UMass Amherst MATH 127 - final practice calc (25 pages)

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final practice calc



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final practice calc

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Pages:
25
School:
University of Massachusetts Amherst
Course:
Math 127 - Calculus for the Life and Social Sciences I
Calculus for the Life and Social Sciences I Documents
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Calculus I Review Problems 1 New York University Exam is on Sections 2 3 2 6 2 8 3 1 3 5 3 7 4 1 4 2 4 3 4 4 4 5 4 7 5 1 5 5 1 Compute the following limits a limx 3 e2 x 3 Solution 1 1 1 b limx 10 ln 100 x2 Solution c limx e x sin x Solution d 1 2x limx 1 2 x Solution You can also apply L Hospital s e limx x3 e x Solution x f limx 1 x 1 1 Solution Page 1 Page 1 Page 1 1 ln x g limx 2 tan xcos x Solution h lim x ln x 5 ln x Solution lim x ln x 5 x lim x 5 x ln 1 x lim 5 x2 x 5 x x12 lim 5x x2 x 5 1 x2 lim 5x x 5 5 2 Calculate y 0 a y 2x x2 1 0 1Solution y 1 b y sin x sin x 1 Solution 2 x2 1 2x2 x2 1 1 2 x2 1 sec 2 c y 1 tan 2 1Solution 1 d xy 4 x2 y x 3y Solution Page 2 e sin xy x2 y Solution f y ln x2 ex Solution g y cos x x Solution 1 1 h y ln 1 x 1 Solution Page 1 1 ln x 1 i y arctan arcsin x 1 Solution j xey y 1 Solution k y tan x 1 x Solution 2 l ex y x y Solution Page Page12 3 The graph of y x3 9x2 16x 1 has a slope of 5 at two points Find the coordinates of the points Solution y0 5 0 3x2 18x 16 3x2 18x 16 0 0 x 3x2 18x 21 x2 6x 7 x 1 x 7 1 or x 7 When x 1 y 7 when x 7 y 209 Thus the two points are 1 7 and 7 209 4 Find the equation of the line tangent to f x at x 2 if f x x3 4 2 3x Solution The slope of the tangent line is the value of the first derivative at x 2 Differentiating gives d x3 4 d 1 3 4 1 x x dx 2 3x dx 2 3 1 4 3x2 1 x 2 2 3 3 2 4 x 2 2 3x For x 2 f 0 2 3 2 4 1 2 6 6 333 2 3 2 2 3 and f 2 23 4 2 4 3 333 2 3 2 3 To find the y intercept for the tangent line equation at the point 2 3 333 we substitute in the general equation y b mx and solve for b 3 333 9 333 b 6 333 2 b The tangent line has the equation y 9 333 6 333x 5 a Find the slope of the graph of f x 1 ex at the point where it crosses the x axis b Find the equation of the tangent line to the curve at this point c Find the equation of the line perpendicular to the tangent line at this point This is the normal line Solution a f x 1 ex crosses the x axis where 0 1 ex which happens when ex 1 so x 0 Since f 0 x ex f 0 0 e0 1 b y x c The negative of the reciprocal of 1 is 1 so the equation of the normal line is y x 6 Suppose f and g are differentiable functions with the values shown in the following table For each of the following functions h find h0 2 a h x f x g x b h x f x g x f x c h x g x x 2 f x 3 f 0 x 5 g x 4 g 0 x 2 Solution a We have h0 2 f 0 2 g 0 2 5 2 3 b We have h0 2 f 0 2 g 2 f 2 g 0 2 5 4 3 2 14 c We have h0 2 f 0 2 g 2 f 2 g 0 2 g 2 2 5 4 3 2 42 26 16 13 8 7 If you invest P dollars in a bank account at an annual interest rate of r then after t years you will have B dollars where r t B P 1 100 a Find dB dt assuming P and r are constant In terms of money what does dB dt represent b Find dB dr assuming P and t are constant In terms of money what does dB dr represent Solution r r t dB dB ln 1 P 1 The expression tells us how fast the amount of money in the bank dt 100 100 dt Job chap3 temp Sheet 39 Page 153 March 8 2012 10 10 ex 6 is changing with respect to time for fixed initial investment P and interest rate r dB r t 1 1 dB b Pt 1 The expression indicates how fast the amount of money changes with dr 100 100 dr respect to the interest rate r assuming fixed initial investment P and time t a 3 6 THE CHAIN RULE AND INVERSE FUNCTIONS 6w51 53 153 8 Use the following graph to calculate the derivative 3 6w59 68 Figure 3 33 shows the number of motor vehicles 7 f t In Problems 60 62 use Figure 3 32 to calculate the derivative in millions registered in the world t years after 1965 With units estimate and interpret 2 1 5 3 a f 20 c f 1 500 f x b d f 20 f 1 500 2 4 6 millions 2 5 800 ins3 6w51 53fig 3 6w51 3 6w52 3 6w53 3 6w54 600 10 400 Figure 3 32 0 8 12 3 a h 2 if h x f x 200 Solution Since the point 2 5 is on the curve we know f 2 5 The point 2 1 5 3 14 is on the tangent 3 60 h 2 if h x f x line so year 3 6w59fig 61 k 2 if k x f x 1 65 70 75 805 850 3 90 95 2000 5 3 16 Slope tangent 3 62 g 5 if g x f 1 x 2 1 Figure 2 3 33 0 1 63 a b c d Given that Thus f x fx03 2 find f3 2 1 Find f x By the chain rule 3 6w60 69 Using Figure 3 34 where f 2 2 1 f 4 3 0 Use your answer from part b to find f 1 8 2 0 2 h0 2 f3 f 4 2 3 5find 3 f 1225 6 2 3 7 ff 2 8 8 How could you have used your answer from part a 1 to find b f k 0 2 8 if k x f x 1 18 20 22 3 6w55 64 a For f x Solution 2x5 3x3Since …


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