MATH 141 Midterm: MATH141H_HERB-R_SPRING2005_0101_MID_SOL_1

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turquoisegnu128

15 Feb 2019

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University of Maryland

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Mathematics

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MATH 141

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All

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$MATH 141 Full Course Notes$

MATH 141 Full Course Notes

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Math 141h - exam 2 - solutions - march 14, 2005: (a) f (x) = 3x2 6x = 3x(x 2) is positive on ( , 0) and (2, ) and negative on (0, 2). But ln(1/x) > 0 when 1/x > 1, that is x < 1. Thus the domain is 0 < x < 1. f (x) = d dx ex ln(ln(1/x)) = ex ln(ln(1/x))[ln(ln(1/x)) + x. 1 x2 ]: this is a limit of the form 0/0. After di erentiating once we still have 0/0. lim x 1 ln x x + 1 x3 3x + 2. = z x dx sin 1 y = x2/2 + c y = sin(x2/2 + c). dy p1 y2 (b) 1/2 = y(0) = sin(0 + c) so c = /6. Thus y = sin(x2/2 + /6): (a) integrate by parts with u = x and dv = 3xdx so du = dx and v = 3x/ ln 3.

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[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

4. (21 pts) Let g be the function given by ( 6 - X 2 + x if x < 2, x + -2 g(x) = x2 – x-1 if x>2. (a) (6 pts) Use the definition of continuity to determine whether the function g is continuous at 2. Show your work. (b) (3 pts) Evaluate the limit. Justify your answer. Show your work. Write "does not exist" only if the limit does not exist and is not + or -0. lim g(x) = lim -2+ = +00 X 2 + X since the form of the limit is NUM: POS, DENOM: POS (c) (2 pts) Write the equation(s) of all vertical asymptotes of g. Write "none" if appropriate. The result in part (b) implies that the line x= -2 is a vertical asymptote. This is the only vertical asymptote, since the function g is continuous on its domain (d) (2 pts) Determine the largest) intervals of continuity of g. Use interval notation to write your answer. Let g be the function given by 12+ if x < 2, x# -2 2 + x g(x)= x2-x-1 x-1 if x>2. (e) (6 pts) Evaluate each limit. Do not use L'Hôpital's Rule. Write "does not exist" only if the limit does not exist and is not too or -00. Show your work. 6-1 i lim g(x) = lim -00 2 + X = lim -002 + x 0-1 + 1 x -00 x x x -+- 2 +1 0 0 + 11 x2 - x-1 1 ii. lim g(x) = lim x++ V x2 – x-1 X-1 x2 - x-1 X - 1 = lim x +oc x +0 - lim Vam M93001 (f) (2 pts) Write the equation(s) of all horizontal asymptotes of g. Write "none" if appropriate.

MATH 141 Midterm: MATH141H_HERB-R_SPRING2005_0101_MID_SOL_1

MATH 141 Full Course Notes

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