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Ways to Represent a Function - Assignment Example

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This assignment "Ways to Represent a Function" presents the curve that 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. Yes, the curve is the graph of a function because it passes the Vertical Line Test…
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Ways to Represent a Function
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Calculus Assignment Chapters 1 Exercise 69 ODD only. Four ways to represent a function a) The point 2) is on graph, therefore, f(-1) = -2. (b) For, x = 2, y is about 2.8, therefore, f(2) ≈ 2.8. (c) f(x) = 2 is equivalent to y = 2. y = 2 is for x = -3, and x = 1. (d) Estimates for x when y = 0 are x = -2.5, and x = 0.3. (e) The domain for f consists of all x-values on the graph of f. The Domain is , or [-3,3]. The range for f consists of all y-values on the graph of f. The range is , 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, The lowest point occurs at about (t,a) = (12, -85). The highest point occurs at about (17, 115). Therefore, the range of the vertical ground acceleration function at USC during the Northridge earthquake is 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 . 9. The person’s weight increases from birth and it is about 160 pounds at age 20 and stayed fairly constant for next 10 years (20 to 30 years). The person’s weight dropped to about 125 pounds for the next 5 years, and then increased rapidly to about 175 pounds. The next 30 years saw a gradual increase to about 200 pounds. Possible reasons for the drop in person’s weight at 30 years of age may be diet, exercise, or 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. This graph will depend on the geographical location. 15. As the amount of a particular brand of coffee increases, the price will also increases, therefore, the graph will be linear as shown below. 17. The grass will grow till Wednesday afternoon when the homeowner mows it. 19. (a) Graph of N as a function of t is given below. (b) The number of cell-phone subscriber at midyear in 1995 and 1999 are approximately 110 million, and 495 million, respectively. 21. 23. 25. 27. is defined for all x except when 3x – 1 = 0 or x = 1/3, therefore, the domain is . 29. is defined when . These values of t give real number result for , whereas any value of t gives areal number result for . Therefore the domain is . 31. is defined when > 0 or x(x-5) > 0. ≠ 0since it would result in division by zero. The expression x(x-5) is positive if x < 0 or x > 5. Therefore, the domain is . 33. is defined for all real numbers, so the domain is R, or . The graph of f is a horizontal line with y-intercept equal to 5. 35. is defined for all real numbers, so the domain is R, or . The graph of f is a parabola opening upward since the coefficient of is positive. The t-intercepts are t = 0 and t = 6 or at points (0, 0) and (6, 0).. The t-coordinate of the vertex is halfway between the t-intercepts, that is at t = 3. Since , therefore the vertex is (3, -6). 37. is defined when or , so the domain is . The graph of the is the top half of a parabola as shown below. 39. is not defined for x = 0. The domain is . Since , therefore, The graph of G is given below. 41. is defined for all real numbers, so the domain is R, or . 43. is defined for all real numbers, so the domain is R, or . 45. The points are (1, -3) and (5, 7) The slope m of a line between the two points, An equation of the line connecting those two points is Therefore, the function is , 47. Solving given equation for y 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, the bottom half of the parabola is . The domain is. 49. For the first part of the line the slope is An equation of the line for the first part is For the second part of the line the slope is An equation of the line for the second part is Therefore, the function is 51. Let the length and width of the rectangle be L and W. Then the perimeter is 2L + 2W = 20 (1) The area is A = LW. (2) From equation (1) W = 10 – L Thus, A = LW = L(10-L) Since lengths are positive, the domain of A is 0 < L < 10. If L is larger than W, then 5 < L < 10 would be 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 satisfies The area of the triangle, The area of an equilateral triangle as a function of the length of a side is, 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, therefore, The surface area is The surface area of the box as a function of the length of a side of the base is , with domain x > 0. 57. The height of the box is x and the length and width are L = 20 – 2x, W = 12 – 2x Therefore, volume is V = LWx The side L, W, and x must be positive. Thus, L > 0 or 20- 2x > 0 or x < 10 W > 0 or 12 – 2x> 0 or x < 6 and x > 0 Combining above restrictions, the domain is 0 < x < 6 and the volume is 59. (a) The graph of the tax rate R as a function of the income I is given below. (b) On $14,000, tax is assessed on $4,000 that is equal to 0.1($4000) = $400. On $26,000, tax is assessed on $16,000 that is equal to 0.1($10000) + 0.15($6000) = $1900. (c) Up to $10,000 income, assessed tax is 0. Over $10,000 and up to $20,000 the assessed tax is $1,000. The tax on $30,000 is $2,500. Therefore, the graph of T for x > 20,000 is the line with initial point (20000, 1000) that passes through (30000, 2500). The graph of total assessed tax T as a function of the income I is given below 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. So f is an odd function. 67. Since, this is neither nor , therefore, the function f is neither odd nor even function. 69. Since, this is neither nor , therefore, the function f is neither odd nor even function. Read More
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