1. 1 FUNCTIONS AND MODELS 1.1 Four Ways to Represent a Function In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here. 1. (a) The point (1, 2) is on the graph of f, so f(1) = 2. (b) When x = 2, y is about 2.8, so f(2) 2.8. (c) f(x) = 2 is equivalent to y = 2. When y = 2, we have x = 3 and x = 1. (d) Reasonable estimates for x when y = 0 are x = 2.5 and x = 0.3. (e) The domain of f consists of all x-values on the graph of f. For this function, the domain is 3 x 3, or [3, 3]. The range of f consists of all y-values on the graph of f. For this function, the range is 2 y 3, or [2, 3]. (f ) As x increases from 1 to 3, y increases from 2 to 3. Thus, f is increasing on the interval [1, 3]. 3. From Figure 1 in the text, the lowest point occurs at about (t, a) = (12, 85). The highest point occurs at about (17, 115). Thus, the range of the vertical ground acceleration is 85 a 115. Written in interval notation, we get [85, 115]. 5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails the Vertical Line Test. 7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [3, 2] and the range is [3, 2) [1, 3]. 9. The persons weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The persons weight dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems. 11. The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the water will slowly warm up to room temperature. 13. Of course, this graph depends strongly on the geographical location! 15. As the price increases, the amount sold decreases. 17. 9 2. 10 CHAPTER 1 FUNCTIONS AND MODELS 19. (a) (b) From the graph, we estimate the number of cell-phone subscribers worldwide to be about 92 million in 1995 and 485 million in 1999. 21. f(x) = 3x2 x + 2. f(2) = 3(2)2 2 + 2 = 12 2 + 2 = 12. f(2) = 3(2)2 (2) + 2 = 12 + 2 + 2 = 16. f(a) = 3a2 a + 2. f(a) = 3(a)2 (a) + 2 = 3a2 + a + 2. f(a + 1) = 3(a + 1)2 (a + 1) + 2 = 3(a2 + 2a + 1) a 1 + 2 = 3a2 + 6a + 3 a + 1 = 3a2 + 5a + 4. 2f(a) = 2 f(a) = 2(3a2 a + 2) = 6a2 2a + 4. f(2a) = 3(2a)2 (2a) + 2 = 3(4a2 ) 2a + 2 = 12a2 2a + 2. f(a2 ) = 3(a2 )2 (a2 ) + 2 = 3(a4 ) a2 + 2 = 3a4 a2 + 2. [f(a)]2 = 3a2 a + 2 2 = 3a2 a + 2 3a2 a + 2 = 9a4 3a3 + 6a2 3a3 + a2 2a + 6a2 2a + 4 = 9a4 6a3 + 13a2 4a + 4. f(a + h) = 3(a + h)2 (a + h) + 2 = 3(a2 + 2ah + h2 ) a h + 2 = 3a2 + 6ah + 3h2 a h + 2. 23. f(x) = 4 + 3x x2 , so f(3 + h) = 4 + 3(3 + h) (3 + h)2 = 4 + 9 + 3h (9 + 6h + h2 ) = 4 3h h2 , and f(3 + h) f(3) h = (4 3h h2 ) 4 h = h(3 h) h = 3 h. 25. f(x) f(a) x a = 1 x 1 a x a = a x xa x a = a x xa(x a) = 1(x a) xa(x a) = 1 ax 27. f(x) = x/(3x 1) is dened for all x except when 0 = 3x 1 x = 1 3 , so the domain is x R | x 6= 1 3 = , 1 3 1 3 , . 29. f(t) = t + 3 t is dened when t 0. These values of t give real number results for t, whereas any value of t gives a real number result for 3 t. The domain is [0, ). 31. h(x) = 1 4 x2 5x is dened when x2 5x > 0 x(x 5) > 0. Note that x2 5x 6= 0 since that would result in division by zero. The expression x(x 5) is positive if x < 0 or x > 5. (See Appendix A for methods for solving inequalities.) Thus, the domain is (, 0) (5, ). 3. SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION 11 33. f(x) = 5 is dened for all real numbers, so the domain is R, or (, ). The graph of f is a horizontal line with y-intercept 5. 35. f(t) = t2 6t is dened for all real numbers, so the domain is R, or (, ). The graph of f is a parabola opening upward since the coefcient of t2 is positive. To nd the t-intercepts, let y = 0 and solve for t. 0 = t2 6t = t(t 6) t = 0 and t = 6. The t-coordinate of the vertex is halfway between the t-intercepts, that is, at t = 3. Since f(3) = 32 6 3 = 9, the vertex is (3, 9). 37. g(x) = x 5 is dened when x 5 0 or x 5, so the domain is [5, ). Since y = x 5 y2 = x 5 x = y2 + 5, we see that g is the top half of a parabola. 39. G(x) = 3x + |x| x . Since |x| = x if x 0 x if x < 0 , we have G(x) = 3x + x x if x > 0 3x x x if x < 0 = 4x x if x > 0 2x x if x < 0 = 4 if x > 0 2 if x < 0 Note that G is not dened for x = 0. The domain is (, 0) (0, ). 41. f(x) = x + 2 if x < 0 1 x if x 0 The domain is R. 43. f(x) = x + 2 if x 1 x2 if x > 1 Note that for x = 1, both x + 2 and x2 are equal to 1. The domain is R. 45. Recall that the slope m of a line between the two points (x1, y1) and (x2, y2) is m = y2 y1 x2 x1 and an equation of the line connecting those two points is y y1 = m(x x1). The slope of this line segment is 7 (3) 5 1 = 5 2 , so an equation is y (3) = 5 2 (x 1). The function is f(x) = 5 2 x 11 2 , 1 x 5. 4. 12 CHAPTER 1 FUNCTIONS AND MODELS 47. We need to solve the given equation for y. x + (y 1)2 = 0 (y 1)2 = x y 1 = x y = 1 x. The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom half. Hence, we want f(x) = 1 x. Note that the domain is x 0. 49. For 0 x 3, the graph is the line with slope 1 and y-intercept 3, that is, y = x + 3. For 3 < x 5, the graph is the line with slope 2 passing through (3, 0); that is, y 0 = 2(x 3), or y = 2x 6. So the function is f(x) = x + 3 if 0 x 3 2x 6 if 3 < x 5 51. Let the length and width of the rectangle be L and W. Then the perimeter is 2L + 2W = 20 and the area is A = LW. Solving the rst equation for W in terms of L gives W = 20 2L 2 = 10 L. Thus, A(L) = L(10 L) = 10L L2 . Since lengths are positive, the domain of A is 0 < L < 10. If we further restrict L to be larger than W, then 5 < L < 10 would be the domain. 53. Let the length of a side of the equilateral triangle be x. Then by the Pythagorean Theorem, the height y of the triangle satises y2 + 1 2 x 2 = x2 , so that y2 = x2 1 4 x2 = 3 4 x2 and y = 3 2 x. Using the formula for the area A of a triangle, A = 1 2 (base)(height), we obtain A(x) = 1 2 (x) 3 2 x = 3 4 x2 , with domain x > 0. 55. Let each side of the base of the box have length x, and let the height of the box be h. Since the volume is 2, we know that 2 = hx2 , so that h = 2/x2 , and the surface area is S = x2 + 4xh. Thus, S(x) = x2 + 4x(2/x2 ) = x2 + (8/x), with domain x > 0. 57. The height of the box is x and the length and width are L = 20 2x, W = 12 2x. Then V = LWx and so V (x) = (20 2x)(12 2x)(x) = 4(10 x)(6 x)(x) = 4x(60 16x + x2 ) = 4x3 64x2 + 240x. The sides L, W, and x must be positive. Thus, L > 0 20 2x > 0 x < 10; W > 0 12 2x > 0 x < 6; and x > 0. Combining these restrictions gives us the domain 0 < x < 6. 59. (a) (b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400. On $26,000, tax is assessed on $16,000, and 10%($10,000) + 15%($6000) = $1000 + $900 = $1900. (c) As in part (b), there is $1000 tax assessed on $20,000 of income, so the graph of T is a line segment from (10,000, 0) to (20,000, 1000). The tax on $30,000 is $2500, so the graph of T for x > 20,000 is the ray with initial point (20,000, 1000) that passes through (30,000, 2500). 5. SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS 13 61. f is an odd function because its graph is symmetric about the origin. g is an even function because its graph is symmetric with respect to the y-axis. 63. (a) Because an even function is symmetric with respect to the y-axis, and the point (5, 3) is on the graph of this even function, the point (5, 3) must also be on its graph. (b) Because an odd function is symmetric with respect to the origin, and the point (5, 3) is on the graph of this odd function, the point (5, 3) must also be on its graph. 65. f(x) = x x2 + 1 . f(x) = x (x)2 + 1 = x x2 + 1 = x x2 + 1 = f(x). So f is an odd function. 67. f(x) = x x + 1 , so f(x) = x x + 1 = x x 1 . Since this is neither f(x) nor f(x), the function f is neither even nor odd. 69. f(x) = 1 + 3x2 x4 . f(x) = 1 + 3(x)2 (x)4 = 1 + 3x2 x4 = f(x). So f is an even function. 1.2 Mathematical Models: A Catalog of Essential Functions 1. (a) f(x) = 5 x is a root function with n = 5. (b) g(x) = 1 x2 is an algebraic function because it is a root of a polynomial. (c) h(x) = x9 + x4 is a polynomial of degree 9. (d) r(x) = x2 + 1 x3 + x is a rational function because it is a ratio of polynomials. (e) s(x) = tan 2x is a trigonometric function. (f ) t(x) = log10 x is a logarithmic function. 3. We notice from the gure that g and h are even functions (symmetric with respect to the y-axis) and that f is an odd function (symmetric with respect to the origin). So (b) y = x5 must be f. Since g is atter than h near the origin, we must have (c) y = x8 matched with g and (a) y = x2 matched with h. 6. 14 CHAPTER 1 FUNCTIONS AND MODELS 5. (a) An equation for the family of linear functions with slope 2 is y = f(x) = 2x + b, where b is the y-intercept. (b) f(2) = 1 means that the point (2, 1) is on the graph of f. We can use the point-slope form of a line to obtain an equation for the family of linear functions through the point (2, 1). y 1 = m(x 2), which is equivalent to y = mx + (1 2m) in slope-intercept form. (c) To belong to both families, an equation must have slope m = 2, so the equation in part (b), y = mx + (1 2m), becomes y = 2x 3. It is the only function that belongs to both families. 7. All members of the family of linear functions f(x) = c x have graphs that are lines with slope 1. The y-intercept is c. 9. Since f(1) = f(0) = f(2) = 0, f has zeros of 1, 0, and 2, so an equation for f is f(x) = a[x (1)](x 0)(x 2), or f(x) = ax(x + 1)(x 2). Because f(1) = 6, well substitute 1 for x and 6 for f(x). 6 = a(1)(2)(1) 2a = 6 a = 3, so an equation for f is f(x) = 3x(x + 1)(x 2). 11. (a) D = 200, so c = 0.0417D(a + 1) = 0.0417(200)(a + 1) = 8.34a + 8.34. The slope is 8.34, which represents the change in mg of the dosage for a child for each change of 1 year in age. (b) For a newborn, a = 0, so c = 8.34 mg. 13. (a) (b) The slope of 9 5 means that F increases 9 5 degrees for each increase of 1 C. (Equivalently, F increases by 9 when C increases by 5 and F decreases by 9 when C decreases by 5.) The F-intercept of 32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0. 7. SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS 15 15. (a) Using N in place of x and T in place of y, we nd the slope to be T2 T1 N2 N1 = 80 70 173 113 = 10 60 = 1 6 . So a linear equation is T 80 = 1 6 (N 173) T 80 = 1 6 N 173 6 T = 1 6 N + 307 6 307 6 = 51.16 . (b) The slope of 1 6 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1 F. (c) When N = 150, the temperature is given approximately by T = 1 6 (150) + 307 6 = 76.16 F 76 F. 17. (a) We are given change in pressure 10 feet change in depth = 4.34 10 = 0.434. Using P for pressure and d for depth with the point (d, P) = (0, 15), we have the slope-intercept form of the line, P = 0.434d + 15. (b) When P = 100, then 100 = 0.434d + 15 0.434d = 85 d = 85 0.434 195.85 feet. Thus, the pressure is 100 lb/in2 at a depth of approximately 196 feet. 19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form f(x) = a cos(bx) + c seems appropriate. (b) The data appear to be decreasing in a linear fashion. A model of the form f(x) = mx + b seems appropriate. Some values are given to many decimal places. These are the results given by several computer algebra systems rounding is left to the reader. 