# test4a - MAC 2233/0001-006 Business Calculus Fall 2011...

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MAC 2233/0001-006 Business Calculus, Fall 2011 Final Exam Name: Date: 12/06/2011 Time: 10:00am-12:00nn Section: Show ALL steps. One hundred points equal 100% Question 1 . (4 points) The following diagram shows the graphs of eight equations. Match each of the following equations with its graph. (Explanation is not needed.) (a) y = | ln( x ) | . Answer: C (b) y = ln(e x + 1). Answer: A (c) y = 4 e - 2 x +2 . Answer: D (d) y = x 2 + 2 x + 5 x 3. Answer: B Question 2 . (2+3+4 points) Mr. Smith opened his first restaurant in 2010, and he then expanded his operations by opening a branch 10 months later. The graph on the right shows the monthly revenue R from his operations, where t is the number of months since he opened his first restaurant. (a) Explain why the discontinuity at t = 10 is not remov- able? Solution. The discontinuity is not removable because the left-sided limit is different from the right-sided limit . (Note. This discontinuity is “intrinsic” in Mr. Smith’s operations because of the addition of a new eatery. The discontinuity is not a measurement error and should not be “removed.”) (b) Estimate the average rate of change of R from t = 0 to t = 7. Solution. Over the period from t = 0 to t = 7, we find that Δ R = 8 2 = 6, and Δ t = 7 0. The average rate of change of R over this period is Δ R Δ t = 6 7 0 . 8571 thousand dollars per month . (c) Estimate the instantaneous rate of change of R at t = 8. Solution. We draw a tangent line at t = 8, and find two convenient points are the tangent line. The points are (0 , 3) and (15 , 14). Thus, the instantaneous rate of change of R at t = 8 is Δ R Δ t = 14 3 15 0 = 0 . 733 thousand dollars per month . Keeping Scores : Question 1 2 3 4 5 6 7 8 9 10 11 12 13 Total Score Out of 4 9 14 19 11 10 6 8 8 8 6 9 10 122 1

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Question 3 . (2+4+4+4 points) Find the limits below analytically. If the limit does not exist, check if it is + or −∞ . Show all work. (a) lim x 1 ( x 3 + ln( x ) e x 1 + x ). Solution. Since the function is continuous at x = 1 (as each term in the function is continuous at x = 1), direct substitution works, and it gives lim x 1 ( x 3 + ln( x ) e x 1 + x ) = 1 3 + ln(1) e 1 1 + 1 = 1 . (b) lim x 2 - x 2 x 2 4 . Solution. Direct substitution (with x = 2) gives 4 0 . Hence the limit does not exist . To check if it is + or −∞ , we see that as x 2 , we have x 2 4 0 . Thus x 2 x 2 4 < 0 as x 2 . Hence, lim x 2 - x 2 x 2 4 = −∞ . (c) lim Δ x 0 x + Δ x x Δ x . Solution. Direct substitution gives 0 / 0, and we need to rationalize the numerator: x + Δ x x Δ x = x + Δ x x Δ x × x + Δ x + x x + Δ x + x = ( x + Δ x ) x Δ x ( x + Δ x + x ) = 1 x + Δ x + x . Thus, lim Δ x 0 x + Δ x x Δ x = lim Δ x 0 1 x + Δ x + x = 1 2 x .
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