Practice Exam 1 Tese problems are meant to approximate wat Exam 1 will be like. You can expect tat problems on te exam will be of similar difficulty. Te actual exam will ave problems from sections 11.1 troug 11.5. Te problems ere are just a sample of te many kinds of problems tat could appear on te exam. You will be expected to sow all your work. Please see me if you ave any questions on tese or oter problems. For te sake of brevity I ave not included room for your work on tis list of problems. On te actual exam you will be expected to sow your work and tere will be sufficient room to do so.. 1. Suppose tat it is known tat: fx) =, x 2 gx) = 4 x 2 Use te it laws from class compute te following it. Sow all your work and cite all rules used explicitly. gx) 2 ) xfx) + 5x x 2 gx) 2 ) xfx) + 5x x 2 gx) 2 ) xfx) + 5x Product Property x 2 x 2 ) 2 ) 1 gx) xfx) + 5x 2 Power Property x 2 x 2 ) 2 ) 1 gx) xfx)) + 5x) 2 Sum Property x 2 x 2 x 2 ) 2 ) 1 gx) x) fx)) + 5) x) 2 Product Property x 2 x 2 x 2 x 2 x 2 = 4) 2 2 + 5 2) 1 2 Identity and Constant Property = 1616) 1 2 = 64 2. Use algebra and properties from class to compute te following its. You do not need to state te properties. If a it does not exist write DNE and indicate wy you know it doesn t exist. 2x 2 14x + 24) a) x x 2. 9 2x 2 14x + 24) x x 2 9 2x 2 7x + 12) x x )x + ) 2 x )x 4) x x )x + ) 2 = 4) + = 2 6 = 1 b) x 2x 2 14x + 24) x 2 9
2 2x 2 14x + 24) x x 2 9 2x 2 7x + 12) x x )x + ) 2 x )x 4) x x )x + ) x 2x 4) x + ) = DNE c) We know tat tis doesn t exist because wile te denominator is going to zero te numerator is not so tere is no way to get any cancelation. We can also sow tat in fact te it on one side is positive infinity and on te oter side it is negative infinity. x 4 x + x 2 x x + 2x 9 d) x 2 + 24x 5 x 4 + 2 x x + 2x 5 x 4 + 2x x 4 x + x 2 ) 1 x x x + 2x 9) 1 x x x 0 0 x + x 2 x 0 0 2 1 + 9 x 2 x x 1 = x 2 + 24x 5 x 4 + 2 x x + 2x 5 x 4 + 2x x = = 24 2 = 12 0 0 x + 2 x 5 0 0 0 1 1 + 2 x 2 x + 2 x 4 0 1 + 24 x
. Use te it definition of te derivative to compute te derivative of te function fx) = 1 2x 1 To receive full credit you must be clear and you must use te it definition. Te it definition says tat, Tus we ave tat: f x) f x) fx + ) fx) = 1 2x+) 1 1 2x 1 2x 1 2x+) 1)2x 1) 2x 1 2x+)+1 2x+) 1)2x 1) 2 2x+) 1)2x 1) 2 2x+) 1)2x 1) 2x+) 1 2x+) 1)2x 1) ) ) 2 2x + ) 1)2x 1) 2 2x 1)2x 1) = 2 2x 1) 2
4 4. Using te sort cut formulas compute te derivative of te following functions: a) fx) = 2 x + x 5 First rewrite te function as: fx) = 2x 1 2 + x 5 Ten applying te power rule for derivatives we ave tat: f x) = 1 2 2)x 1 2 5)x 6 or f x) = x 1 2 15x 6 b) gx) = x + 2x 2 x First we rewrite te equation as: gx) = x + 2x 2 x 1 2 Ten we apply te power rule to get: = x x 1 2 + 2 x 2 x 1 2 = x 1 2 + 2x g x) = 1 2 x 1 2 + 2 5. Given tat te derivative of te function fx) = e x 2) is f x) = e x 2), use your knowledge of calculus to find te equation of te tangent line to te grap of y = e x 2) wen x = 2. Te slope of te tangent line is given by f 2) = e 2 2) =. Te tangent line must also pass troug te point 2, f2)) = 2, 1). Tus using te point slope formula we get: or y 1) = x 2) y = x 5