21. (a) A linear model does seem appropriate. (b) Using the points (4000, 14.1) and (60,000, 8.2), we obtain y 14.1 = 8.2 14.1 60,000 4000 (x 4000) or, equivalently, y 0.000105357x + 14.521429. (c) Using a computing device, we obtain the least squares regression line y = 0.0000997855x + 13.950764. The following commands and screens illustrate how to nd the least squares regression line on a TI-83 Plus. Enter the data into list one (L1) and list two (L2). Press to enter the editor. Find the regession line and store it in Y1. Press . 8. 16 CHAPTER 1 FUNCTIONS AND MODELS Note from the last gure that the regression line has been stored in Y1 and that Plot1 has been turned on (Plot1 is highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing or by pressing . Now press to produce a graph of the data and the regression line. Note that choice 9 of the ZOOM menu automatically selects a window that displays all of the data. (d) When x = 25,000, y 11.456; or about 11.5 per 100 population. (e) When x = 80,000, y 5.968; or about a 6% chance. (f ) When x = 200,000, y is negative, so the model does not apply. 23. (a) A linear model does seem appropriate. (b) Using a computing device, we obtain the least squares regression line y = 0.089119747x 158.2403249, where x is the year and y is the height in feet. (c) When x = 2000, the model gives y 20.00 ft. Note that the actual winning height for the 2000 Olympics is less than the winning height for 1996so much for that prediction. (d) When x = 2100, y 28.91 ft. This would be an increase of 9.49 ft from 1996 to 2100. Even though there was an increase of 8.59 ft from 1900 to 1996, it is unlikely that a similar increase will occur over the next 100 years. 25. Using a computing device, we obtain the cubic function y = ax3 + bx2 + cx + d with a = 0.0012937, b = 7.06142, c = 12,823, and d = 7,743,770. When x = 1925, y 1914 (million). 9. SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS 17 1.3 New Functions from Old Functions 1. (a) If the graph of f is shifted 3 units upward, its equation becomes y = f(x) + 3. (b) If the graph of f is shifted 3 units downward, its equation becomes y = f(x) 3. (c) If the graph of f is shifted 3 units to the right, its equation becomes y = f(x 3). (d) If the graph of f is shifted 3 units to the left, its equation becomes y = f(x + 3). (e) If the graph of f is reected about the x-axis, its equation becomes y = f(x). (f ) If the graph of f is reected about the y-axis, its equation becomes y = f(x). (g) If the graph of f is stretched vertically by a factor of 3, its equation becomes y = 3f(x). (h) If the graph of f is shrunk vertically by a factor of 3, its equation becomes y = 1 3 f(x). 3. (a) (graph 3) The graph of f is shifted 4 units to the right and has equation y = f(x 4). (b) (graph 1) The graph of f is shifted 3 units upward and has equation y = f(x) + 3. (c) (graph 4) The graph of f is shrunk vertically by a factor of 3 and has equation y = 1 3 f(x). (d) (graph 5) The graph of f is shifted 4 units to the left and reected about the x-axis. Its equation is y = f(x + 4). (e) (graph 2) The graph of f is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is y = 2f(x + 6). 5. (a) To graph y = f(2x) we shrink the graph of f horizontally by a factor of 2. The point (4, 1) on the graph of f corresponds to the point 1 2 4, 1 = (2, 1). (b) To graph y = f 1 2 x we stretch the graph of f horizontally by a factor of 2. The point (4, 1) on the graph of f corresponds to the point (2 4, 1) = (8, 1). (c) To graph y = f(x) we reect the graph of f about the y-axis. The point (4, 1) on the graph of f corresponds to the point (1 4, 1) = (4, 1). (d) To graph y = f(x) we reect the graph of f about the y-axis, then about the x-axis. The point (4, 1) on the graph of f corresponds to the point (1 4, 1 1) = (4, 1). 10. 18 CHAPTER 1 FUNCTIONS AND MODELS 7. The graph of y = f(x) = 3x x2 has been shifted 4 units to the left, reected about the x-axis, and shifted downward 1 unit. Thus, a function describing the graph is y = 1 reect about x-axis f (x + 4) shift 4 units left 1 shift 1 unit left This function can be written as y = f(x + 4) 1 = 3(x + 4) (x + 4)2 1 = 3x + 12 (x2 + 8x + 16) 1 = x2 5x 4 1 9. y = x3 : Start with the graph of y = x3 and reect about the x-axis. Note: Reecting about the y-axis gives the same result since substituting x for x gives us y = (x)3 = x3 . 11. y = (x + 1)2 : Start with the graph of y = x2 and shift 1 unit to the left. 13. y = 1 + 2 cos x: Start with the graph of y = cos x, stretch vertically by a factor of 2, and then shift 1 unit upward. 15. y = sin(x/2): Start with the graph of y = sin x and stretch horizontally by a factor of 2. 11. SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS 19 17. y = x + 3 : Start with the graph of y = x and shift 3 units to the left. 19. y = 1 2 (x2 + 8x) = 1 2 (x2 + 8x + 16) 8 = 1 2 (x + 4)2 8: Start with the graph of y = x2 , compress vertically by a factor of 2, shift 4 units to the left, and then shift 8 units downward. 0 0 0 0 21. y = 2/(x + 1): Start with the graph of y = 1/x, shift 1 unit to the left, and then stretch vertically by a factor of 2. 23. y = |sin x|: Start with the graph of y = sin x and reect all the parts of the graph below the x-axis about the x-axis. 25. This is just like the solution to Example 4 except the amplitude of the curve (the 30 N curve in Figure 9 on June 21) is 14 12 = 2. So the function is L(t) = 12 + 2 sin 2 365 (t 80) . March 31 is the 90th day of the year, so the model gives L(90) 12.34 h. The daylight time (5:51 AM to 6:18 PM) is 12 hours and 27 minutes, or 12.45 h. The model value differs from the actual value by 12.4512.34 12.45 0.009, less than 1%. 27. (a) To obtain y = f(|x|), the portion of the graph of y = f(x) to the right of the y-axis is reected about the y-axis. (b) y = sin |x| (c) y = |x| 12. 20 CHAPTER 1 FUNCTIONS AND MODELS 29. f(x) = x3 + 2x2 ; g(x) = 3x2 1. D = R for both f and g. (f + g)(x) = (x3 + 2x2 ) + (3x2 1) = x3 + 5x2 1, D = R. (f g)(x) = (x3 + 2x2 ) (3x2 1) = x3 x2 + 1, D = R. (fg)(x) = (x3 + 2x2 )(3x2 1) = 3x5 + 6x4 x3 2x2 , D = R. f g (x) = x3 + 2x2 3x2 1 , D = x | x 6= 1 3 since 3x2 1 6= 0. 31. f(x) = x2 1, D = R; g(x) = 2x + 1, D = R. (a) (f g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 1 = (4x2 + 4x + 1) 1 = 4x2 + 4x, D = R. (b) (g f)(x) = g(f(x)) = g(x2 1) = 2(x2 1) + 1 = (2x2 2) + 1 = 2x2 1, D = R. (c) (f f)(x) = f(f(x)) = f(x2 1) = (x2 1)2 1 = (x4 2x2 + 1) 1 = x4 2x2 , D = R. (d) (g g)(x) = g(g(x)) = g(2x + 1) = 2(2x + 1) + 1 = (4x + 2) + 1 = 4x + 3, D = R. 33. f(x) = 1 3x; g(x) = cos x. D = R for both f and g, and hence for their composites. (a) (f g)(x) = f(g(x)) = f(cos x) = 1 3 cos x. (b) (g f)(x) = g(f(x)) = g(1 3x) = cos(1 3x). (c) (f f)(x) = f(f(x)) = f(1 3x) = 1 3(1 3x) = 1 3 + 9x = 9x 2. (d) (g g)(x) = g(g(x)) = g(cos x) = cos(cos x) [Note that this is not cos x cos x.] 35. f(x) = x + 1 x , D = {x | x 6= 0}; g(x) = x + 1 x + 2 , D = {x | x 6= 2} (a) (f g)(x) = f(g(x)) = f x + 1 x + 2 = x + 1 x + 2 + 1 x + 1 x + 2 = x + 1 x + 2 + x + 2 x + 1 = (x + 1)(x + 1) + (x + 2)(x + 2) (x + 2)(x + 1) = x2 + 2x + 1 + x2 + 4x + 4 (x + 2)(x + 1) = 2x2 + 6x + 5 (x + 2)(x + 1) Since g(x) is not dened for x = 2 and f(g(x)) is not dened for x = 2 and x = 1, the domain of (f g)(x) is D = {x | x 6= 2, 1}. (b) (g f)(x) = g(f(x)) = g x + 1 x = x + 1 x + 1 x + 1 x + 2 = x2 + 1 + x x x2 + 1 + 2x x = x2 + x + 1 x2 + 2x + 1 = x2 + x + 1 (x + 1)2 Since f(x) is not dened for x = 0 and g(f(x)) is not dened for x = 1, the domain of (g f)(x) is D = {x | x 6= 1, 0}. (c) (f f)(x) = f(f(x)) = f x + 1 x = x + 1 x + 1 x + 1 x = x + 1 x + 1 x2 +1 x = x + 1 x + x x2 + 1 = x(x) x2 + 1 + 1 x2 + 1 + x(x) x(x2 + 1) = x4 + x2 + x2 + 1 + x2 x(x2 + 1) = x4 + 3x2 + 1 x(x2 + 1) , D = {x | x 6= 0} 13. SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS 21 (d) (g g)(x) = g(g(x)) = g x + 1 x + 2 = x + 1 x + 2 + 1 x + 1 x + 2 + 2 = x + 1 + 1(x + 2) x + 2 x + 1 + 2(x + 2) x + 2 = x + 1 + x + 2 x + 1 + 2x + 4 = 2x + 3 3x + 5 Since g(x) is not dened for x = 2 and g(g(x)) is not dened for x = 5 3 , the domain of (g g)(x) is D = x | x 6= 2, 5 3 . 37. (f g h)(x) = f(g(h(x))) = f(g(x 1)) = f(2(x 1)) = 2(x 1) + 1 = 2x 1 39. (f g h)(x) = f(g(h(x))) = f(g(x3 + 2)) = f[(x3 + 2)2 ] = f(x6 + 4x3 + 4) = (x6 + 4x3 + 4) 3 = x6 + 4x3 + 1 41. Let g(x) = x2 + 1 and f(x) = x10 . Then (f g)(x) = f(g(x)) = f(x2 + 1) = (x2 + 1)10 = F(x). 43. Let g(x) = 3 x and f(x) = x 1 + x . Then (f g)(x) = f(g(x)) = f( 3 x ) = 3 x 1 + 3 x = F(x). 45. Let g(t) = cos t and f(t) = t. Then (f g)(t) = f(g(t)) = f(cos t) = cos t = u(t). 47. Let h(x) = x2 , g(x) = 3x , and f(x) = 1 x. Then (f g h)(x) = f(g(h(x))) = f(g(x2 )) = f 3x2 = 1 3x2 = H(x). 49. Let h(x) = x, g(x) = sec x, and f(x) = x4 . Then (f g h)(x) = f(g(h(x))) = f(g( x )) = f(sec x ) = (sec x ) 4 = sec4 ( x ) = H(x). 51. (a) g(2) = 5, because the point (2, 5) is on the graph of g. Thus, f(g(2)) = f(5) = 4, because the point (5, 4) is on the graph of f. (b) g(f(0)) = g(0) = 3 (c) (f g)(0) = f(g(0)) = f(3) = 0 (d) (g f)(6) = g(f(6)) = g(6). This value is not dened, because there is no point on the graph of g that has x-coordinate 6. (e) (g g)(2) = g(g(2)) = g(1) = 4 (f ) (f f)(4) = f(f(4)) = f(2) = 2 53. (a) Using the relationship distance = rate time with the radius r as the distance, we have r(t) = 60t. (b) A = r2 (A r)(t) = A(r(t)) = (60t)2 = 3600t2 . This formula gives us the extent of the rippled area (in cm2 ) at any time t. 55. (a) From the gure, we have a right triangle with legs 6 and d, and hypotenuse s. By the Pythagorean Theorem, d2 + 62 = s2 s = f(d) = d2 + 36. (b) Using d = rt, we get d = (30 km/hr)(t hr) = 30t (in km). Thus, d = g(t) = 30t. (c) (f g)(t) = f(g(t)) = f(30t) = (30t)2 + 36 = 900t2 + 36. This function represents the distance between the lighthouse and the ship as a function of the time elapsed since noon. 14. 22 CHAPTER 1 FUNCTIONS AND MODELS 57. (a) H(t) = 0 if t < 0 1 if t 0 (b) V (t) = 0 if t < 0 120 if t 0 so V (t) = 120H(t). (c) Starting with the formula in part (b), we replace 120 with 240 to reect the different voltage. Also, because we are starting 5 units to the right of t = 0, we replace t with t 5. Thus, the formula is V (t) = 240H(t 5). 