Review for Test 2 Calculus I

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1 Review for Test Calculus I Find the absolute etreme values of the function on the interval. ) f() = -, - ) g() = , ) F() = -,.5 ) F() =, - 6 5) g() = 7-8, - Find the absolute etreme values of the function on the interval. 6) f() = /, - 7 Determine all critical points for the function. 7) f() = + + Find the etreme values of the function and where the occur. 8) = + - Find the derivative at each critical point and determine the local etreme values. 9) = (9 - ) Find the value or values of c that satisf the equation the function and interval. ) f() = + +, [-, ] f(b) - f(a) b - a = f (c) in the conclusion of the Mean Value Theorem for Solve the problem. ) Given the velocit and initial position of a bod moving along a coordinate line at time t, find the bodʹs position at time t. v = -9t +, s() = 8 Find the largest open interval where the function is changing as requested. ) Increasing = 7-5 Identif the functionʹs local and absolute etreme values, if an, saing where the occur. ) f() =

2 Graph the equation. Include the coordinates of an local and absolute etreme points and inflection points. 6 ) = + 9 Sketch the graph and show all local etrema and inflection points. 5) = ) =

3 Graph the equation. Include the coordinates of an local and absolute etreme points and inflection points. 7) = - - Identif the functionʹs local and absolute etreme values, if an, saing where the occur. 8) f() = ) f() = ) f(r) = (r - 8) - ) h() = Find the largest open interval where the function is changing as requested. ) Increasing f() = - ) Decreasing f() = - ) Decreasing f() = - 8 5) Decreasing = + 7 Solve the problem. 6) Given the acceleration, initial velocit, and initial position of a bod moving along a coordinate line at time t, find the bodʹs position at time t. a =, v() =, s() = Find the value or values of c that satisf the equation the function and interval. 7) f() = + 75, [, 5] f(b) - f(a) b - a = f (c) in the conclusion of the Mean Value Theorem for

4 Find the derivative at each critical point and determine the local etreme values. 8) = 5 - Find the etreme values of the function and where the occur. 9) = - + ) = - 5 ) = + ) = ( + )/ ) = Determine all critical points for the function. ) f() = - + 5) f() = ) f() = - 5 7) f() = + 8 8) f() = ( - ) 9) = - 6 Find the absolute etreme values of the function on the interval. ) f() = 5/, -7 8

5 Graph the rational function. - ) = Solve the problem. ) A compan is constructing an open-top, square-based, rectangular metal tank that will have a volume of 6 ft. What dimensions ield the minimum surface area? Round to the nearest tenth, if necessar. ) Find the number of units that must be produced and sold in order to ield the maimum profit, given the following equations for revenue and cost: R = C = Find the most general antiderivative. ) 7t + t 5 dt 5) ( ) d 6) d 7) ( t - 6 t) dt 8) + d A) C B) - + C C) - + C D) C 9) (-8 cos t) dt 5) (-7 sec ) d

6 Solve the problem. 5) Suppose c() = - 8 +, is the cost of manufacturing items. Find a production level that will minimize the average cost of making items. 5) Suppose that c() = is the cost of manufacturing items. Find a production level that will minimize the average cost of making items. 5) You are planning to close off a corner of the first quadrant with a line segment units long running from (, ) to (, ). Show that the area of the triangle enclosed b the segment is largest when =. 5) A private shipping compan will accept a bo for domestic shipment onl if the sum of its length and girth (distance around) does not eceed 8 in. What dimensions will give a bo with a square end the largest possible volume?

7 55) A trough is to be made with an end of the dimensions shown. The length of the trough is to be feet long. Onl the angle θ can be varied. What value of θ will maimize the troughʹs volume? θ θ Graph the rational function. 56) = Solve the initial value problem. 57) d d = +, > ; () = Write the sum without sigma notation and evaluate it. 8k 58) k + 7 k = Epress the sum in sigma notation. 59) Evaluate the sum. 6) k k =

8 6) 8 k = k 6) 6 k = k - 6) k = 5 Epress the sum in sigma notation. 6) ) Write the sum without sigma notation and evaluate it. k + 66) k k = 67) k = (-)k (k - 5) 68) k = k Solve the initial value problem. 69) d d = -/, () = 7) ds dt = cos t - sin t, s π = 5 Evaluate the integral. 7) d Use a definite integral to find an epression that represents the area of the region between the given curve and the -ais on the interval [, b]. 7) =

9 Evaluate the integral. 9 7) 5 d 7) 75) 65 d - ( + ) d 76) t + t d 77) t + t dt 78) 79) π/ sin d π/ 8 cos d π/ Use a definite integral to find an epression that represents the area of the region between the given curve and the -ais on the interval [, b]. 8) = 8π 8) = 8) = + Evaluate the integral. π/ 8) θ dθ π 8) /9 t dt

10 85) 9 d 86) 87) 88) 5 π/ θ dθ 7 d 7 6 d 89) z- dz 9) d 9) d Find the derivative. d 9) d sin t dt Find the total area of the region between the curve and the -ais. 9) = + 7; 5 Find the area of the shaded region. 9)

11 95) Find the total area of the region between the curve and the -ais. 96) = ; 97) = ; 98) = - + 9; 5 99) = ; Find the derivative. ) d d 8t5 dt ) d dt sin t 9 - u du ) = dt ) = 8 cos t dt ) = tan t dt Evaluate the integral. d 5) (7 + )5

