MATH 021 Midterm: math21_2006f_exam1_soln

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turquoisegnu128

9 Jan 2019

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Uc merced, fall 2006 (e) consider the logarithmic function f (x) = c ln(kx), where c < 0 and k > 0 are constants. The graph of f (x) is: increasing and concave up, decreasing and concave down, decreasing and concave up, increasing and concave down. Answers: (a) e. (b) b or c. (c) e. (d) c. (e) b: (10 points) consider the piecewise function f (x) de ned below. Can you nd a value for b such that f (x) is continuous at x = 2. If not, explain why. f (x) =(cos(cid:0)(x 1) b, 2(cid:1) x 2 for x 6= 2 for x = 2. Solutions: f (x) is continuous at x = 2 if. = f (2) = lim x 2 f (x) = lim x 2 cos(cid:16)(x 1) Because of the absolute value sign, we need to discuss two one sided limits. lim x 2+ cos(cid:16)(x 1)

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Problem 4 Circle your answer. a) (6 points) If f(x) = 2x + 3x2 - 361 + 5, then the tangent line to the graph y= f(x) is horizontal when i) 1 = -2,3 ii) I = 2,-3 iii) r = 0 iv) not listed b) [6 points] Suppose Given h(x) = f(g(T)), where g(2) = 3, g'(2) = 7, f'(2) = -1 and 1'(3) = -2. Then h'(2) is equal to i) - 1 ii) -2 iii) - 14 iv) not listed (x+3)(1 - 1)3 c) (6 points) The function f(1) == (x2 - 2x + 1)(x + 3) wa has vertical asymptotes at i) x = -3,0 ii) = -3,0,1 iii) x = 0,1 iv) not listed d) (6 points] The solution to the initial value problem v'(x) = 473 – cos(x) +e", y(0) = 3 is i) y(x) = x + sin(x) + e* + 2 ii) y(I) = x4 – sin(x) +et+2 iii) y(x) = 2* – sin(1) + et + 3 iv) not listed

Problem 6 Circle your answer. pas du d a) 6 points) The value of da is inşa wa mã nv) not listed iv) not listed b) [6 pointsThe integral / sec (2) dx equals 1) sep (s) + C i seca) 2+C ii) in con+C ii) tan(a) +C iii) tan(2) +C iv) not listed iv) not listed c) (6 points) If f(x) dx = 4 and g(x) dx = 3, then 2f() - 4g(x) dx equals i) 4 ile ii) -4 iii) 10 mi) 10 iv) not listed d) (6 points) If h(2) is an odd function on the interval (-5,5), then h(x) dic equals LM(s) do ni) [ n(e) de ii- f*n(x) dx iv) not listed

MATH 021 Midterm: math21_2006f_exam1_soln

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