interval. 13. |x| < 4 solution (−4, 4) 14. |x| ≤ 9 solution [−9, 9] 15. |x − 4| < 2 solution The expression |x − 4| < 2 is equivalent to −2 < x − 4 < 2. Therefore, 2 < x < 6, which represents the interval (2, 6). 16. |x + 7| < 2 solution The expression |x + 7| < 2 is equivalent to −2 < x + 7 < 2. Therefore, −9 < x < −5, which represents the interval (−9, −5). March 30, 2011 S E C T I O N 1.1 Real Numbers, Functions, and Graphs 3 17. |4x − 1| ≤ 8 solution The expression |4x − 1| ≤ 8 is equivalent to −8 ≤ 4x − 1 ≤ 8 or −7 ≤ 4x ≤ 9. Therefore, − 74 ≤ x ≤ 94 , which represents the interval [− 74 , 94 ]. 18. |3x + 5| < 1 solution The expression |3x + 5| < 1 is equivalent to −1 < 3x + 5 < 1 or −6 < 3x < −4. Therefore, −2 < x < − 43 which represents the interval (−2, − 43 ) In Exercises 19–22, describe the set as a union of finite or infinite intervals. 19. {x : |x − 4| > 2} solution x − 4 > 2 or x − 4 < −2 ⇒ x > 6 or x < 2 ⇒ (−∞, 2) ∪ (6, ∞) 20. {x : |2x + 4| > 3} solution 2x + 4 > 3 or 2x + 4 < −3 ⇒ 2x > −1 or 2x < −7 ⇒ (−∞, − 72 ) ∪ (− 12 , ∞) 21. {x : |x 2 − 1| > 2} √ √ solution x 2 − 1 > 2 or x 2 − 1 < −2 ⇒ x 2 > 3 or x 2 < −1 (this will never happen) ⇒ x > 3 or x < − 3 ⇒ √ √ (−∞, − 3) ∪ ( 3, ∞). 22. {x : |x 2 + 2x| > 2} solution x 2 + 2x > 2 or x 2 + 2x < −2 ⇒ x 2 + 2x − 2 > 0 or x 2 + 2x + 2 < 0. For the first case, the zeroes are x = −1 ± √ 3 ⇒ (−∞, −1 − √ √ 3) ∪ (−1 + 3, ∞). For the second case, note there are no real zeros. Because the parabola opens upward and its vertex √ is located above √ the x-axis, there are no values of x for which x 2 + 2x + 2 < 0. Hence, the solution set is (−∞, −1 − 3) ∪ (−1 + 3, ∞). 23. Match (a)–(f) with (i)–(vi). (a) a > 3    1   (c) a −  < 5 3 (b) |a − 5| < (e) |a − 4| < 3 (f) 1 ≤ a ≤ 5 1 3 (d) |a| > 5 (i) a lies to the right of 3. (ii) a lies between 1 and 7. (iii) The distance from a to 5 is less than 13 . (iv) The distance from a to 3 is at most 2. (v) a is less than 5 units from 13 . (vi) a lies either to the left of −5 or to the right of 5. solution (a) On the number line, numbers greater than 3 appear to the right; hence, a > 3 is equivalent to the numbers to the right of 3: (i). (b) |a − 5| measures the distance from a to 5; hence, |a − 5| < 13 is satisfied by those numbers less than 13 of a unit from 5: (iii). (c) |a − 13 | measures the distance from a to 31 ; hence, |a − 13 | < 5 is satisfied by those numbers less than 5 units from 1 : (v). 3 (d) The inequality |a| > 5 is equivalent to a > 5 or a < −5; that is, either a lies to the right of 5 or to the left of −5: (vi). (e) The interval described by the inequality |a − 4| < 3 has a center at 4 and a radius of 3; that is, the interval consists of those numbers between 1 and 7: (ii). (f) The interval described by the inequality 1 < x < 5 has a center at 3 and a radius of 2; that is, the interval consists of those numbers less than 2 units from 3: (iv).   