5 6. Below is a grap tat sows bot te median and average sales prices of new omes sold in te United States from 196-2008. If you are not viewing tis in color, te lower curve is te median prices.) In indsigt looking at tese graps we can see te so-called ousing bubbles occurring in te late 1980 s and te more recent bubble starting in 200. Use te above grap to answer te following questions as accurately as possible. a) Use te grap to calculate te average rate of cange of te Median ome price from 2002 to 2007. Make sure to include te units in your answer. If we tink of te median ome sales price as a function Mx) were x is te year ten we ave: avg. rate of cange = M2007) M2002) 2007 2002 = 250000 190000 5 = 60000 5 = 12, 000 dollars year b) Use te grap to calculate te average rate of cange of te Median ome price from 2002 to 2008. Make sure to include te units in your answer. If we tink of te median ome sales price as a function Mx) were x is te year ten we ave: avg. rate of cange = M2008) M2002) 2008 2002 = 240000 190000 6 = 50000 6 8,. dollars year c) Explain grapically wat te numbers computed in a) and b) represent. If you used te values in a) to predict te median ome price in 2009 would it be iger or lower ten if you used te answer in b)?
6 Grapically, tese two numbers represent te slopes of te secant line tat would connect te points at year 2002 and 2007 for a)) and for 2002 and 2008 part b)). Te secant lines are drawn on te grap. Since te slope in part a) is larger tan tat in part b), it would predict a iger value in 2009, owever since te interval of 2002-2007 as artificially growing median ousing prices tat value is going to most likely be less accurate ten using te average over te period 2002-2008 wen ome prices ave declined some and te effects of te bubble would be less. 7. A company knows tat tat tey can sell x items at a price of px) = 00.01x dollars per item. a) Find te total revenue for selling x items. Hint: revenue is price multiplied times te number of items sold.) To find te total revenue you need to multiply te price px) by te number of items sold x, tis gives te revenue rx) or: rx) = x00.01x) = 00x.01x 2 b) Find te marginal revenue for selling x items. To find te marginal revenue we compute te derivative of te function found in a). Tis gives: r x) = 00.02x c) Use your answer from b) to determine te marginal revenue wen 100 items are being sold. Include te units wit your answer. To find te marginal revenue wen 100 items are being sold we just evaluate r 100) or r 10) = 00.02100) = 298 dollars item d) Give a brief interpretation in words wat tis number from part c) means. Use complete sentences. Te number 298 computed in c) means tat wen 100 items are being sold, revenue will increase by approximately 298 dollars wen one more item is sold.
7 8. Sketc one grap of a function fx) tat satisfies all te following properties: a) f) = 2. b) f x) is not defined at x =. c) f x) = 0 for x <. d) f x) < 0 for x <. e) fx) = 1. x 2 1 4 2 2 4 6 8 1 2 Te blue grap above represents a possible plot of fx). Te dased line represents te asymptote at y = 1.
8 9. Below is te grap of a function fx). Use te grap to answer te questions tat follow. 5 1-5 - -1 1 5 2-1 - a) List all values a suc tat 5 < a < 5 and fx) does not exist. x a Te values of a are: -2, 1,, 4. b) List all values a suc tat 5 < a < 5 and f a) does not exist. 4-4 Te values of a are: 4,, 1, 0, 1,, 4-5 c) List all values of a suc tat 5 < a < 5 and x a fx) exists but is different tan fa). Te values of a are: 4. d) Wat is f 2)? Te value of f 2) is kist te slope of te line at tat point wic is 1/2. e) Wat is fx)? x 1 Te it from te left at 1 is 2 f) Wat is x + fx)? Te it from te rigt at is -2 4