59. If f(x) = m1x + b1 and g(x) = m2x + b2, then (f g)(x) = f(g(x)) = f(m2x + b2) = m1(m2x + b2) + b1 = m1m2x + m1b2 + b1. So f g is a linear function with slope m1m2. 61. (a) By examining the variable terms in g and h, we deduce that we must square g to get the terms 4x2 and 4x in h. If we let f(x) = x2 + c, then (f g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 + c = 4x2 + 4x + (1 + c). Since h(x) = 4x2 + 4x + 7, we must have 1 + c = 7. So c = 6 and f(x) = x2 + 6. (b) We need a function g so that f(g(x)) = 3(g(x)) + 5 = h(x). But h(x) = 3x2 + 3x + 2 = 3(x2 + x) + 2 = 3(x2 + x 1) + 5, so we see that g(x) = x2 + x 1. 63. (a) If f and g are even functions, then f(x) = f(x) and g(x) = g(x). (i) (f + g)(x) = f(x) + g(x) = f(x) + g(x) = (f + g)(x), so f + g is an even function. (ii) (fg)(x) = f(x) g(x) = f(x) g(x) = (fg)(x), so fg is an even function. (b) If f and g are odd functions, then f(x) = f(x) and g(x) = g(x). (i) (f + g)(x) = f(x) + g(x) = f(x) + [g(x)] = [f(x) + g(x)] = (f + g)(x), so f + g is an odd function. (ii) (fg)(x) = f(x) g(x) = f(x) [g(x)] = f(x) g(x) = (fg)(x), so fg is an even function. 65. We need to examine h(x). h(x) = (f g)(x) = f(g(x)) = f(g(x)) [because g is even] = h(x) Because h(x) = h(x), h is an even function. 15. SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS 23 1.4 Graphing Calculators and Computers 1. f(x) = x3 5x2 (a) [5, 5] by [5, 5] (There is no graph shown.) (b) [0, 10] by [0, 2] (c) [0, 10] by [0, 10] The most appropriate graph is produced in viewing rectangle (c). 3. Since the graph of f(x) = 5 + 20x x2 is a parabola opening downward, an appropriate viewing rectangle should include the maximum point. 5. f(x) = 4 81 x4 is dened when 81 x4 0 x4 81 |x| 3, so the domain of f is [3, 3]. Also 0 4 81 x4 4 81 = 3, so the range is [0, 3]. 7. The graph of f(x) = x3 225x is symmetric with respect to the origin. Since f(x) = x3 225x = x(x2 225) = x(x + 15)(x 15), there are x-intercepts at 0, 15, and 15. f(20) = 3500. 9. The period of g(x) = sin(1000x) is 2 1000 0.0063 and its range is [1, 1]. Since f(x) = sin2 (1000x) is the square of g, its range is [0, 1] and a viewing rectangle of [0.01, 0.01] by [0, 1.1] seems appropriate. 11. The domain of y = x is x 0, so the domain of f(x) = sin x is [0, ) and the range is [1, 1]. With a little trial-and-error experimentation, we nd that an Xmax of 100 illustrates the general shape of f, so an appropriate viewing rectangle is [0, 100] by [1.5, 1.5]. 16. 24 CHAPTER 1 FUNCTIONS AND MODELS 13. The rst term, 10 sin x, has period 2 and range [10, 10]. It will be the dominant term in any large graph of y = 10 sin x + sin 100x, as shown in the rst gure. The second term, sin 100x, has period 2 100 = 50 and range [1, 1]. It causes the bumps in the rst gure and will be the dominant term in any small graph, as shown in the view near the origin in the second gure. 15. We must solve the given equation for y to obtain equations for the upper and lower halves of the ellipse. 4x2 + 2y2 = 1 2y2 = 1 4x2 y2 = 1 4x2 2 y = 1 4x2 2 17. From the graph of y = 3x2 6x + 1 and y = 0.23x 2.25 in the viewing rectangle [1, 3] by [2.5, 1.5], it is difcult to see if the graphs intersect. If we zoom in on the fourth quadrant, we see the graphs do not intersect. 19. From the graph of f(x) = x3 9x2 4, we see that there is one solution of the equation f(x) = 0 and it is slightly larger than 9. By zooming in or using a root or zero feature, we obtain x 9.05. 21. We see that the graphs of f(x) = x2 and g(x) = sin x intersect twice. One solution is x = 0. The other solution of f = g is the x-coordinate of the point of intersection in the rst quadrant. Using an intersect feature or zooming in, we nd this value to be approximately 0.88. Alternatively, we could nd that value by nding the positive zero of h(x) = x2 sin x. Note: After producing the graph on a TI-83 Plus, we can nd the approximate value 0.88 by using the following keystrokes: . The 1 is just a guess for 0.88. 17. SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS 25 23. g(x) = x3 /10 is larger than f(x) = 10x2 whenever x > 100. 25. We see from the graphs of y = |sin x x| and y = 0.1 that there are two solutions to the equation |sin x x| = 0.1: x 0.85 and x 0.85. The condition |sin x x| < 0.1 holds for any x lying between these two values, that is, 0.85 < x < 0.85. 27. (a) The root functions y = x, y = 4 x and y = 6 x (b) The root functions y = x, y = 3 x and y = 5 x (c) The root functions y = x, y = 3 x, y = 4 x and y = 5 x (d) For any n, the nth root of 0 is 0 and the nth root of 1 is 1; that is, all nth root functions pass through the points (0, 0) and (1, 1). For odd n, the domain of the nth root function is R, while for even n, it is {x R | x 0}. Graphs of even root functions look similar to that of x, while those of odd root functions resemble that of 3 x. As n increases, the graph of n x becomes steeper near 0 and atter for x > 1. 29. f(x) = x4 + cx2 + x. If c < 1.5, there are three humps: two minimum points and a maximum point. These humps get atter as c increases, until at c = 1.5 two of the humps disappear and there is only one minimum point. This single hump then moves to the right and approaches the origin as c increases. 31. y = xn 2x . As n increases, the maximum of the function moves further from the origin, and gets larger. Note, however, that regardless of n, the function approaches 0 as x . 18. 26 CHAPTER 1 FUNCTIONS AND MODELS 33. y2 = cx3 + x2 . If c < 0, the loop is to the right of the origin, and if c is positive, it is to the left. In both cases, the closer c is to 0, the larger the loop is. (In the limiting case, c = 0, the loop is innite, that is, it doesnt close.) Also, the larger |c| is, the steeper the slope is on the loopless side of the origin. 35. The graphing window is 95 pixels wide and we want to start with x = 0 and end with x = 2. Since there are 94 gaps between pixels, the distance between pixels is 20 94 . Thus, the x-values that the calculator actually plots are x = 0 + 2 94 n, where n = 0, 1, 2, . . . , 93, 94. For y = sin 2x, the actual points plotted by the calculator are 2 94 n, sin 2 2 94 n for n = 0, 1, . . . , 94. For y = sin 96x, the points plotted are 2 94 n, sin 96 2 94 n for n = 0, 1, . . . , 94. But sin 96 2 94 n = sin 94 2 94 n + 2 2 94 n = sin 2n + 2 2 94 n = sin 2 2 94 n [by periodicity of sine], n = 0, 1, . . . , 94 So the y-values, and hence the points, plotted for y = sin 96x are identical to those plotted for y = sin 2x. Note: Try graphing y = sin 94x. Can you see why all the y-values are zero? 1.5 Exponential Functions 1. (a) f(x) = ax , a > 0 (b) R (c) (0, ) (d) See Figures 4(c), 4(b), and 4(a), respectively. 3. All of these graphs approach 0 as x , all of them pass through the point (0, 1), and all of them are increasing and approach as x . The larger the base, the faster the function increases for x > 0, and the faster it approaches 0 as x . Note: The notation x can be thought of as x becomes large at this point. More details on this notation are given in Chapter 2. 5. The functions with bases greater than 1 (3x and 10x ) are increasing, while those with bases less than 1 1 3 x and 1 10 x are decreasing. The graph of 1 3 x is the reection of that of 3x about the y-axis, and the graph of 1 10 x is the reection of that of 10x about the y-axis. The graph of 10x increases more quickly than that of 3x for x > 0, and approaches 0 faster as x . 7. We start with the graph of y = 4x (Figure 3) and then shift 3 units downward. This shift doesnt affect the domain, but the range of y = 4x 3 is (3, ) . There is a horizontal asymptote of y = 3. y = 4x y = 4x 3 19. SECTION 1.5 EXPONENTIAL FUNCTIONS 27 9. We start with the graph of y = 2x (Figure 3), reect it about the y-axis, and then about the x-axis (or just rotate 180 to handle both reections) to obtain the graph of y = 2x . In each graph, y = 0 is the horizontal asymptote. y = 2x y = 2x y = 2x 11. We start with the graph of y = ex (Figure 13) and reect about the y-axis to get the graph of y = ex . Then we compress the graph vertically by a factor of 2 to obtain the graph of y = 1 2 ex and then reect about the x-axis to get the graph of y = 1 2 ex . Finally, we shift the graph upward one unit to get the graph of y = 1 1 2 ex . 13. (a) To nd the equation of the graph that results from shifting the graph of y = ex 2 units downward, we subtract 2 from the original function to get y = ex 2. (b) To nd the equation of the graph that results from shifting the graph of y = ex 2 units to the right, we replace x with x 2 in the original function to get y = e(x2) . (c) To nd the equation of the graph that results from reecting the graph of y = ex about the x-axis, we multiply the original function by 1 to get y = ex . (d) To nd the equation of the graph that results from reecting the graph of y = ex about the y-axis, we replace x with x in the original function to get y = ex . (e) To nd the equation of the graph that results from reecting the graph of y = ex about the x-axis and then about the y-axis, we rst multiply the original function by 1 (to get y = ex ) and then replace x with x in this equation to get y = ex . 15. (a) The denominator 1 + ex is never equal to zero because ex > 0, so the domain of f(x) = 1/(1 + ex ) is R. (b) 1 ex = 0 ex = 1 x = 0, so the domain of f(x) = 1/(1 ex ) is (, 0) (0, ). 17. Use y = Cax with the points (1, 6) and (3, 24). 6 = Ca1 C = 6 a and 24 = Ca3 24 = 6 a a3 4 = a2 a = 2 [since a > 0] and C = 6 2 = 3. The function is f(x) = 3 2x . 19. If f(x) = 5x , then f(x + h) f(x) h = 5x+h 5x h = 5x 5h 5x h = 5x 5h 1 h = 5x 5h 1 h . 20. 28 CHAPTER 1 FUNCTIONS AND MODELS 21. 2 ft = 24 in, f(24) = 242 in = 576 in = 48 ft. g(24) = 224 in = 224 /(12 5280) mi 265 mi 23. The graph of g nally surpasses that of f at x 35.8. 25. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours). 100 25 = 3200 (b) In t hours, there will be t/3 doubling periods. The initial population is 100, so the population y at time t is y = 100 2t/3 . (c) t = 20 y = 100 220/3 10,159 (d) We graph y1 = 100 2x/3 and y2 = 50,000. The two curves intersect at x 26.9, so the population reaches 50,000 in about 26.9 hours. 27. An exponential model is y = abt , where a = 3.154832569 1012 and b = 1.017764706. This model gives y(1993) 5498 million and y(2010) 7417 million. 