x 24. Describe x : < 0 as an interval. x+1 solution Case 1: x < 0 and x + 1 > 0. This implies that x < 0 and x > −1 ⇒ −1 < x < 0. Case 2: x > 0 and x < −1 for which there is no such x. Thus, solution set is therefore (−1, 0). March 30, 2011 4 CHAPTER 1 PRECALCULUS REVIEW 25. Describe {x : x 2 + 2x < 3} as an interval. Hint: Plot y = x 2 + 2x − 3. solution The inequality x 2 + 2x < 3 is equivalent to x 2 + 2x − 3 < 0. In the figure below, we see that the graph of y = x 2 + 2x − 3 falls below the x-axis for −3 < x < 1. Thus, the set {x : x 2 + 2x < 3} corresponds to the interval −3 < x < 1. y y = x2 + 2x − 3 −4 −3 −2 10 8 6 4 2 x −2 1 2 26. Describe the set of real numbers satisfying |x − 3| = |x − 2| + 1 as a half-infinite interval. solution We will break the problem into three cases: x ≥ 3, 2 ≤ x < 3 and x < 2. For x ≥ 3, both x − 3 and x − 2 are greater than or equal to 0, so |x − 3| = x − 3 and |x − 2| = x − 2. The equation |x − 3| = |x − 2| + 1 then becomes x − 3 = x − 2 + 1, which is equivalent to −1 = 1. Thus, for x ≥ 3, there are no solutions. Next, we consider 2 ≤ x < 3. Now, x − 3 < 0, so |x − 3| = 3 − x, but x − 2 ≥ 0, so |x − 2| = x − 2. The equation |x − 3| = |x − 2| + 1 then becomes 3 − x = x − 2 + 1, which is equivalent to x = 2. Thus, x = 2 is a solution. Finally, consider x < 2. Both x − 3 and x − 2 are negative, so |x − 3| = 3 − x and |x − 2| = 2 − x. The equation |x − 3| = |x − 2| + 1 then becomes 3 − x = 2 − x + 1, which is equivalent to 1 = 1. Hence, every x < 2 is a solution. Bringing all three cases together, it follows that |x − 3| = |x − 2| + 1 is satisfied for all x ≤ 2, or for all x on the half-infinite interval (−∞, 2]. 27. Show that if a > b, then b−1 > a −1 , provided that a and b have the same sign. What happens if a > 0 and b < 0? solution Case 1a: If a and b are both positive, then a > b ⇒ 1 > ab ⇒ b1 > a1 . Case 1b: If a and b are both negative, then a > b ⇒ 1 < ab (since a is negative) ⇒ b1 > a1 (again, since b is negative). Case 2: If a > 0 and b < 0, then a1 > 0 and b1 < 0 so b1 < a1 . (See Exercise 2f for an example of this). 28. Which x satisfy both |x − 3| < 2 and |x − 5| < 1? solution |x − 3| < 2 ⇒ −2 < x − 3 < 2 ⇒ 1 < x < 5. Also |x − 5| < 1 ⇒ 4 < x < 6. Since we want an x that satisfies both of these, we need the intersection of the two solution sets, that is, 4 < x < 5. 29. Show that if |a − 5| < 12 and |b − 8| < 12 , then |(a + b) − 13| < 1. Hint: Use the triangle inequality. solution |a + b − 13| = |(a − 5) + (b − 8)| ≤ |a − 5| + |b − 8| (by the triangle inequality) 1 1 < + = 1. 2 2 30. Suppose that |x − 4| ≤ 1. (a) What is the maximum possible value of |x + 4|? (b) Show that |x 2 − 16| ≤ 9. solution (a) |x − 4| ≤ 1 guarantees 3 ≤ x ≤ 5. Thus, 7 ≤ x + 4 ≤ 9, so |x + 4| ≤ 9. (b) |x 2 − 16| = |x − 4| · |x + 4| ≤ 1 · 9 = 9. 31. Suppose that |a − 6| ≤ 2 and |b| ≤ 3. (a) What is the largest possible value of |a + b|? (b) What is the smallest possible value of |a + b|? solution |a − 6| ≤ 2 guarantees that 4 ≤ a ≤ 8, while |b| ≤ 3 guarantees that −3 ≤ b ≤ 3. Therefore 1 ≤ a + b ≤ 11. It follows that (a) the largest possible value of |a + b| is 11; and (b) the smallest possible value of |a + b| is 1. 32. Prove that |x| − |y| ≤ |x − y|. Hint: Apply the triangle inequality to y and x − y. solution First note |x| = |x − y + y| ≤ |x − y| + |y| by the triangle inequality. Subtracting |y| from both sides of this inequality yields |x| − |y| ≤ |x − y|. March 30, 2011 S E C T I O N 1.1 Real Numbers, Functions, and Graphs 5 33. Express r1 = 0.27 as a fraction. Hint: 100r1 − r1 is an integer. Then express r2 = 0.2666 . . . as a fraction. solution Let r1 = .27. We observe that 100r1 = 27.27. Therefore, 100r1 − r1 = 27.27 − .27 = 27 and r1 = 3 27 = . 