12 Solve the initial value problem. 6) d d = 8(6-5) -6, () = 7) d d = (5 - ) 5, () = 5 Evaluate the integral. 8) ( - 6) d 9) + d ) d ln 6 ) 8 + d ) sin (9 - ) d Use the substitution formula to evaluate the integral. ) + d Find the area of the shaded region. ) f() = (, 8) 5 (, ) (-, -) -5 - g() = 6 Find the area enclosed b the given curves. 5) = -, = - 6) =, =

13 7) =, = 8) =, = ) Find the area of the region in the first quadrant bounded b the line = 8, the line =, the curve = and the -ais., ) Find the area of the region in the first quadrant bounded on the left b the -ais, below b the line =, above left b = +, and above right b = - +. Find the area of the shaded region. ) f() = g() = 5 5 (, 6) 5 (, ) (-, -) ) = = -

14 ) - = - - A) 7 5 = - B) 5 C) D) 76 5 ) = = A) 9 B) 6 C) 9 D) 8 Use the substitution formula to evaluate the integral. t 5) - + t dt A) B) 7 88 C) - 7 D) ) 6 r dr 6 + r A) 9 - B) 9 - C) 9-8 D) ) (8 - + ) -/ (6 - ) d A) 9 B) C) 8 D) 8

15 8) 9 - A) 5 d B) 5 C) D) - 5 9) π cos sin d π/ A) B) C) - 5 D) 5 6

16 Answer Ke Testname: REVIEW ) absolute maimum is at = ; absolute minimum is - 7 at = - ) absolute maimum is at = ; absolute minimum is at and at = ) absolute maimum is - at = ; absolute minimum is - at = ) absolute maimum is at = 6; absolute minimum is at = 5) absolute maimum is 7 at = ; absolute minimum is -65 at = 6) absolute maimum is 9 at = 7; absolute minimum is at = 7) = - 8) The minimum is - at = -. 9) Critical Pt. derivative Etremum Value =.7 local ma.9 = -.7 local min -.9 ) - ) s = - 9 t + t + 8 ) (-, ) ) local maimum at = ; local minimum at = ) local minimum: (-, -) local maimum: (, ) inflection points: (, ), (-, - ), (, )

17 Answer Ke Testname: REVIEW 5) Absolute maima: (-, -8), (, -8) Local minimum: (, -9) Inflection points: -, 8 9,, ) Local maimum: 6 5,,6 5 Local minimum: (6, ) Inflection point: 5, 78, ) local minimum: (, -) local maimum: (-, 7) inflection point:,

18 Answer Ke Testname: REVIEW 8) local maimum at = -; local minimum at = - 9) no local etrema ) no local etrema ) local minimum at = -; local maimum at = ) (, ) ) (-, ) ) (-, 8) 5) (, ) 6) s = t + t + 7) 5 8) Critical Pt. derivative Etremum Value min = undefined min = 5 = local ma 5 9) Local maimum at (, ), local minimum at (, -). ) Local maimum at (, -). ) The minimum value is - 5 at = -. The maimum value is 5 at =. ) The minimum value is at = -. ) None ) = - and = 5) = and = 6) =, = -, and = 7) = -8 8) = 9) = and = ) absolute maimum is 5 at = -7 ; absolute minimum is at = ) ) 5 ft 5 ft.5 ft ) 65 units ) 7 t + t + C

19 Answer Ke Testname: REVIEW 5) C 6) C 7) t / t 7/6 + C 8) C 9) -8sin t + C 5) -7 tan + C 5) 9 items 5) There is not a production level that will minimize average cost. 5) If, represent the legs of the triangle, then + =. Solving for, = - A() = = Aʹ() = Solving Aʹ() =, = ± Substitute and solve for : ( ) + = ; = 5) 8 in. 8 in. 6 in. 55) 56).5 = ) = ) = ) (-)k k k = 6) 66 6) 96 6) 67 6) 95

20 Answer Ke Testname: REVIEW 6) 5 k + 5 k = 65) k = 66) + k = 67) -( - 5) + ( - 5) - ( - 5) = - 68) = 5 69) = 8/ + 7) s = sin t + cos t + 7) 7) 7b 7) 9 7) 75) 65 76) ) ) 79) -6 8) 6πb 8) 5b 8) b 8 + b 8) 5π 8 8) 87 85) 86) π 87) 88) 6 89) ) ) - 5 9) sin () 9) 5 9)

21 Answer Ke Testname: REVIEW 95) 6 96) 97) 98) 99) ) 9 cos t ) 9 - sin t ) ) 87 cos () ) sec tan 5) - 56 (7 + )- + C 6) = - 5 (6-5) ) = ( 5 - ) ) C 9) 9 + / + C ) 6 ln ln 6 + C ) 8 + 5/ + C ) - 9 cos (9 - ) + C ) - ) 97 5) 6) 8 7) 6 8) 6 9) 5 ) 7 6

22 Answer Ke Testname: REVIEW ) 97 ) 9 ) A ) C 5) D 6) C 7) D 8) B 9) A

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Chapter Practice Dicsclaimer: The actual eam is different. On the actual eam ou must show the correct reasoning to receive credit for the question. SHORT ANSWER. Write the word or phrase that best completes