29. From the graph, it appears that f is an odd function (f is undened for x = 0). To prove this, we must show that f(x) = f(x). f(x) = 1 e1/(x) 1 + e1/(x) = 1 e(1/x) 1 + e(1/x) = 1 1 e1/x 1 + 1 e1/x e1/x e1/x = e1/x 1 e1/x + 1 = 1 e1/x 1 + e1/x = f(x) so f is an odd function. 1.6 Inverse Functions and Logarithms 1. (a) See Denition 1. (b) It must pass the Horizontal Line Test. 3. f is not one-to-one because 2 6= 6, but f(2) = 2.0 = f(6). 5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one. 21. SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS 29 7. The horizontal line y = 0 (the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line Test, f is not one-to-one. 9. The graph of f(x) = x2 2x is a parabola with axis of symmetry x = b 2a = 2 2(1) = 1. Pick any x-values equidistant from 1 to nd two equal function values. For example, f(0) = 0 and f(2) = 0, so f is not one-to-one. 11. g(x) = 1/x. x1 6= x2 1/x1 6= 1/x2 g (x1) 6= g (x2), so g is one-to-one. Geometric solution: The graph of g is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test, so g is one-to-one. 13. A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus, even if t1 does not equal t2, f(t1) may equal f(t2), so f is not 1-1. 15. Since f(2) = 9 and f is 1-1, we know that f1 (9) = 2. Remember, if the point (2, 9) is on the graph of f, then the point (9, 2) is on the graph of f1 . 17. First, we must determine x such that g(x) = 4. By inspection, we see that if x = 0, then g(x) = 4. Since g is 1-1 (g is an increasing function), it has an inverse, and g1 (4) = 0. 19. We solve C = 5 9 (F 32) for F: 9 5 C = F 32 F = 9 5 C + 32. This gives us a formula for the inverse function, that is, the Fahrenheit temperature F as a function of the Celsius temperature C. F 459.67 9 5 C + 32 459.67 9 5 C 491.67 C 273.15, the domain of the inverse function. 21. f(x) = 10 3x y = 10 3x (y 0) y2 = 10 3x 3x = 10 y2 x = 1 3 y2 + 10 3 . Interchange x and y: y = 1 3 x2 + 10 3 . So f1 (x) = 1 3 x2 + 10 3 . Note that the domain of f1 is x 0. 23. y = f(x) = ex3 ln y = x3 x = 3 ln y. Interchange x and y: y = 3 ln x. So f1 (x) = 3 ln x. 25. y = f(x) = ln (x + 3) x + 3 = ey x = ey 3. Interchange x and y: y = ex 3. So f1 (x) = ex 3. 27. y = f(x) = x4 + 1 y 1 = x4 x = 4 y 1 (not since x 0). Interchange x and y: y = 4 x 1. So f1 (x) = 4 x 1. The graph of y = 4 x 1 is just the graph of y = 4 x shifted right one unit. From the graph, we see that f and f1 are reections about the line y = x. 29. Reect the graph of f about the line y = x. The points (1, 2), (1, 1), (2, 2), and (3, 3) on f are reected to (2, 1), (1, 1), (2, 2), and (3, 3) on f1 . 22. 30 CHAPTER 1 FUNCTIONS AND MODELS 31. (a) It is dened as the inverse of the exponential function with base a, that is, loga x = y ay = x. (b) (0, ) (c) R (d) See Figure 11. 33. (a) log5 125 = 3 since 53 = 125. (b) log3 1 27 = 3 since 33 = 1 33 = 1 27 . 35. (a) log2 6 log2 15 + log2 20 = log2( 6 15 ) + log2 20 [by Law 2] = log2( 6 15 20) [by Law 1] = log2 8, and log2 8 = 3 since 23 = 8. (b) log3 100 log3 18 log3 50 = log3 100 18 log3 50 = log3 100 1850 = log3(1 9 ), and log3 1 9 = 2 since 32 = 1 9 . 37. ln 5 + 5 ln 3 = ln 5 + ln 35 [by Law 3] = ln(5 35 ) [by Law 1] = ln 1215 39. ln(1 + x2 ) + 1 2 ln x ln sin x = ln(1 + x2 ) + ln x1/2 ln sin x = ln[(1 + x2 ) x ] ln sin x = ln (1 + x2 ) x sin x 41. To graph these functions, we use log1.5 x = ln x ln 1.5 and log50 x = ln x ln 50 . These graphs all approach as x 0+ , and they all pass through the point (1, 0). Also, they are all increasing, and all approach as x . The functions with larger bases increase extremely slowly, and the ones with smaller bases do so somewhat more quickly. The functions with large bases approach the y-axis more closely as x 0+ . 43. 3 ft = 36 in, so we need x such that log2 x = 36 x = 236 = 68,719,476,736. In miles, this is 68,719,476,736 in 1 ft 12 in 1 mi 5280 ft 1,084,587.7 mi. 45. (a) Shift the graph of y = log10 x ve units to the left to obtain the graph of y = log10(x + 5). Note the vertical asymptote of x = 5. y = log10 x y = log10(x + 5) (b) Reect the graph of y = ln x about the x-axis to obtain the graph of y = ln x. y = ln x y = ln x 47. (a) 2 ln x = 1 ln x = 1 2 x = e1/2 = e (b) ex = 5 x = ln 5 x = ln 5 49. (a) 2x5 = 3 log2 3 = x 5 x = 5 + log2 3. Or: 2x5 = 3 ln 2x5 = ln 3 (x 5) ln 2 = ln 3 x 5 = ln 3 ln 2 x = 5 + ln 3 ln 2 23. SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS 31 (b) ln x + ln(x 1) = ln(x(x 1)) = 1 x(x 1) = e1 x2 x e = 0. The quadratic formula (with a = 1, b = 1, and c = e) gives x = 1 2 1 1 + 4e , but we reject the negative root since the natural logarithm is not dened for x < 0. So x = 1 2 1 + 1 + 4e . 51. (a) ex < 10 ln ex < ln 10 x < ln 10 x (, ln 10) (b) ln x > 1 eln x > e1 x > e1 x (1/e, ) 53. (a) For f(x) = 3 e2x, we must have 3 e2x 0 e2x 3 2x ln 3 x 1 2 ln 3. Thus, the domain of f is (, 1 2 ln 3]. (b) y = f(x) = 3 e2x [note that y 0] y2 = 3 e2x e2x = 3 y2 2x = ln(3 y2 ) x = 1 2 ln(3 y2 ). Interchange x and y: y = 1 2 ln(3 x2 ). So f1 (x) = 1 2 ln(3 x2 ). For the domain of f1 , we must have 3 x2 > 0 x2 < 3 |x| < 3 3 < x < 3 0 x < 3 since x 0. Note that the domain of f1 , [0, 3 ), equals the range of f. 55. We see that the graph of y = f(x) = x3 + x2 + x + 1 is increasing, so f is 1-1. Enter x = y3 + y2 + y + 1 and use your CAS to solve the equation for y. Using Derive, we get two (irrelevant) solutions involving imaginary expressions, as well as one which can be simplied to the following: y = f1 (x) = 3 4 6 3 D 27x2 + 20 3 D + 27x2 20 + 3 2 where D = 3 3 27x4 40x2 + 16. Maple and Mathematica each give two complex expressions and one real expression, and the real expression is equivalent to that given by Derive. For example, Maples expression simplies to 1 6 M2/3 8 2M1/3 2M1/3 , where M = 108x2 + 12 48 120x2 + 81x4 80. 57. (a) n = 100 2t/3 n 100 = 2t/3 log2 n 100 = t 3 t = 3 log2 n 100 . Using formula (10), we can write this as t = f1 (n) = 3 ln(n/100) ln 2 . This function tells us how long it will take to obtain n bacteria (given the number n). (b) n = 50,000 t = f1 (50,000) = 3 ln 50,000 100 ln 2 = 3 ln 500 ln 2 26.9 hours 59. (a) sin1 3 2 = 3 since sin 3 = 3 2 and 3 is in 2 , 2 . (b) cos1 (1) = since cos = 1 and is in [0, ]. 61. (a) arctan 1 = 4 since tan 4 = 1 and 4 is in 2 , 2 . (b) sin1 1 2 = 4 since sin 4 = 1 2 and 4 is in 2 , 2 . 24. 32 CHAPTER 1 FUNCTIONS AND MODELS 63. (a) In general, tan(arctan x) = x for any real number x. Thus, tan(arctan 10) = 10. (b) sin1 sin 7 3 = sin1 sin 3 = sin1 3 2 = 3 since sin 3 = 3 2 and 3 is in 2 , 2 . [Recall that 7 3 = 3 + 2 and the sine function is periodic with period 2.] 65. Let y = sin1 x. Then 2 y 2 cos y 0, so cos(sin1 x) = cos y = 1 sin2 y = 1 x2. 67. Let y = tan1 x. Then tan y = x, so from the triangle we see that sin(tan1 x) = sin y = x 1 + x2 . 69. The graph of sin1 x is the reection of the graph of sin x about the line y = x. 71. g(x) = sin1 (3x + 1). Domain (g) = {x | 1 3x + 1 1} = {x | 2 3x 0} = x | 2 3 x 0 = 2 3 , 0 . Range (g) = y | 2 y 2 = 2 , 2 . 73. (a) If the point (x, y) is on the graph of y = f(x), then the point (x c, y) is that point shifted c units to the left. Since f is 1-1, the point (y, x) is on the graph of y = f1 (x) and the point corresponding to (x c, y) on the graph of f is (y, x c) on the graph of f1 . Thus, the curves reection is shifted down the same number of units as the curve itself is shifted to the left. So an expression for the inverse function is g1 (x) = f1 (x) c. (b) If we compress (or stretch) a curve horizontally, the curves reection in the line y = x is compressed (or stretched) vertically by the same factor. Using this geometric principle, we see that the inverse of h(x) = f(cx) can be expressed as h1 (x) = (1/c) f1 (x). 25. CHAPTER 1 REVIEW 33 1 Review 1. (a) A function f is a rule that assigns to each element x in a set A exactly one element, called f(x), in a set B. The set A is called the domain of the function. The range of f is the set of all possible values of f(x) as x varies throughout the domain. (b) If f is a function with domain A, then its graph is the set of ordered pairs {(x, f(x)) | x A}. (c) Use the Vertical Line Test on page 16. 2. The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is given below. Verbally: An assignment of students to chairs in a classroom (a description in words) Numerically: A tax table that assigns an amount of tax to an income (a table of values) Visually: A graphical history of the Dow Jones average (a graph) Algebraically: A relationship between distance, rate, and time: d = rt (an explicit formula) 3. (a) An even function f satises f(x) = f(x) for every number x in its domain. It is symmetric with respect to the y-axis. (b) An odd function g satises g(x) = g(x) for every number x in its domain. It is symmetric with respect to the origin. 4. A function f is called increasing on an interval I if f(x1) < f(x2) whenever x1 < x2 in I. 5. A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world phenomenon. 6. (a) Linear function: f(x) = 2x + 1, f(x) = ax + b 7. (b) Power function: f(x) = x2 , f(x) = xa (c) Exponential function: f(x) = 2x , f(x) = ax (d) Quadratic function: f(x) = x2 + x + 1, f(x) = ax2 + bx + c (e) Polynomial of degree 5: f(x) = x5 + 2 (f ) Rational function: f(x) = x x + 2 , f(x) = P(x) Q(x) where P(x) and Q(x) are polynomials 8. (a) (b) 26. 34 CHAPTER 1 FUNCTIONS AND MODELS (c) (d) (e) (f ) (g) (h) 9. (a) The domain of f + g is the intersection of the domain of f and the domain of g; that is, A B. (b) The domain of fg is also A B. (c) The domain of f/g must exclude values of x that make g equal to 0; that is, {x A B | g(x) 6= 0}. 10. Given two functions f and g, the composite function f g is dened by (f g) (x) = f(g (x)). The domain of f g is the set of all x in the domain of g such that g(x) is in the domain of f. 11. (a) If the graph of f is shifted 2 units upward, its equation becomes y = f(x) + 2. (b) If the graph of f is shifted 2 units downward, its equation becomes y = f(x) 2. (c) If the graph of f is shifted 2 units to the right, its equation becomes y = f(x 2). (d) If the graph of f is shifted 2 units to the left, its equation becomes y = f(x + 2). (e) If the graph of f is reected about the x-axis, its equation becomes y = f(x). (f ) If the graph of f is reected about the y-axis, its equation becomes y = f(x). (g) If the graph of f is stretched vertically by a factor of 2, its equation becomes y = 2f(x). (h) If the graph of f is shrunk vertically by a factor of 2, its equation becomes y = 1 2 f(x). (i) If the graph of f is stretched horizontally by a factor of 2, its equation becomes y = f 1 2 x . (j) If the graph of f is shrunk horizontally by a factor of 2, its equation becomes y = f(2x). 