99 11 Now, let r2 = 0.2666. Then 10r2 = 2.666 and 100r2 = 26.666. Therefore, 100r2 − 10r2 = 26.666 − 2.666 = 24 and r2 = 24 4 = . 90 15 34. Represent 1/7 and 4/27 as repeating decimals. solution 4 1 = 0.142857; = 0.148 7 27 35. The text states: If the decimal expansions of numbers a and b agree to k places, then |a − b| ≤ 10−k . Show that the converse is false: For all k there are numbers a and b whose decimal expansions do not agree at all but |a − b| ≤ 10−k . solution Let a = 1 and b = 0.9 (see the discussion before Example 1). The decimal expansions of a and b do not agree, but |1 − 0.9| < 10−k for all k. 36. Plot each pair of points and compute the distance between them: (a) (1, 4) and (3, 2) (b) (2, 1) and (2, 4) (c) (0, 0) and (−2, 3) (d) (−3, −3) and (−2, 3) solution (a) The points (1, 4) and (3, 2) are plotted in the figure below. The distance between the points is d=   √ √ (3 − 1)2 + (2 − 4)2 = 22 + (−2)2 = 8 = 2 2. y 4 3 2 1 x 1 2 3 (b) The points (2, 1) and (2, 4) are plotted in the figure below. The distance between the points is d=  √ (2 − 2)2 + (4 − 1)2 = 9 = 3. y 4 3 2 1 x 1 2 3 (c) The points (0, 0) and (−2, 3) are plotted in the figure below. The distance between the points is d=  √ √ (−2 − 0)2 + (3 − 0)2 = 4 + 9 = 13. y 4 3 2 1 −2 March 30, 2011 −1 x 6 CHAPTER 1 PRECALCULUS REVIEW (d) The points (−3, −3) and (−2, 3) are plotted in the figure below. The distance between the points is  √ √ d = (−3 − (−2))2 + (−3 − 3)2 = 1 + 36 = 37. y 3 2 1 −3 −2 −1 x −1 −2 −3 37. Find the equation of the circle with center (2, 4): (a) with radius r = 3. (b) that passes through (1, −1). solution (a) The equation of the indicated circle is (x − 2)2 + (y − 4)2 = 32 = 9. (b) First determine the radius as the distance from the center to the indicated point on the circle:  √ r = (2 − 1)2 + (4 − (−1))2 = 26. Thus, the equation of the circle is (x − 2)2 + (y − 4)2 = 26. 38. Find all points with integer coordinates located at a distance 5 from the origin. Then find all points with integer coordinates located at a distance 5 from (2, 3). solution • To be located a distance 5 from the origin, the points must lie on the circle x 2 + y 2 = 25. This leads to 12 points with integer coordinates: (5, 0) (3, 4) (4, 3) (−5, 0) (−3, 4) (−4, 3) (0, 5) (3, −4) (4, −3) (0, −5) (−3, −4) (−4, −3) • To be located a distance 5 from the point (2, 3), the points must lie on the circle (x − 2)2 + (y − 3)2 = 25, which implies that we must shift the points listed above two units to the right and three units up. This gives the 12 points: (7, 3) (5, 7) (6, 6) (−3, 3) (−1, 7) (−2, 6) (2, 8) (5, −1) (6, 0) (2, −2) (−1, −1) (−2, 0) 39. Determine the domain and range of the function f : {r, s, t, u} → {A, B, C, D, E} defined by f (r) = A, f (s) = B, f (t) = B, f (u) = E. solution The domain is the set D = {r, s, t, u}; the range is the set R = {A, B, E}. 40. Give an example of a function whose domain D has three elements and whose range R has two elements. Does a function exist whose domain D has two elements and whose range R has three elements? solution Define f by f : {a, b, c} → {1, 2} where f (a) = 1, f (b) = 1, f (c) = 2. There is no function whose domain has two elements and range has three elements. If that happened, one of the domain elements would get assigned to more than one element of the range, which would contradict the definition of a function. In Exercises 41–48, find the domain and range of the function. 