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by: Vivien Bradtke V


Vivien Bradtke V
Texas A&M
GPA 3.6


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This 3 page Class Notes was uploaded by Vivien Bradtke V on Wednesday October 21, 2015. The Class Notes belongs to MATH 151 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/226044/math-151-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
Fall 2004 Math 151 Exam 1A Solutions Mon 040ct 1 N L Lquot gt1 2004 Art Belmonte a Wehave 2a 7 3b 22 3 7 3 1 71 4 6 7 3 73 19 c From the gure below we have a v b from which we conclude v b 7 a A d With a 72 3 and b 1 2 the scalar projection of bontoais ab 7271767 4 compab 7 W Hall m a Since u and v are unit vectors we have v2u73v 2vu73vv 2mwmmtwwwmw 7 unama73manu 17372 a ForO I g we havex sint andy cos2 I Thus sinzlcoszl 1 x2y 1 y 17x2 This is part of a parabola d With f de ned on an open interval containing 2 and f2 3 it is always true that if lim2 fx 3 f2 x7 then f is continuous at x 2 c Letfc c3 07 1 772 Thenf2 97712 lt 0 and f3 29 7 n2 gt 0 Now f is a polynomial and thus continuous everywhere Therefore by the Intermediate Value Theorem fc 0 for some 0 e 2 3 So for some number ce23wehavec c7l7r2 c Let s divide numerator and denominator of the limiting expression by x2 Vx4 Then 2x43x21 l1m l1m x7gtoltgt 3x22x4 x7gtoltgt 3 2 4 7 2 7 2 9 b Note thaty x7 7 7mwm candidates for vertical asymptotes are x 2 an x 72 Now comes the election Therefore d o Asx 7gt 2 we have x72 1 1 mm m y Thus x 2 is not a vertical asymptote c As x 7 72 we see that 7 x 7 2 7 1 7 1 7 y7x72x27x2 0 039 Hence x 72 is a vertical asymptote 9 Here is a plot which corroborates these assertions y H I x2 4 1 N I 4 I 10 c Resolve the absolute value 2x73 2x7330 fx 2x 3 72x73 2x73lt0 7 2x73 x332 7 372x xlt32 Now draw a rough sketch to see that f x does not exist for x since the graph is sharp or kinked thereat y2X 3 393 u 2x 7 1 11 b Use the quotient rule to differentiate f x 72 x 139 fx 62 1 2 7 2x27 1 2x x2 1 7 2x2274x22x 7 x2 12 7 2 1 x 7 x2 x2 1f 12 b Velocity is the derivative of position no sI 211 Whenl 2 s the velocity is v2 5 ms 13 a With fO 0 andxl39in0 71 we have f 0 limo fmifw lim amp 7139thatis x7gt x 7 x7gt0 x f 0 71 Since gx 2x 71fxwe have gx 2fx 2x 71fx gO 2f0 2071f0 2071711 14 Let the positive xaxis point east and the positive yaxis point north Then the velocity of the woman relative to the water is v 73i 1 22j with components in mih The woman s speed is the magnitude of the velocity HvH 732 222 49 w 2220mih Her direction 0 points west of north into Quadrant 2 and satis es tan0 whence 0 8 or N8 W from the north 8 toward the west 25 75 75 0 5 15 The vector from A1 1 to B2 71 is wEF72172 A vector perpendicular to w is v wi 2 1 a direction vector for our line A vector equation of the line through P 3 2 in this direction is La Iv 7 32r21 213 12 2czxzcxc x lt1 16 Letfx 1 x 1 1norder for cx 1 x gt1 lim f x to exist we must ensure that the lefthand and x7 righthand limits are equal lim f x lim f x x7gt1 x7gt1 c From the left we have lim fx 2c2 1 20 whereas 6 lim f x c 1 for the righthand limit Set these x7gt1 onesided limits equal 202 2c c 1 o Equivalently 2c2 1 c 7 1 0 Now 2c7lc10 Nlquot whence c 71 17 i The derivative of f at a is de ned by 7 fx7fa f a 7ggnlI x 7a or equivalently h 7 fa 1 fw f a 1130 h ii With fx we have 1 2x3 f 1 lim x7gt1 fx7f1 x71 572x3 1 gt 11m x7gt1 52x3 x71 2 7 2x 11m x7gt1 5 2x 3 x 71 2 hm 05 1 x7gt1 5 2x 3 x 71 7 2 11m 7 77 x7gt1 5 2x 3 25 1 111 A p01nt on the tangent line to the curve y 7 at 2x 3 x 1 is 1 f1 lt1 The slope ofthe tangent line from part a is f 1 7 Hence an equation of the tangentline isy 7 72275 x 7 1 or 2 7 y 73x 3 18 i Here is a sketch of the graphs of f and g on the same gure We label the slopes of the piecewiselinear components Problem 18 ii Clearly f is not differentiable at x 1 due to the sharp kink in its graph thereat iii Similarly g is not differentiable at x g for the same reason iv Atx We have fg f g fg lt0 86gt 1 9 7 V Atx 2 we have 8 am2 e 2 30 77 702479 121


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