12. (a) A function f is called a one-to-one function if it never takes on the same value twice; that is, if f(x1) 6= f(x2) whenever x1 6= x2. (Or, f is 1-1 if each output corresponds to only one input.) Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than once. (b) If f is a one-to-one function with domain A and range B, then its inverse function f1 has domain B and range A and is dened by f1 (y) = x f(x) = y for any y in B. The graph of f1 is obtained by reecting the graph of f about the line y = x. 27. CHAPTER 1 REVIEW 35 13. (a) The inverse sine function f(x) = sin1 x is dened as follows: sin1 x = y sin y = x and 2 y 2 Its domain is 1 x 1 and its range is 2 y 2 . (b) The inverse cosine function f(x) = cos1 x is dened as follows: cos1 x = y cos y = x and 0 y Its domain is 1 x 1 and its range is 0 y . (c) The inverse tangent function f(x) = tan1 x is dened as follows: tan1 x = y tan y = x and 2 < y < 2 Its domain is R and its range is 2 < y < 2 . 1. False. Let f(x) = x2 , s = 1, and t = 1. Then f(s + t) = (1 + 1)2 = 02 = 0, but f(s) + f(t) = (1)2 + 12 = 2 6= 0 = f(s + t). 3. False. Let f(x) = x2 . Then f(3x) = (3x)2 = 9x2 and 3f(x) = 3x2 . So f(3x) 6= 3f(x). 5. True. See the Vertical Line Test. 7. False. Let f(x) = x3 . Then f is one-to-one and f1 (x) = 3 x. But 1/f(x) = 1/x3 , which is not equal to f1 (x). 9. True. The function ln x is an increasing function on (0, ). 11. False. Let x = e2 and a = e. Then ln x ln a = ln e2 ln e = 2 ln e ln e = 2 and ln x a = ln e2 e = ln e = 1, so in general the statement is false. What is true, however, is that ln x a = ln x ln a. 13. False. For example, tan1 20 is dened; sin1 20 and cos1 20 are not. 1. (a) When x = 2, y 2.7. Thus, f(2) 2.7. (b) f(x) = 3 x 2.3, 5.6 (c) The domain of f is 6 x 6, or [6, 6]. (d) The range of f is 4 y 4, or [4, 4]. (e) f is increasing on [4, 4], that is, on 4 x 4. (f) f is not one-to-one since it fails the Horizontal Line Test. (g) f is odd since its graph is symmetric about the origin. 28. 36 CHAPTER 1 FUNCTIONS AND MODELS 3. f(x) = x2 2x + 3, so f(a + h) = (a + h)2 2(a + h) + 3 = a2 + 2ah + h2 2a 2h + 3, and f(a + h) f(a) h = (a2 + 2ah + h2 2a 2h + 3) (a2 2a + 3) h = h(2a + h 2) h = 2a + h 2. 5. f(x) = 2/(3x 1). Domain: 3x 1 6= 0 3x 6= 1 x 6= 1 3 . D = , 1 3 1 3 , Range: all reals except 0 (y = 0 is the horizontal asymptote for f.) R = (, 0) (0, ) 7. h(x) = ln(x + 6). Domain: x + 6 > 0 x > 6. D = (6, ) Range: x + 6 > 0, so ln(x + 6) takes on all real numbers and, hence, the range is R. R = (, ) 9. (a) To obtain the graph of y = f(x) + 8, we shift the graph of y = f(x) up 8 units. (b) To obtain the graph of y = f(x + 8), we shift the graph of y = f(x) left 8 units. (c) To obtain the graph of y = 1 + 2f(x), we stretch the graph of y = f(x) vertically by a factor of 2, and then shift the resulting graph 1 unit upward. (d) To obtain the graph of y = f(x 2) 2, we shift the graph of y = f(x) right 2 units (for the 2 inside the parentheses), and then shift the resulting graph 2 units downward. (e) To obtain the graph of y = f(x), we reect the graph of y = f(x) about the x-axis. (f) To obtain the graph of y = f1 (x), we reect the graph of y = f(x) about the line y = x (assuming f is oneto-one). 11. y = sin 2x: Start with the graph of y = sin x, compress horizontally by a factor of 2, and reect about the x-axis. 13. y = 1 2 (1 + ex ): Start with the graph of y = ex , shift 1 unit upward, and compress vertically by a factor of 2. 15. f(x) = 1 x + 2 : Start with the graph of f(x) = 1/x and shift 2 units to the left. 29. CHAPTER 1 REVIEW 37 17. (a) The terms of f are a mixture of odd and even powers of x, so f is neither even nor odd. (b) The terms of f are all odd powers of x, so f is odd. (c) f(x) = e(x)2 = ex2 = f(x), so f is even. (d) f(x) = 1 + sin(x) = 1 sin x. Now f(x) 6= f(x) and f(x) 6= f(x), so f is neither even nor odd. 19. f(x) = ln x, D = (0, ); g(x) = x2 9, D = R. (a) (f g)(x) = f(g(x)) = f(x2 9) = ln(x2 9). Domain: x2 9 > 0 x2 > 9 |x| > 3 x (, 3) (3, ) (b) (g f)(x) = g(f(x)) = g(ln x) = (ln x)2 9. Domain: x > 0, or (0, ) (c) (f f)(x) = f(f(x)) = f(ln x) = ln(ln x). Domain: ln x > 0 x > e0 = 1, or (1, ) (d) (g g)(x) = g(g(x)) = g(x2 9) = (x2 9)2 9. Domain: x R, or (, ) 21. Many models appear to be plausible. Your choice depends on whether you think medical advances will keep increasing life expectancy, or if there is bound to be a natural leveling-off of life expectancy. A linear model, y = 0.2493x 423.4818, gives us an estimate of 77.6 years for the year 2010. 23. We need to know the value of x such that f(x) = 2x + ln x = 2. Since x = 1 gives us y = 2, f1 (2) = 1. 25. (a) e2 ln 3 = eln 3 2 = 32 = 9 (b) log10 25 + log10 4 = log10(25 4) = log10 100 = log10 102 = 2 (c) tan arcsin 1 2 = tan 6 = 1 3 (d) Let = cos1 4 5 , so cos = 4 5 . Then sin cos1 4 5 = sin = 1 cos2 = 1 4 5 2 = 9 25 = 3 5 . 27. (a) The population would reach 900 in about 4.4 years. (b) P = 100,000 100 + 900et 100P + 900Pet = 100,000 900Pet = 100,000 100P et = 100,000 100P 900P t = ln 1000 P 9P t = ln 1000 P 9P , or ln 9P 1000 P ; this is the time required for the population to reach a given number P. (c) P = 900 t = ln 9 900 1000 900 = ln 81 4.4 years, as in part (a). 30. PRINCIPLES OF PROBLEM SOLVING 1. By using the area formula for a triangle, 1 2 (base) (height), in two ways, we see that 1 2 (4) (y) = 1 2 (h) (a), so a = 4y h . Since 42 + y2 = h2 , y = h2 16, and a = 4 h2 16 h . 3. |2x 1| = 2x 1 if x 1 2 1 2x if x < 1 2 and |x + 5| = x + 5 if x 5 x 5 if x < 5 Therefore, we consider the three cases x < 5, 5 x < 1 2 , and x 1 2 . If x < 5, we must have 1 2x (x 5) = 3 x = 3, which is false, since we are considering x < 5. If 5 x < 1 2 , we must have 1 2x (x + 5) = 3 x = 7 3 . If x 1 2 , we must have 2x 1 (x + 5) = 3 x = 9. So the two solutions of the equation are x = 7 3 and x = 9. 5. f(x) = x2 4 |x| + 3 . If x 0, then f(x) = x2 4x + 3 = |(x 1)(x 3)|. Case (i): If 0 < x 1, then f(x) = x2 4x + 3. Case (ii): If 1 < x 3, then f(x) = (x2 4x + 3) = x2 + 4x 3. Case (iii): If x > 3, then f(x) = x2 4x + 3. This enables us to sketch the graph for x 0. Then we use the fact that f is an even function to reect this part of the graph about the y-axis to obtain the entire graph. Or, we could consider also the cases x < 3, 3 x < 1, and 1 x < 0. 7. Remember that |a| = a if a 0 and that |a| = a if a < 0. Thus, x + |x| = 2x if x 0 0 if x < 0 and y + |y| = 2y if y 0 0 if y < 0 We will consider the equation x + |x| = y + |y| in four cases. (1) x 0, y 0 2x = 2y x = y (2) x 0, y < 0 2x = 0 x = 0 (3) x < 0, y 0 0 = 2y 0 = y (4) x < 0, y < 0 0 = 0 Case 1 gives us the line y = x with nonnegative x and y. Case 2 gives us the portion of the y-axis with y negative. Case 3 gives us the portion of the x-axis with x negative. Case 4 gives us the entire third quadrant. 39 31. 40 CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING 9. |x| + |y| 1. The boundary of the region has equation |x| + |y| = 1. In quadrants I, II, III, and IV, this becomes the lines x + y = 1, x + y = 1, x y = 1, and x y = 1 respectively. 11. (log2 3)(log3 4)(log4 5) (log31 32) = ln 3 ln 2 ln 4 ln 3 ln 5 ln 4 ln 32 ln 31 = ln 32 ln 2 = ln 25 ln 2 = 5 ln 2 ln 2 = 5 13. ln x2 2x 2 0 x2 2x 2 e0 = 1 x2 2x 3 0 (x 3)(x + 1) 0 x [1, 3]. Since the argument must be positive, x2 2x 2 > 0 x 1 3 x 1 + 3 > 0 x , 1 3 1 + 3, . The intersection of these intervals is 1, 1 3 1 + 3, 3 . 15. Let d be the distance traveled on each half of the trip. Let t1 and t2 be the times taken for the rst and second halves of the trip. For the rst half of the trip we have t1 = d/30 and for the second half we have t2 = d/60. Thus, the average speed for the entire trip is total distance total time = 2d t1 + t2 = 2d d 30 + d 60 60 60 = 120d 2d + d = 120d 3d = 40. The average speed for the entire trip is 40 mi/h. 17. Let Sn be the statement that 7n 1 is divisible by 6. S1 is true because 71 1 = 6 is divisible by 6. Assume Sk is true, that is, 7k 1 is divisible by 6. In other words, 7k 1 = 6m for some positive integer m. Then 7k+1 1 = 7k 7 1 = (6m + 1) 7 1 = 42m + 6 = 6(7m + 1), which is divisible by 6, so Sk+1 is true. Therefore, by mathematical induction, 7n 1 is divisible by 6 for every positive integer n. 19. f0(x) = x2 and fn+1(x) = f0(fn(x)) for n = 0, 1, 2, . . .. f1(x) = f0(f0(x)) = f0 x2 = x2 2 = x4 , f2(x) = f0(f1(x)) = f0(x4 ) = (x4 )2 = x8 , f3(x) = f0(f2(x)) = f0(x8 ) = (x8 )2 = x16 , . . .. Thus, a general formula is fn(x) = x2n+1 . 32. 2 LIMITS AND DERIVATIVES 2.1 The Tangent and Velocity Problems 1. (a) Using P(15, 250), we construct the following table: t Q slope = mP Q 5 (5, 694) 694250 515 = 444 10 = 44.4 10 (10, 444) 444250 1015 = 194 5 = 38.8 20 (20, 111) 111250 2015 = 139 5 = 27.8 25 (25, 28) 28250 2515 = 222 10 = 22.2 30 (30, 0) 0250 3015 = 250 15 = 16.6 (b) Using the values of t that correspond to the points closest to P (t = 10 and t = 20), we have 38.8 + (27.8) 2 = 33.3 (c) From the graph, we can estimate the slope of the tangent line at P to be 300 9 = 33.3. 3. (a) x Q mP Q (i) 0.5 (0.5, 0.333333) 0.333333 (ii) 0.9 (0.9, 0.473684) 0.263158 (iii) 0.99 (0.99, 0.497487) 0.251256 (iv) 0.999 (0.999, 0.499750) 0.250125 (v) 1.5 (1.5, 0.6) 0.2 (vi) 1.1 (1.1, 0.523810) 0.238095 (vii) 1.01 (1.01, 0.502488) 0.248756 (viii) 1.001 (1.001, 0.500250) 0.249875 (b) The slope appears to be 1 4 . (c) y 1 2 = 1 4 (x 1) or y = 1 4 x + 1 4 . 5. (a) y = y(t) = 40t 16t2 . At t = 2, y = 40(2) 16(2)2 = 16. The average velocity between times 2 and 2 + h is vave = y(2 + h) y(2) (2 + h) 2 = 40(2 + h) 16(2 + h)2 16 h = 24h 16h2 h = 24 16h, if h 6= 0. (i) [2, 2.5]: h = 0.5, vave = 32 ft/s (ii) [2, 2.1]: h = 0.1, vave = 25.6 ft/s (iii) [2, 2.05]: h = 0.05, vave = 24.8 ft/s (iv) [2, 2.01]: h = 0.01, vave = 24.16 ft/s (b) The instantaneous velocity when t = 2 (h approaches 0) is 24 ft/s. 41 33. 42 CHAPTER 2 LIMITS AND DERIVATIVES 7. (a) (i) On the interval [1, 3], vave = s(3) s(1) 3 1 = 10.7 1.4 2 = 9.3 2 = 4.65 m/s. (ii) On the interval [2, 3], vave = s(3) s(2) 3 2 = 10.7 5.1 1 = 5.6 m/s. (iii) On the interval [3, 5], vave = s(5) s(3) 5 3 = 25.8 10.7 2 = 15.1 2 = 7.55 m/s. (iv) On the interval [3, 4], vave = s(4) s(3) 4 3 = 17.7 10.7 1 = 7 m/s. (b) Using the points (2, 4) and (5, 23) from the approximate tangent line, the instantaneous velocity at t = 3 is about 23 4 5 2 6.3 m/s. 9. (a) For the curve y = sin(10/x) and the point P(1, 0): x Q mP Q 2 (2, 0) 0 1.5 (1.5, 0.8660) 1.7321 1.4 (1.4, 0.4339) 1.0847 1.3 (1.3, 0.8230) 2.7433 1.2 (1.2, 0.8660) 4.3301 1.1 (1.1, 0.2817) 2.8173 x Q mP Q 0.5 (0.5, 0) 0 0.6 (0.6, 0.8660) 2.1651 0.7 (0.7, 0.7818) 2.6061 0.8 (0.8, 1) 5 0.9 (0.9, 0.3420) 3.4202 As x approaches 1, the slopes do not appear to be approaching any particular value. (b) We see that problems with estimation are caused by the frequent oscillations of the graph. The tangent is so steep at P that we need to take x-values much closer to 1 in order to get accurate estimates of its slope. (c) If we choose x = 1.001, then the point Q is (1.001, 0.0314) and mP Q 31.3794. If x = 0.999, then Q is (0.999, 0.0314) and mP Q = 31.4422. The average of these slopes is 31.4108. So we estimate that the slope of the tangent line at P is about 31.4. 