41. f (x) = −x solution D : all reals; R : all reals 42. g(t) = t 4 solution D : all reals; R : {y : y ≥ 0} March 30, 2011 S E C T I O N 1.1 Real Numbers, Functions, and Graphs 7 43. f (x) = x 3 solution D : all reals; R : all reals √ 44. g(t) = 2 − t solution D : {t : t ≤ 2}; R : {y : y ≥ 0} 45. f (x) = |x| solution 46. h(s) = solution D : all reals; R : {y : y ≥ 0} 1 s D : {s : s = 0}; R : {y : y = 0} 1 47. f (x) = 2 x solution D : {x : x = 0}; R : {y : y > 0} 48. g(t) = cos solution 1 t D : {t : t = 0}; R : {y : −1 ≤ y ≤ 1} In Exercises 49–52, determine where f (x) is increasing. 49. f (x) = |x + 1| solution A graph of the function y = |x + 1| is shown below. From the graph, we see that the function is increasing on the interval (−1, ∞). y 2 1 −3 −2 x −1 1 50. f (x) = x 3 solution A graph of the function y = x 3 is shown below. From the graph, we see that the function is increasing for all real numbers. y 5 −2 −1 x 1 2 −5 51. f (x) = x 4 solution A graph of the function y = x 4 is shown below. From the graph, we see that the function is increasing on the interval (0, ∞). y 12 8 4 −2 March 30, 2011 −1 x 1 2 8 CHAPTER 1 PRECALCULUS REVIEW 1 52. f (x) = 4 x + x2 + 1 solution A graph of the function y = increasing on the interval (−∞, 0). 1 is shown below. From the graph, we see that the function is x4 + x2 + 1 y 1 −3 −2 x −1 1 2 3 In Exercises 53–58, find the zeros of f (x) and sketch its graph by plotting points. Use symmetry and increase/decrease information where appropriate. 53. f (x) = x 2 − 4 solution Zeros: ±2 Increasing: x > 0 Decreasing: x < 0 Symmetry: f (−x) = f (x) (even function). So, y-axis symmetry. y 4 2 −2 −1 x 1 −2 2 −4 54. f (x) = 2x 2 − 4 √ solution Zeros: ± 2 Increasing: x > 0 Decreasing: x < 0 Symmetry: f (−x) = f (x) (even function). So, y-axis symmetry. y 10 5 −2 −1 1 2 x 55. f (x) = x 3 − 4x solution Zeros: 0, ±2; Symmetry: f (−x) = −f (x) (odd function). So origin symmetry. y 10 5 x −2 −1 −5 1 2 −10 56. f (x) = x 3 solution Zeros: 0; Increasing for all x; Symmetry: f (−x) = −f (x) (odd function). So origin symmetry. y 20 10 −3 −2 −1 −10 −20 March 30, 2011 1 2 3 x Real Numbers, Functions, and Graphs S E C T I O N 1.1 9 57. f (x) = 2 − x 3 solution This is an x-axis reflection of x 3 translated up 2 units. There is one zero at x = √ 3 2. y 20 10 x −2 −1−10 −20 58. f (x) = 1 2 1 (x − 1)2 + 1 1 translated to the right 1 unit. The function has no zeros. solution This is the graph of 2 x +1 y 1 0.8 0.6 0.4 0.2 −4 x −2 2 4 59. Which of the curves in Figure 26 is the graph of a function? y y x x (A) (B) y y x x (C) (D) FIGURE 26 solution (B) is the graph of a function. (A), (C), and (D) all fail the vertical line test. 60. Determine whether the function is even, odd, or neither. (a) f (x) = x 5 (b) g(t) = t 3 − t 2 1 (c) F (t) = 4 t + t2 solution (a) f (−x) = (−x)5 = −x 5 = −f (x), so this function is odd. (b) g(−t) = (−t)3 − (−t)2 = −t 3 − t 2 which is equal to neither g(t) nor −g(t), so this function is neither odd nor even. (c) This function is even because F (−t) = March 30, 2011 1 1 = 4 = F (t). (−t)4 + (−t)2 t + t2 10 CHAPTER 1 PRECALCULUS REVIEW 61. Determine whether the function is even, odd, or neither. 