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
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