34. SECTION 2.2 THE LIMIT OF A FUNCTION 43 2.2 The Limit of a Function 1. As x approaches 2, f(x) approaches 5. [Or, the values of f(x) can be made as close to 5 as we like by taking x sufciently close to 2 (but x 6= 2).] Yes, the graph could have a hole at (2, 5) and be dened such that f(2) = 3. 3. (a) lim x3 f(x) = means that the values of f(x) can be made arbitrarily large (as large as we please) by taking x sufciently close to 3 (but not equal to 3). (b) lim x4+ f(x) = means that the values of f(x) can be made arbitrarily large negative by taking x sufciently close to 4 through values larger than 4. 5. (a) f(x) approaches 2 as x approaches 1 from the left, so lim x1 f(x) = 2. (b) f(x) approaches 3 as x approaches 1 from the right, so lim x1+ f(x) = 3. (c) lim x1 f(x) does not exist because the limits in part (a) and part (b) are not equal. (d) f(x) approaches 4 as x approaches 5 from the left and from the right, so lim x5 f(x) = 4. (e) f(5) is not dened, so it doesnt exist. 7. (a) lim t0 g(t) = 1 (b) lim t0+ g(t) = 2 (c) lim t0 g(t) does not exist because the limits in part (a) and part (b) are not equal. (d) lim t2 g(t) = 2 (e) lim t2+ g(t) = 0 (f) lim t2 g(t) does not exist because the limits in part (d) and part (e) are not equal. (g) g(2) = 1 (h) lim t4 g(t) = 3 9. (a) lim x7 f(x) = (b) lim x3 f(x) = (c) lim x0 f(x) = (d) lim x6 f(x) = (e) lim x6+ f(x) = (f) The equations of the vertical asymptotes are x = 7, x = 3, x = 0, and x = 6. 11. (a) lim x0 f(x) = 1 (b) lim x0+ f(x) = 0 (c) lim x0 f(x) does not exist because the limits in part (a) and part (b) are not equal. 13. lim x1 f(x) = 2, lim x1+ f(x) = 2, f(1) = 2 15. lim x3+ f(x) = 4, lim x3 f(x) = 2, lim x2 f(x) = 2, f(3) = 3, f(2) = 1 35. 44 CHAPTER 2 LIMITS AND DERIVATIVES 17. For f(x) = x2 2x x2 x 2 : x f(x) 2.5 0.714286 2.1 0.677419 2.05 0.672131 2.01 0.667774 2.005 0.667221 2.001 0.666778 x f(x) 1.9 0.655172 1.95 0.661017 1.99 0.665552 1.995 0.666110 1.999 0.666556 It appears that lim x2 x2 2x x2 x 2 = 0.6 = 2 3 . 19. For f(x) = ex 1 x x2 : x f(x) 1 0.718282 0.5 0.594885 0.1 0.517092 0.05 0.508439 0.01 0.501671 x f(x) 1 0.367879 0.5 0.426123 0.1 0.483742 0.05 0.491770 0.01 0.498337 It appears that lim x0 ex 1 x x2 = 0.5 = 1 2 . 21. For f(x) = x + 4 2 x : x f(x) 1 0.236068 0.5 0.242641 0.1 0.248457 0.05 0.249224 0.01 0.249844 x f(x) 1 0.267949 0.5 0.258343 0.1 0.251582 0.05 0.250786 0.01 0.250156 It appears that lim x0 x + 4 2 x = 0.25 = 1 4 . 23. For f(x) = x6 1 x10 1 : x f(x) 0.5 0.985337 0.9 0.719397 0.95 0.660186 0.99 0.612018 0.999 0.601200 x f(x) 1.5 0.183369 1.1 0.484119 1.05 0.540783 1.01 0.588022 1.001 0.598800 It appears that lim x1 x6 1 x10 1 = 0.6 = 3 5 . 25. lim x3+ x + 2 x + 3 = since the numerator is negative and the denominator approaches 0 from the positive side as x 3+ . 27. lim x1 2 x (x 1)2 = since the numerator is positive and the denominator approaches 0 through positive values as x 1. 29. Let t = x2 9. Then as x 3+ , t 0+ , and lim x3+ ln(x2 9) = lim t0+ ln t = by (3). 31. lim x2 x csc x = lim x2 x sin x = since the numerator is positive and the denominator approaches 0 through negative values as x 2 . 33. (a) f(x) = 1 x3 1 . From these calculations, it seems that lim x1 f(x) = and lim x1+ f(x) = . x f(x) 0.5 1.14 0.9 3.69 0.99 33.7 0.999 333.7 0.9999 3333.7 0.99999 33,333.7 x f(x) 1.5 0.42 1.1 3.02 1.01 33.0 1.001 333.0 1.0001 3333.0 1.00001 33,333.3 36. SECTION 2.2 THE LIMIT OF A FUNCTION 45 (b) If x is slightly smaller than 1, then x3 1 will be a negative number close to 0, and the reciprocal of x3 1, that is, f(x), will be a negative number with large absolute value. So lim x1 f(x) = . If x is slightly larger than 1, then x3 1 will be a small positive number, and its reciprocal, f(x), will be a large positive number. So lim x1+ f(x) = . (c) It appears from the graph of f that lim x1 f(x) = and lim x1+ f(x) = . 35. (a) Let h(x) = (1 + x)1/x . x h(x) 0.001 2.71964 0.0001 2.71842 0.00001 2.71830 0.000001 2.71828 0.000001 2.71828 0.00001 2.71827 0.0001 2.71815 0.001 2.71692 It appears that lim x0 (1 + x)1/x 2.71828, which is approximately e. In Section 3.6 we will see that the value of the limit is exactly e. (b) 37. For f(x) = x2 (2x /1000): (a) x f(x) 1 0.998000 0.8 0.638259 0.6 0.358484 0.4 0.158680 0.2 0.038851 0.1 0.008928 0.05 0.001465 It appears that lim x0 f(x) = 0. (b) x f(x) 0.04 0.000572 0.02 0.000614 0.01 0.000907 0.005 0.000978 0.003 0.000993 0.001 0.001000 It appears that lim x0 f(x) = 0.001. 37. 46 CHAPTER 2 LIMITS AND DERIVATIVES 39. No matter how many times we zoom in toward the origin, the graphs of f(x) = sin(/x) appear to consist of almost-vertical lines. This indicates more and more frequent oscillations as x 0. 41. There appear to be vertical asymptotes of the curve y = tan(2 sin x) at x 0.90 and x 2.24. To nd the exact equations of these asymptotes, we note that the graph of the tangent function has vertical asymptotes at x = 2 + n. Thus, we must have 2 sin x = 2 + n, or equivalently, sin x = 4 + 2 n. Since 1 sin x 1, we must have sin x = 4 and so x = sin1 4 (corresponding to x 0.90). Just as 150 is the reference angle for 30 , sin1 4 is the reference angle for sin1 4 . So x = sin1 4 are also equations of vertical asymptotes (corresponding to x 2.24). 2.3 Calculating Limits Using the Limit Laws 1. (a) lim x2 [f(x) + 5g(x)] = lim x2 f(x) + lim x2 [5g(x)] [Limit Law 1] = lim x2 f(x) + 5 lim x2 g(x) [Limit Law 3] = 4 + 5(2) = 6 (b) lim x2 [g(x)]3 = lim x2 g(x) 3 [Limit Law 6] = ( 2)3 = 8 (c) lim x2 f(x) = lim x2 f(x) [Limit Law 11] = 4 = 2 38. SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS 47 (d) lim x2 3f(x) g(x) = lim x2 [3f(x)] lim x2 g(x) [Limit Law 5] = 3 lim x2 f(x) lim x2 g(x) [Limit Law 3] = 3(4) 2 = 6 (e) Because the limit of the denominator is 0, we cant use Limit Law 5. The given limit, lim x2 g(x) h(x) , does not exist because the denominator approaches 0 while the numerator approaches a nonzero number. (f ) lim x2 g(x) h(x) f(x) = lim x2 [g(x) h(x)] lim x2 f(x) [Limit Law 5] = lim x2 g(x) lim x2 h(x) lim x2 f(x) [Limit Law 4] = 2 0 4 = 0 3. lim x2 (3x4 + 2x2 x + 1) = lim x2 3x4 + lim x2 2x2 lim x2 x + lim x2 1 [Limit Laws 1 and 2] = 3 lim x2 x4 + 2 lim x2 x2 lim x2 x + lim x2 1 [3] = 3(2)4 + 2(2)2 (2) + (1) [9, 8, and 7] = 48 + 8 + 2 + 1 = 59 5. lim x8 (1 + 3 x ) (2 6x2 + x3 ) = lim x8 (1 + 3 x ) lim x8 (2 6x2 + x3 ) [Limit Law 4] = lim x8 1 + lim x8 3 x lim x8 2 6 lim x8 x2 + lim x8 x3 [1, 2, and 3] = 1 + 3 8 2 6 82 + 83 [7, 10, 9] = (3)(130) = 390 7. lim x1 1 + 3x 1 + 4x2 + 3x4 3 = lim x1 1 + 3x 1 + 4x2 + 3x4 3 [6] = lim x1 (1 + 3x) lim x1 (1 + 4x2 + 3x4) 3 [5] = lim x1 1 + 3 lim x1 x lim x1 1 + 4 lim x1 x2 + 3 lim x1 x4 3 [2, 1, and 3] = 1 + 3(1) 1 + 4(1)2 + 3(1)4 3 = 4 8 3 = 1 2 3 = 1 8 [7, 8, and 9] 9. lim x4 16 x2= lim x4 (16 x2) [11] = lim x4 16 lim x4 x2 [2] = 16 (4)2 = 0 [7 and 9] 39. 48 CHAPTER 2 LIMITS AND DERIVATIVES 11. lim x2 x2 + x 6 x 2 = lim x2 (x + 3)(x 2) x 2 = lim x2 (x + 3) = 2 + 3 = 5 13. lim x2 x2 x + 6 x 2 does not exist since x 2 0 but x2 x + 6 8 as x 2. 15. lim t3 t2 9 2t2 + 7t + 3 = lim t3 (t + 3)(t 3) (2t + 1)(t + 3) = lim t3 t 3 2t + 1 = 3 3 2(3) + 1 = 6 5 = 6 5 17. lim h0 (4 + h)2 16 h = lim h0 (16 + 8h + h2 ) 16 h = lim h0 8h + h2 h = lim h0 h(8 + h) h = lim h0 (8 + h) = 8 + 0 = 8 19. By the formula for the sum of cubes, we have lim x2 x + 2 x3 + 8 = lim x2 x + 2 (x + 2)(x2 2x + 4) = lim x2 1 x2 2x + 4 = 1 4 + 4 + 4 = 1 12 . 21. lim t9 9 t 3 t = lim t9 3 + t 3 t 3 t = lim t9 3 + t = 3 + 9 = 6 23. lim x7 x + 2 3 x 7 = lim x7 x + 2 3 x 7 x + 2 + 3 x + 2 + 3 = lim x7 (x + 2) 9 (x 7) x + 2 + 3 = lim x7 x 7 (x 7) x + 2 + 3 = lim x7 1 x + 2 + 3 = 1 9 + 3 = 1 6 25. lim x4 1 4 + 1 x 4 + x = lim x4 x + 4 4x 4 + x = lim x4 x + 4 4x(4 + x) = lim x4 1 4x = 1 4(4) = 1 16 27. lim x16 4 x 16x x2 = lim x16 (4 x )(4 + x ) (16x x2)(4 + x ) = lim x16 16 x x(16 x)(4 + x ) = lim x16 1 x(4 + x ) = 1 16 4 + 16 = 1 16(8) = 1 128 29. lim t0 1 t 1 + t 1 t = lim t0 1 1 + t t 1 + t = lim t0 1 1 + t 1 + 1 + t t t + 1 1 + 1 + t = lim t0 t t 1 + t 1 + 1 + t = lim t0 1 1 + t 1 + 1 + t = 1 1 + 0 1 + 1 + 0 = 1 2 31. (a) lim x0 x 1 + 3x 1 2 3 (b) x f(x) 0.001 0.6661663 0.0001 0.6666167 0.00001 0.6666617 0.000001 0.6666662 0.000001 0.6666672 0.00001 0.6666717 0.0001 0.6667167 0.001 0.6671663 The limit appears to be 2 3 . 40. SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS 49 (c) lim x0 x 1 + 3x 1 1 + 3x + 1 1 + 3x + 1 = lim x0 x 1 + 3x + 1 (1 + 3x) 1 = lim x0 x 1 + 3x + 1 3x = 1 3 lim x0 1 + 3x + 1 [Limit Law 3] = 1 3 lim x0 (1 + 3x) + lim x0 1 [1 and 11] = 1 3 lim x0 1 + 3 lim x0 x + 1 [1, 3, and 7] = 1 3 1 + 3 0 + 1 [7 and 8] = 1 3 (1 + 1) = 2 3 33. Let f(x) = x2 , g(x) = x2 cos 20x and h(x) = x2 . Then 1 cos 20x 1 x2 x2 cos 20x x2 f(x) g(x) h(x). So since lim x0 f(x) = lim x0 h(x) = 0, by the Squeeze Theorem we have lim x0 g(x) = 0. 35. We have lim x4 (4x 9) = 4(4) 9 = 7 and lim x4 x2 4x + 7 = 42 4(4) + 7 = 7. Since 4x 9 f(x) x2 4x + 7 for x 0, lim x4 f(x) = 7 by the Squeeze Theorem. 37. 1 cos(2/x) 1 x4 x4 cos(2/x) x4 . Since lim x0 x4 = 0 and lim x0 x4 = 0, we have lim x0 x4 cos(2/x) = 0 by the Squeeze Theorem. 39. |x 3| = x 3 if x 3 0 (x 3) if x 3 < 0 = x 3 if x 3 3 x if x < 3 Thus, lim x3+ (2x + |x 3|) = lim x3+ (2x + x 3) = lim x3+ (3x 3) = 3(3) 3 = 6 and lim x3 (2x + |x 3|) = lim x3 (2x + 3 x) = lim x3 (x + 3) = 3 + 3 = 6. Since the left and right limits are equal, lim x3 (2x + |x 3|) = 6. 41. 2x3 x2 = x2 (2x 1) = x2 |2x 1| = x2 |2x 1| |2x 1| = 2x 1 if 2x 1 0 (2x 1) if 2x 1 < 0 = 2x 1 if x 0.5 (2x 1) if x < 0.5 So 2x3 x2 = x2 [(2x 1)] for x < 0.5. Thus, lim x0.5 2x 1 |2x3 x2| = lim x0.5 2x 1 x2[(2x 1)] = lim x0.5 1 x2 = 1 (0.5)2 = 1 0.25 = 4. 41. 50 CHAPTER 2 LIMITS AND DERIVATIVES 43. Since |x| = x for x < 0, we have lim x0 1 x 1 |x| = lim x0 1 x 1 x = lim x0 2 x , which does not exist since the denominator approaches 0 and the numerator does not. 45. (a) (b) (i) Since sgn x = 1 for x > 0, lim x0+ sgn x = lim x0+ 1 = 1. (ii) Since sgn x = 1 for x < 0, lim x0 sgn x = lim x0 1 = 1. (iii) Since lim x0 sgn x 6= lim x0+ sgn x, lim x0 sgn x does not exist. (iv) Since |sgn x| = 1 for x 6= 0, lim x0 |sgn x| = lim x0 1 = 1. 