1 1 (a) f (t) = 4 − (b) g(t) = 2t − 2−t t + t + 1 t4 − t + 1 (c) G(θ) = sin θ + cos θ (d) H (θ) = sin(θ 2 ) solution (a) This function is odd because f (−t) = 1 (−t)4 + (−t) + 1 − 1 (−t)4 − (−t) + 1 1 1 − = −f (t). = 4 t − t + 1 t4 + t + 1 (b) g(−t) = 2−t − 2−(−t) = 2−t − 2t = −g(t), so this function is odd. (c) G(−θ ) = sin(−θ ) + cos(−θ) = − sin θ + cos θ which is equal to neither G(θ) nor −G(θ ), so this function is neither odd nor even. (d) H (−θ ) = sin((−θ )2 ) = sin(θ 2 ) = H (θ), so this function is even. 62. Write f (x) = 2x 4 − 5x 3 + 12x 2 − 3x + 4 as the sum of an even and an odd function. solution Let g(x) = 2x 4 + 12x 2 + 4 and h(x) = −5x 3 − 3x, so that f (x) = g(x) + h(x). Observe g(−x) = 2(−x)4 + 12(−x)2 + 4 = 2x 4 + 12x 2 + 4 = g(x), while h(−x) = −5(−x)3 − 3(−x) = 5x 3 + 3x = −h(x). Thus, g(x) is an even function, and h(x) is an odd function.   1−x is an odd function. 63. Show that f (x) = ln 1+x solution   1 − (−x) 1 + (−x)     1−x 1+x = − ln = −f (x), = ln 1−x 1+x f (−x) = ln so this is an odd function. 64. State whether the function is increasing, decreasing, or neither. (a) Surface area of a sphere as a function of its radius (b) Temperature at a point on the equator as a function of time (c) Price of an airline ticket as a function of the price of oil (d) Pressure of the gas in a piston as a function of volume solution (a) Increasing (b) Neither (c) Increasing In Exercises 65–70, let f (x) be the function shown in Figure 27. y 4 3 2 1 x 0 1 2 3 4 FIGURE 27 65. Find the domain and range of f (x)? solution D : [0, 4]; R : [0, 4] March 30, 2011 (d) Decreasing Real Numbers, Functions, and Graphs S E C T I O N 1.1 11 66. Sketch the graphs of f (x + 2) and f (x) + 2. solution The graph of y = f (x + 2) is obtained by shifting the graph of y = f (x) two units to the left (see the graph below on the left). The graph of y = f (x) + 2 is obtained by shifting the graph of y = f (x) two units up (see the graph below on the right). y y 4 6 5 4 3 2 1 3 2 1 −2 x −1 1 x 2 1 f (x + 2) 2 3 4 f (x) + 2  67. Sketch the graphs of f (2x), f 12 x , and 2f (x). solution The graph of y = f (2x) is obtained by compressing the graph of y = f (x) horizontally by a factor of 2 (see the graph below on the left). The graph of y = f ( 12 x) is obtained by stretching the graph of y = f (x) horizontally by a factor of 2 (see the graph below in the middle). The graph of y = 2f (x) is obtained by stretching the graph of y = f (x) vertically by a factor of 2 (see the graph below on the right). y y y 4 4 8 3 3 6 2 2 4 1 1 2 x x 1 2 3 2 4 4 6 x 8 1 f (x/2) f(2x) 2 3 4 2f(x) 68. Sketch the graphs of f (−x) and −f (−x). solution The graph of y = f (−x) is obtained by reflecting the graph of y = f (x) across the y-axis (see the graph below on the left). The graph of y = −f (−x) is obtained by reflecting the graph of y = f (x) across both the x- and y-axes, or equivalently, about the origin (see the graph below on the right). y y −4 4 −3 −2 −3 −2 −1 x 3 −1 2 −2 −3...
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