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MATH 226: Calculus I MIDTERM II Fall 2015 NAME : SOLUTIONS NOTE : There are 5 problems on this midterm (total of 6 pages). Use of calculators will NOT be permitted. In order to receive full credit for any problem, you must show work leading to your answer. You have 50 minutes to complete this test. Problem Possible points 1 20 2 20 3 20 4 20 5 20 Total 100 Score BONUS (5pts): What is the anti-derivative of f (x) = tan−1 x + C 1 ? 1 + x2 MATH 226 – MIDTERM II Page 2 Problem 1. (20pts) Find the absolute maximum and minimum values of f (x) = x2 − 2 ln x on the interval 1 ≤ x ≤ 3. Solving 2 =0 x 2(x2 − 1) =0 x 2(x − 1)(x + 1) =0 x f 0 (x) = 2x − gives x = 1 is the only critical point of f (x), since −1 and 0 are not in the domain. Since f 0 (2) = 3 > 0, f is increasing on [1, 3]. This means f (1) = 1 is the absolute minimum of f while f (3) = 9 − 2 ln 3 is the absolute maximum. MATH 226 – MIDTERM II Page 3 Problem 2. (20pts) Find all local extrema, intervals of increase and decrease, and identify any points of inflection. f (x) = x4 − 4x2 = x2 (x2 − 4) Solving f 0 (x) = 4x3 − 8x = 0 4x(x2 − 2) = 0 √ √ 4x(x − 2)(x + 2) = 0 √ gives 3 critical points at x = 0, ± 2. Since f 0 (−2) = −16 < 0, f (1) = −4 < 0, f (−1) = 4 > 0, f (2) = 16 > 0, √ √ 2] and [0, 2] f is decreasing on (−∞, − √ √ and increasing on [− 2, 0] and [ 2, ∞). p Since f 00 (x) = 12x2 − 8 = 0 when x = ± 2/3, and f 00 (−1) = 4 = f 00 (1) > 0 and f 00 (0) = −8 < 0, both points where f 00 (x) = 0 are inflection points. p They have a common value of f (± 2/3) = −20/9. p Hence, there are 2 inflection points (± 2/3, −20/9). MATH 226 – MIDTERM II Page 4 Problem 3. (20pts) A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions? The equation of the semicircle maybe be represented as x2 + y 2 = 4, y ≥ 0. Let (x, y) be the corner of the inscribed √ rectangle, with 0 < x < 2. Since its base is 2x and height is y = 4 − x2 , its area is √ A = 2x 4 − x2 . Since, √ 1 dA = 2 4 − x2 + (2x) √ · (−2x) dx 2 4 − x2 √ 2x2 = 2 4 − x2 − √ 4 − x2 2(4 − x2 ) − 2x2 √ = 4 − x2 8 − 4x2 =√ 4 − x2 4(2 − x2 ) = √ =0 4 − x2 √ √ when x = ± 2, the only critical point with x > 0 is x = 2. At this critical point, the area √ A = 4, which is the absolute maximum √ 0 0 because A (x) > 0 for 0 < x < 2 and A (x) < 0 for 2 < x < 2. Alternatively, one can deduce A = 4 is an absolute maximum by checking √ −x 4 − x2 (−8x) − (8 − 4x2 ) √4−x 2 A00 (x) = 2 4−x 2 (4 − x )(−8x) − (8 − 4x2 )(−x) = (4 − x2 )3/2 4x3 − 24x 4x(x2 − 6) = = <0 (4 − x2 )3/2 (4 − x2 )3/2 √ when x = 2. MATH 226 – MIDTERM II Page 5 Problem 4. (20pts) Use L’Hopital’s rule to compute the following limit. t2 − et t→∞ 1 + et lim t2 − et 2t − et = lim t→∞ 1 + et t→∞ et 2 − et = lim t→∞ et −et = lim t t→∞ e = −1 lim MATH 226 – MIDTERM II Page 6 √ Problem 5. (15pts) Use Newton’s method to estimate 10 by using the function f (x) = x2 − 10 with an initial guess x0 = 3 and finding x2 . Since f 0 (x) = 2x, f (3) f 0 (3) −1 =3− 6 19 = 6 x1 = 3 − and x2 = = = = = 19 f (19/6) − 0 6 f (19/6) 361 − 10 19 − 36 19 6 2· 6 19 1 3 − · 6 36 19 2 · 361 − 1 12 · 19 721 228