47. (a) (i) lim x1+ F(x) = lim x1+ x2 1 |x 1| = lim x1+ x2 1 x 1 = lim x1+ (x + 1) = 2 (c) (ii) lim x1 F(x) = lim x1 x2 1 |x 1| = lim x1 x2 1 (x 1) = lim x1 (x + 1) = 2 (b) No, lim x1 F(x) does not exist since lim x1+ F(x) 6= lim x1 F(x). 49. (a) (i) [[x]] = 2 for 2 x < 1, so lim x2+ [[x]] = lim x2+ (2) = 2 (ii) [[x]] = 3 for 3 x < 2, so lim x2 [[x]] = lim x2 (3) = 3. The right and left limits are different, so lim x2 [[x]] does not exist. (iii) [[x]] = 3 for 3 x < 2, so lim x2.4 [[x]] = lim x2.4 (3) = 3. (b) (i) [[x]] = n 1 for n 1 x < n, so lim xn [[x]] = lim xn (n 1) = n 1. (ii) [[x]] = n for n x < n + 1, so lim xn+ [[x]] = lim xn+ n = n. (c) lim xa [[x]] exists a is not an integer. 51. The graph of f(x) = [[x]] + [[x]] is the same as the graph of g(x) = 1 with holes at each integer, since f(a) = 0 for any integer a. Thus, lim x2 f(x) = 1 and lim x2+ f (x) = 1, so lim x2 f(x) = 1. However, f(2) = [[2]] + [[2]] = 2 + (2) = 0, so lim x2 f(x) 6= f(2). 53. Since p(x) is a polynomial, p(x) = a0 + a1x + a2x2 + + anxn . Thus, by the Limit Laws, lim xa p(x) = lim xa a0 + a1x + a2x2 + + anxn = a0 + a1 lim xa x + a2 lim xa x2 + + an lim xa xn = a0 + a1a + a2a2 + + anan = p(a) Thus, for any polynomial p, lim xa p(x) = p(a). 42. SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 51 55. lim x1 [f(x) 8] = lim x1 f(x) 8 x 1 (x 1) = lim x1 f(x) 8 x 1 lim x1 (x 1) = 10 0 = 0. Thus, lim x1 f(x) = lim x1 {[f(x) 8] + 8} = lim x1 [f(x) 8] + lim x1 8 = 0 + 8 = 8. Note: The value of lim x1 f(x) 8 x 1 does not affect the answer since its multiplied by 0. Whats important is that lim x1 f(x) 8 x 1 exists. 57. Observe that 0 f(x) x2 for all x, and lim x0 0 = 0 = lim x0 x2 . So, by the Squeeze Theorem, lim x0 f(x) = 0. 59. Let f(x) = H(x) and g(x) = 1 H(x), where H is the Heaviside function dened in Exercise 1.3.57. Thus, either f or g is 0 for any value of x. Then lim x0 f(x) and lim x0 g(x) do not exist, but lim x0 [f(x)g(x)] = lim x0 0 = 0. 61. Since the denominator approaches 0 as x 2, the limit will exist only if the numerator also approaches 0 as x 2. In order for this to happen, we need lim x2 3x2 + ax + a + 3 = 0 3(2)2 + a(2) + a + 3 = 0 12 2a + a + 3 = 0 a = 15. With a = 15, the limit becomes lim x2 3x2 + 15x + 18 x2 + x 2 = lim x2 3(x + 2)(x + 3) (x 1)(x + 2) = lim x2 3(x + 3) x 1 = 3(2 + 3) 2 1 = 3 3 = 1. 2.4 The Precise Definition of a Limit 1. On the left side of x = 2, we need |x 2| < 10 7 2 = 4 7 . On the right side, we need |x 2| < 10 3 2 = 4 3 . For both of these conditions to be satised at once, we need the more restrictive of the two to hold, that is, |x 2| < 4 7 . So we can choose = 4 7 , or any smaller positive number. 3. The leftmost question mark is the solution of x = 1.6 and the rightmost, x = 2.4. So the values are 1.62 = 2.56 and 2.42 = 5.76. On the left side, we need |x 4| < |2.56 4| = 1.44. On the right side, we need |x 4| < |5.76 4| = 1.76. To satisfy both conditions, we need the more restrictive condition to holdnamely, |x 4| < 1.44. Thus, we can choose = 1.44, or any smaller positive number. 5. From the graph, we nd that tan x = 0.8 when x 0.675, so 4 1 0.675 1 4 0.675 0.1106. Also, tan x = 1.2 when x 0.876, so 4 + 2 0.876 2 = 0.876 4 0.0906. Thus, we choose = 0.0906 (or any smaller positive number) since this is the smaller of 1 and 2. 43. 52 CHAPTER 2 LIMITS AND DERIVATIVES 7. For = 1, the denition of a limit requires that we nd such that 4 + x 3x3 2 < 1 1 < 4 + x 3x3 < 3 whenever 0 < |x 1| < . If we plot the graphs of y = 1, y = 4 + x 3x3 and y = 3 on the same screen, we see that we need 0.86 x 1.11. So since |1 0.86| = 0.14 and |1 1.11| = 0.11, we choose = 0.11 (or any smaller positive number). For = 0.1, we must nd such that 4 + x 3x3 2 < 0.1 1.9 < 4 + x 3x3 < 2.1 whenever 0 < |x 1| < . From the graph, we see that we need 0.988 x 1.012. So since |1 0.988| = 0.012 and |1 1.012| = 0.012, we choose = 0.012 (or any smaller positive number) for the inequality to hold. 9. (a) From the graph, we nd that y = tan2 x = 1000 when x 1.539 and x 1.602 for x near 2 . Thus, we get 1.602 2 0.031 for M = 1000. (b) From the graph, we nd that y = tan2 x = 10,000 when x 1.561 and x 1.581 for x near 2 . Thus, we get 1.581 2 0.010 for M = 10,000. 11. (a) A = r2 and A = 1000 cm2 r2 = 1000 r2 = 1000 r = 1000 (r > 0) 17.8412 cm. (b) |A 1000| 5 5 r2 1000 5 1000 5 r2 1000 + 5 995 r 1005 17.7966 r 17.8858. 1000 995 0.04466 and 1005 1000 0.04455. So if the machinist gets the radius within 0.0445 cm of 17.8412, the area will be within 5 cm2 of 1000. (c) x is the radius, f(x) is the area, a is the target radius given in part (a), L is the target area (1000), is the tolerance in the area (5), and is the tolerance in the radius given in part (b). 13. (a) |4x 8| = 4 |x 2| < 0.1 |x 2| < 0.1 4 , so = 0.1 4 = 0.025. (b) |4x 8| = 4 |x 2| < 0.01 |x 2| < 0.01 4 , so = 0.01 4 = 0.0025. 44. SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 53 15. Given > 0, we need > 0 such that if 0 < |x 1| < , then |(2x + 3) 5| < . But |(2x + 3) 5| < |2x 2| < 2 |x 1| < |x 1| < /2. So if we choose = /2, then 0 < |x 1| < |(2x + 3) 5| < . Thus, lim x1 (2x + 3) = 5 by the denition of a limit. 17. Given > 0, we need > 0 such that if 0 < |x (3)| < , then |(1 4x) 13| < . But |(1 4x) 13| < |4x 12| < |4| |x + 3| < |x (3)| < /4. So if we choose = /4, then 0 < |x (3)| < |(1 4x) 13| < . Thus, lim x3 (1 4x) = 13 by the denition of a limit. x 19. Given > 0, we need > 0 such that if 0 < |x 3| < , then x 5 3 5 < 1 5 |x 3| < |x 3| < 5. So choose = 5. Then 0 < |x 3| < |x 3| < 5 |x 3| 5 < x 5 3 5 < . By the denition of a limit, lim x3 x 5 = 3 5 . 21. Given > 0, we need > 0 such that if 0 < |x 2| < , then x2 + x 6 x 2 5 < (x + 3)(x 2) x 2 5 < |x + 3 5| < [x 6= 2] |x 2| < . So choose = . Then 0 < |x 2| < |x 2| < |x + 3 5| < (x + 3)(x 2) x 2 5 < [x 6= 2] x2 + x 6 x 2 5 < . By the denition of a limit, lim x2 x2 + x 6 x 2 = 5. 23. Given > 0, we need > 0 such that if 0 < |x a| < , then |x a| < . So = will work. 25. Given > 0, we need > 0 such that if 0 < |x 0| < , then x2 0 < x2 < |x| < . Take = . Then 0 < |x 0| < x2 0 < . Thus, lim x0 x2 = 0 by the denition of a limit. 27. Given > 0, we need > 0 such that if 0 < |x 0| < , then |x| 0 < . But |x| = |x|. So this is true if we pick = . Thus, lim x0 |x| = 0 by the denition of a limit. 45. 54 CHAPTER 2 LIMITS AND DERIVATIVES 29. Given > 0, we need > 0 such that if 0 < |x 2| < , then x2 4x + 5 1 < x2 4x + 4 < (x 2)2 < . So take = . Then 0 < |x 2| < |x 2| < (x 2)2 < . Thus, lim x2 x2 4x + 5 = 1 by the denition of a limit. 31. Given > 0, we need > 0 such that if 0 < |x (2)| < , then x2 1 3 < or upon simplifying we need x2 4 < whenever 0 < |x + 2| < . Notice that if |x + 2| < 1, then 1 < x + 2 < 1 5 < x 2 < 3 |x 2| < 5. So take = min {/5, 1}. Then 0 < |x + 2| < |x 2| < 5 and |x + 2| < /5, so x2 1 3 = |(x + 2)(x 2)| = |x + 2| |x 2| < (/5)(5) = . Thus, by the denition of a limit, lim x2 (x2 1) = 3. 33. Given > 0, we let = min 2, 8 . If 0 < |x 3| < , then |x 3| < 2 2 < x 3 < 2 4 < x + 3 < 8 |x + 3| < 8. Also |x 3| < 8 , so x2 9 = |x + 3| |x 3| < 8 8 = . Thus, lim x3 x2 = 9. 35. (a) The points of intersection in the graph are (x1, 2.6) and (x2, 3.4) with x1 0.891 and x2 1.093. Thus, we can take to be the smaller of 1 x1 and x2 1. So = x2 1 0.093. (b) Solving x3 + x + 1 = 3 + gives us two nonreal complex roots and one real root, which is x() = 216 + 108 + 12 336 + 324 + 812 2/3 12 6 216 + 108 + 12 336 + 324 + 812 1/3 . Thus, = x() 1. (c) If = 0.4, then x() 1.093 272 342 and = x() 1 0.093, which agrees with our answer in part (a). 37. 1. Guessing a value for Given > 0, we must nd > 0 such that | x a| < whenever 0 < |x a| < . But | x a| = |x a| x + a < (from the hint). Now if we can nd a positive constant C such that x + a > C then |x a| x + a < |x a| C < , and we take |x a| < C. We can nd this number by restricting x to lie in some interval centered at a. If |x a| < 1 2 a, then 1 2 a < x a < 1 2 a 1 2 a < x < 3 2 a x + a > 1 2 a + a, and so C = 1 2 a + a is a suitable choice for the constant. So |x a| < 1 2 a + a . This suggests that we let = min 1 2 a, 1 2 a + a . 2. Showing that works Given > 0, we let = min 1 2 a, 1 2 a + a . If 0 < |x a| < , then |x a| < 1 2 a x + a > 1 2 a + a (as in part 1). Also |x a| < 1 2 a + a , so | x a | = |x a| x + a < a/2 + a a/2 + a = . Therefore, lim xa x = a by the denition of a limit. 46. SECTION 2.5 CONTINUITY 55 39. Suppose that lim x0 f(x) = L. Given = 1 2 , there exists > 0 such that 0 < |x| < |f(x) L| < 1 2 . Take any rational number r with 0 < |r| < . Then f(r) = 0, so |0 L| < 1 2 , so L |L| < 1 2 . Now take any irrational number s with 0 < |s| < . Then f(s) = 1, so |1 L| < 1 2 . Hence, 1 L < 1 2 , so L > 1 2 . This contradicts L < 1 2 , so lim x0 f(x) does not exist. 41. 1 (x + 3)4 > 10,000 (x + 3)4 < 1 10,000 |x + 3| < 1 4 10,000 |x (3)| < 1 10 43. Given M < 0 we need > 0 so that ln x < M whenever 0 < x < ; that is, x = eln x < eM whenever 0 < x < . This suggests that we take = eM . If 0 < x < eM , then ln x < ln eM = M. By the denition of a limit, lim x0+ ln x = . 2.5 Continuity 1. From Denition 1, lim x4 f(x) = f(4). 3. (a) The following are the numbers at which f is discontinuous and the type of discontinuity at that number: 4 (removable), 2 ( jump), 2 (jump), 4 (innite). (b) f is continuous from the left at 2 since lim x2 f(x) = f(2). f is continuous from the right at 2 and 4 since lim x2+ f(x) = f(2) and lim x4+ f(x) = f(4). It is continuous from neither side at 4 since f(4) is undened. 5. The graph of y = f(x) must have a discontinuity at x = 3 and must show that lim x3 f(x) = f(3). 7. (a) (b) There are discontinuities at times t = 1, 2, 3, and 4. A person parking in the lot would want to keep in mind that the charge will jump at the beginning of each hour. 9. Since f and g are continuous functions, lim x3 [2f(x) g(x)] = 2 lim x3 f(x) lim x3 g(x) [by Limit Laws 2 and 3] = 2f(3) g(3) [by continuity of f and g at x = 3] = 2 5 g(3) = 10 g(3) Since it is given that lim x3 [2f(x) g(x)] = 4, we have 10 g(3) = 4, so g(3) = 6. 47. 56 CHAPTER 2 LIMITS AND DERIVATIVES 11. lim x1 f(x) = lim x1 x + 2x3 4 = lim x1 x + 2 lim x1 x3 4 = 1 + 2(1)3 4 = (3)4 = 81 = f(1). By the denition of continuity, f is continuous at a = 1. 