A b 0 and 0 16 b 16 c only c 0 16 only d

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Unformatted text preview: ge is positive. (A) (b, 0) and (0, 16) (B) (16, c) only (C) (0, 16) only (D) (b, 0) and (16, c) (E) (0, 16) and (16, c) (ii) Choose the interval(s) on which the rate of change is negative. (A) (b, 0) and (0, 16) (B) (16, c) only (C) (0, 16) only (D) (b, 0) and (16, c) (E) (0, 16) and (16, c) (iii) Choose the values at which the rate of change is 0. (A) x = 0 and x = 19 (B) x = 0 and x = 16 (D) x = 19 and x = 16 (C) x = 16 and x = −2 (E) x = 0 and x = −2 62. For f (x) = x2 + x, find the equation of the tangent line when x = −4. The tangent line is (A) y = −4(x − 7) + 12 (B) y = −7(x − 4) − 20 (D) y = −7(x + 4) + 12 (E) does not exist (C) y = −4(x + 7) − 20 63. Suppose the demand for a certain item is given by D(p) = −3p2 − 5p + 200, where p represents the price of the item in dollars. Answer parts (i) and (ii) (i) Find the rate of change of demand with respect to price. The rate of change with respect to price is (A) −6p − 5 (B) −3p2 − 5p + 200 (D) −6p + 195 (E) None of these (C) −3p (ii) The rate of change of demand when the price is $11 is -71. Choose the correct interpretation below. (A) When the price is $11, demand is decreasing at a rate of about 71 items for each increase in price of $11. (B) When the price is $11, demand is increasing at a rate of about 71 items for each increase in price of $11. (C) When the price is $11, demand is increasing at a rate of about 71 items for each increase in price of $1. (D) When the price is $11, demand is decreasing at a rate of about 71 items for each increase in price of $1. 64. The cost of recycling q tons of paper is given in the following table. q (tons) 1000 1500 2000 2500 3000 3500 C (q ) (dollars) 2500 3200 3640 4060 4270 4415 Estimate the marginal cost at q = 2000. (A) less than 1 (B) between 1 and 3 (D) between 5 and 7 (E) more than 7 (C) between 3 and 5 65. If f (x) = x2 + 3, use the definition of the derivative to find f ￿ (x). Formulas You Might Find Useful I = P rt A = ￿ r ￿mt P 1+ m A = P ert rE = ￿ rE = er − 1 f ￿ ( x) = f ( x + h) − f ( x ) h→0 h 1+ lim r ￿m −1 m...
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This note was uploaded on 01/14/2014 for the course MATH 116 taught by Professor Jess during the Fall '09 term at University of Arizona- Tucson.

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