13. For a > 2, we have lim xa f(x) = lim xa 2x + 3 x 2 = lim xa (2x + 3) lim xa (x 2) [Limit Law 5] = 2 lim xa x + lim xa 3 lim xa x lim xa 2 [1, 2, and 3] = 2a + 3 a 2 [7 and 8] = f(a) Thus, f is continuous at x = a for every a in (2, ); that is, f is continuous on (2, ). 15. f(x) = ln |x 2| is discontinuous at 2 since f(2) = ln 0 is not dened. 17. f(x) = ex if x < 0 x2 if x 0 The left-hand limit of f at a = 0 is lim x0 f(x) = lim x0 ex = 1. The right-hand limit of f at a = 0 is lim x0+ f(x) = lim x0+ x2 = 0. Since these limits are not equal, lim x0 f(x) does not exist and f is discontinuous at 0. 19. f(x) = cos x if x < 0 0 if x = 0 1 x2 if x > 0 lim x0 f(x) = 1, but f(0) = 0 6= 1, so f is discontinuous at 0. 21. F(x) = x x2 + 5x + 6 is a rational function. So by Theorem 5 (or Theorem 7), F is continuous at every number in its domain, x | x2 + 5x + 6 6= 0 = {x | (x + 3)(x + 2) 6= 0} = {x | x 6= 3, 2} or (, 3) (3, 2) (2, ). 23. By Theorem 5, the polynomials x2 and 2x 1 are continuous on (, ). By Theorem 7, the root function x is continuous on [0, ). By Theorem 9, the composite function 2x 1 is continuous on its domain, [1 2 , ). By part 1 of Theorem 4, the sum R(x) = x2 + 2x 1 is continuous on [1 2 , ). 48. SECTION 2.5 CONTINUITY 57 25. By Theorem 7, the exponential function e5t and the trigonometric function cos 2t are continuous on (, ). By part 4 of Theorem 4, L(t) = e5t cos 2t is continuous on (, ). 27. By Theorem 5, the polynomial t4 1 is continuous on (, ). By Theorem 7, ln x is continuous on its domain, (0, ). By Theorem 9, ln t4 1 is continuous on its domain, which is t | t4 1 > 0 = t | t4 > 1 = {t | |t| > 1} = (, 1) (1, ) 29. The function y = 1 1 + e1/x is discontinuous at x = 0 because the left- and right-hand limits at x = 0 are different. 31. Because we are dealing with root functions, 5 + x is continuous on [0, ), x + 5 is continuous on [5, ), so the quotient f(x) = 5 + x 5 + x is continuous on [0, ). Since f is continuous at x = 4, lim x4 f(x) = f(4) = 7 3 . 33. Because x2 x is continuous on R, the composite function f(x) = ex2x is continuous on R, so lim x1 f(x) = f(1) = e1 1 = e0 = 1. 35. f(x) = x2 if x < 1 x if x 1 By Theorem 5, since f(x) equals the polynomial x2 on (, 1), f is continuous on (, 1). By Theorem 7, since f(x) equals the root function x on (1, ), f is continuous on (1, ). At x = 1, lim x1 f(x) = lim x1 x2 = 1 and lim x1+ f(x) = lim x1+ x = 1. Thus, lim x1 f(x) exists and equals 1. Also, f(1) = 1 = 1. Thus, f is continuous at x = 1. We conclude that f is continuous on (, ). 37. f(x) = 1 + x2 if x 0 2 x if 0 < x 2 (x 2)2 if x > 2 f is continuous on (, 0), (0, 2), and (2, ) since it is a polynomial on each of these intervals. Now lim x0 f(x) = lim x0 (1 + x2 ) = 1 and lim x0+ f(x) = lim x0+ (2 x) = 2, so f is discontinuous at 0. Since f(0) = 1, f is continuous from the left at 0. Also, lim x2 f(x) = lim x2 (2 x) = 0, lim x2+ f(x) = lim x2+ (x 2)2 = 0, and f(2) = 0, so f is continuous at 2. The only number at which f is discontinuous is 0. 49. 58 CHAPTER 2 LIMITS AND DERIVATIVES 39. f(x) = x + 2 if x < 0 ex if 0 x 1 2 x if x > 1 f is continuous on (, 0) and (1, ) since on each of these intervals it is a polynomial; it is continuous on (0, 1) since it is an exponential. Now lim x0 f(x) = lim x0 (x + 2) = 2 and lim x0+ f(x) = lim x0+ ex = 1, so f is discontinuous at 0. Since f(0) = 1, f is continuous from the right at 0. Also lim x1 f(x) = lim x1 ex = e and lim x1+ f(x) = lim x1+ (2 x) = 1, so f is discontinuous at 1. Since f(1) = e, f is continuous from the left at 1. 41. f(x) = cx2 + 2x if x < 2 x3 cx if x 2 f is continuous on (, 2) and (2, ). Now lim x2 f(x) = lim x2 cx2 + 2x = 4c + 4 and lim x2+ f(x) = lim x2+ x3 cx = 8 2c. So f is continuous 4c + 4 = 8 2c 6c = 4 c = 2 3 . Thus, for f to be continuous on (, ), c = 2 3 . 43. (a) f(x) = x4 1 x 1 = (x2 + 1)(x2 1) x 1 = (x2 + 1)(x + 1)(x 1) x 1 = (x2 + 1)(x + 1) [or x3 + x2 + x + 1] for x 6= 1. The discontinuity is removable and g(x) = x3 + x2 + x + 1 agrees with f for x 6= 1 and is continuous on R. (b) f(x) = x3 x2 2x x 2 = x(x2 x 2) x 2 = x(x 2)(x + 1) x 2 = x(x + 1) [or x2 + x] for x 6= 2. The discontinuity is removable and g(x) = x2 + x agrees with f for x 6= 2 and is continuous on R. (c) lim x f(x) = lim x [[sin x]] = lim x 0 = 0 and lim x+ f(x) = lim x+ [[sin x]] = lim x+ (1) = 1, so lim x f(x) does not exist. The discontinuity at x = is a jump discontinuity. 45. f(x) = x2 + 10 sin x is continuous on the interval [31, 32], f(31) 957, and f(32) 1030. Since 957 < 1000 < 1030, there is a number c in (31, 32) such that f(c) = 1000 by the Intermediate Value Theorem. Note: There is also a number c in (32, 31) such that f(c) = 1000. 47. f(x) = x4 + x 3 is continuous on the interval [1, 2], f(1) = 1, and f(2) = 15. Since 1 < 0 < 15, there is a number c in (1, 2) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation x4 + x 3 = 0 in the interval (1, 2). 49. f(x) = cos x x is continuous on the interval [0, 1], f(0) = 1, and f(1) = cos 1 1 0.46. Since 0.46 < 0 < 1, there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x x = 0, or cos x = x, in the interval (0, 1). 50. SECTION 2.5 CONTINUITY 59 51. (a) f(x) = cos x x3 is continuous on the interval [0, 1], f(0) = 1 > 0, and f(1) = cos 1 1 0.46 < 0. Since 1 > 0 > 0.46, there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x x3 = 0, or cos x = x3 , in the interval (0, 1). (b) f(0.86) 0.016 > 0 and f(0.87) 0.014 < 0, so there is a root between 0.86 and 0.87, that is, in the interval (0.86, 0.87). 53. (a) Let f(x) = 100ex/100 0.01x2 . Then f(0) = 100 > 0 and f(100) = 100e1 100 63.2 < 0. So by the Intermediate Value Theorem, there is a number c in (0, 100) such that f(c) = 0. This implies that 100ec/100 = 0.01c2 . (b) Using the intersect feature of the graphing device, we nd that the root of the equation is x = 70.347, correct to three decimal places. 55. () If f is continuous at a, then by Theorem 8 with g(h) = a + h, we have lim h0 f(a + h) = f lim h0 (a + h) = f(a). () Let > 0. Since lim h0 f(a + h) = f(a), there exists > 0 such that 0 < |h| < |f(a + h) f(a)| < . So if 0 < |x a| < , then |f(x) f(a)| = |f(a + (x a)) f(a)| < . Thus, lim xa f(x) = f(a) and so f is continuous at a. 57. As in the previous exercise, we must show that lim h0 cos(a + h) = cos a to prove that the cosine function is continuous. lim h0 cos(a + h) = lim h0 (cos a cos h sin a sin h) = lim h0 (cos a cos h) lim h0 (sin a sin h) = lim h0 cos a lim h0 cos h lim h0 sin a lim h0 sin h = (cos a)(1) (sin a)(0) = cos a 59. f(x) = 0 if x is rational 1 if x is irrational is continuous nowhere. For, given any number a and any > 0, the interval (a , a + ) contains both innitely many rational and innitely many irrational numbers. Since f(a) = 0 or 1, there are innitely many numbers x with 0 < |x a| < and |f(x) f(a)| = 1. Thus, lim xa f(x) 6= f(a). [In fact, lim xa f(x) does not even exist.] 61. If there is such a number, it satises the equation x3 + 1 = x x3 x + 1 = 0. Let the left-hand side of this equation be called f(x). Now f(2) = 5 < 0, and f(1) = 1 > 0. Note also that f(x) is a polynomial, and thus continuous. So by the Intermediate Value Theorem, there is a number c between 2 and 1 such that f(c) = 0, so that c = c3 + 1. 63. f(x) = x4 sin(1/x) is continuous on (, 0) (0, ) since it is the product of a polynomial and a composite of a trigonometric function and a rational function. Now since 1 sin(1/x) 1, we have x4 x4 sin(1/x) x4 . Because 51. 60 CHAPTER 2 LIMITS AND DERIVATIVES lim x0 (x4 ) = 0 and lim x0 x4 = 0, the Squeeze Theorem gives us lim x0 (x4 sin(1/x)) = 0, which equals f(0). Thus, f is continuous at 0 and, hence, on (, ). 65. Dene u(t) to be the monks distance from the monastery, as a function of time, on the rst day, and dene d(t) to be his distance from the monastery, as a function of time, on the second day. Let D be the distance from the monastery to the top of the mountain. From the given information we know that u(0) = 0, u(12) = D, d(0) = D and d(12) = 0. Now consider the function u d, which is clearly continuous. We calculate that (u d)(0) = D and (u d)(12) = D. So by the Intermediate Value Theorem, there must be some time t0 between 0 and 12 such that (u d)(t0) = 0 u(t0) = d(t0). So at time t0 after 7:00 AM, the monk will be at the same place on both days. 2.6 Limits at Infinity; Horizontal Asymptotes 1. (a) As x becomes large, the values of f(x) approach 5. (b) As x becomes large negative, the values of f(x) approach 3. 3. (a) lim x2 f(x) = (b) lim x1 f(x) = (c) lim x1+ f(x) = (d) lim x f(x) = 1 (e) lim x f(x) = 2 (f ) Vertical: x = 1, x = 2; Horizontal: y = 1, y = 2 5. f(0) = 0, f(1) = 1, lim x f(x) = 0, f is odd 7. lim x2 f(x) = , lim x f(x) = , lim x f(x) = 0, lim x0+ f(x) = , lim x0 f(x) = 9. f(0) = 3, lim x0 f(x) = 4, lim x0+ f(x) = 2, lim x f(x) = , lim x4 f(x) = , lim x4+ f(x) = , lim x f(x) = 3 11. If f(x) = x2 /2x , then a calculator gives f(0) = 0, f(1) = 0.5, f(2) = 1, f(3) = 1.125, f(4) = 1, f(5) = 0.78125, f(6) = 0.5625, f(7) = 0.3828125, f(8) = 0.25, f(9) = 0.158203125, f(10) = 0.09765625, f(20) 0.00038147, f(50) 2.2204 1012 , f(100) 7.8886 1027 . It appears that lim x x2 /2x = 0. 52. SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 61 13. lim x 3x2 x + 4 2x2 + 5x 8 = lim x (3x2 x + 4)/x2 (2x2 + 5x 8)/x2 [divide both the numerator and denominator by x2 (the highest power of x thatappears in the denominator)] = lim x (3 1/x + 4/x2 ) lim x (2 + 5/x 8/x2) [Limit Law 5] = lim x 3 lim x (1/x) + lim x (4/x2 ) lim x 2 + lim x (5/x) lim x (8/x2) [Limit Laws 1 and 2] = 3 lim x (1/x) + 4 lim x (1/x2 ) 2 + 5 lim x (1/x) 8 lim x (1/x2) [Limit Laws 7 and 3] = 3 0 + 4(0) 2 + 5(0) 8(0) [Theorem 5 of Section 2.5] = 3 2 15. lim x 1 2x + 3 = lim x 1/x (2x + 3)/x = lim x (1/x) lim x (2 + 3/x) = lim x (1/x) lim x 2 + 3 lim x (1/x) = 0 2 + 3(0) = 0 2 = 0 17. lim x 1 x x2 2x2 7 = lim x (1 x x2 )/x2 (2x2 7)/x2 = lim x (1/x2 1/x 1) lim x (2 7/x2) = lim x (1/x2 ) lim x (1/x) lim x 1 lim x 2 7 lim x (1/x2) = 0 0 1 2 7(0) = 1 2 19. Divide both the numerator and denominator by x3 (the highest power of x that occurs in the denominator). lim x x3 + 5x 2x3 x2 + 4 = lim x x3 + 5x x3 2x3 x2 + 4 x3 = lim x 1 + 5 x2 2 1 x + 4 x3 = lim x 1 + 5 x2 lim x 2 1 x + 4 x3 = lim x 1 + 5 lim x 1 x2 lim x 2 lim x 1 x + 4 lim x 1 x3 = 1 + 5(0) 2 0 + 4(0) = 1 2 21. First, multiply the factors in the denominator. Then divide both the numerator and denominator by u4 . lim u 4u4 + 5 (u2 2)(2u2 1) = lim u 4u4 + 5 2u4 5u2 + 2 = lim u 4u4 + 5 u4 2u4 5u2 + 2 u4 = lim u 4 + 5 u4 2 5 u2 + 2 u4 = lim u 4 + 5 u4 lim u 2 5 u2 + 2 u4 = lim u 4 + 5 lim u 1 u4 lim u 2 5 lim u 1 u2 + 2 lim u 1 u4 = 4 + 5(0) 2 5(0) + 2(0) = 4 2 = 2