AP Calc AB First Semester Review

Transcription

1 AP Calc AB First Semester Review MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the limit. 1) lim (7-7) 7 A) -4 B) -56 C) 4 D) 56 1) Determine the limit b sketching an appropriate graph. ) lim f(), where f() = < < 3 3 = 3 ) A) 1 B) 0 C) Does not eist D) 3 Find the limit, if it eists. 3) lim A) 0 B) C) Does not eist D) 4 3) 4) lim ) A) B) 6 5 C) Does not eist D) ) lim h 0 ( + h)3-3 h A) 0 B) 3 C) Does not eist D) 3 + 3h + h 5) 1

2 Find the limit. 6) lim -1-7) lim A) -1 B) 0 C) - D) 1 + A) 1/ B) - C) Does not eist D) 8) lim A) B) 0 C) - D) -1 6) 7) 8) 3 9) lim - A) -5 B) 1 C) - D) 9) ) lim A) 5 B) 1 C) D) ) 11) lim ) A) 0 B) 7 4 C) 1 D) Divide numerator and denominator b the highest power of in the denominator to find the limit. 1) lim ( - 13)( + 1) A) 0 B) 6 C) 36 D) 1) Find all points where the function is discontinuous. 13) 13) A) = 1 B) None C) = -, = 1 D) = -

3 Find the intervals on which the function is continuous. 3 14) = - 9 A) discontinuous onl when = -3 B) discontinuous onl when = 9 C) discontinuous onl when = -9 or = 9 D) discontinuous onl when = -3 or = 3 14) SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response. 15) Use the Intermediate Value Theorem to prove that = 0 has a solution between and 3. 15) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find numbers a and b, or k, so that f is continuous at ever point. ), if 6 f() = + k, if > 6 A) k = 30 B) k = 4 C) k = -6 D) Impossible ) Calculate the derivative of the function. Then find the value of the derivative as specified. 17) f() = 8 ; f (-1) 17) A) f () = - 8 ; f (-1) = -8 B) f () = - 8 ; f (-1) = - 8 C) f () = 8 ; f (-1) = 8 D) f () = 8; f (-1) = 8 3

4 The figure shows the graph of a function. At the given value of, does the function appear to be differentiable, continuous but not differentiable, or neither continuous nor differentiable? 18) = -1 18) A) Differentiable B) Continuous but not differentiable C) Neither continuous nor differentiable Compare the right-hand and left-hand derivatives to determine whether or not the function is differentiable at the point whose coordinates are given. 19) 19) (-1, -1) = 1 = -1 A) Since lim -1 + f () = -1 while lim -1 - f () = 0, f() is not differentiable at = -1. B) Since lim -1 + f () = 0 while lim -1 - f () = 1, f() is not differentiable at = -1. C) Since lim -1 + f () = 0 while lim -1- f () = -1, f() is not differentiable at = -1. D) Since lim -1 + f () = 0 while lim -1- f () = 0, f() is differentiable at = -1. 4

5 Find. 0) = (3-5)( ) A) B) C) D) ) Find an equation of the tangent line at = a. 1) = ; a = A) = 8-5 B) = 8-1 C) = -5 D) = 3-1 1) Suppose u and v are differentiable functions of. Use the given values of the functions and their derivatives to find the value of the indicated derivative. ) u(1) = 4, u (1) = -7, v(1) = 7, v (1) = -4. ) d v d u at = 1 A) - 65 B) 33 C) 33 4 D) ) u() = 7, u () = 3, v() = -1, v () = -5. d (uv) at = d A) 6 B) -38 C) 38 D) -3 3) Solve the problem. 4) The power P (in W) generated b a particular windmill is given b P = 0.015V3 where V is the velocit of the wind (in mph). Find the instantaneous rate of change of power with respect to velocit when the velocit is 7.3 mph. A) 5.3 W/mph B) 0.3 W/mph C) 11.7 W/mph D).4 W/mph 4) Find the derivative of the function. 5) = A) = C) = + 8 3/ B) = + 8 D) = / 5) Provide an appropriate response. 10 6) Find an equation for the tangent to the curve = at the point (1, 5). + 1 A) = 5 B) = 0 C) = 5 D) = + 5 6) 5

6 Find the second derivative of the function. 7) = ( - 8)( + 3) 3 A) d d = C) d d = B) d d = D) d d = ) Find the derivative. 8) = sec A) = sec tan B) = - 5 sec tan C) = tan D) = csc 8) 9) = sin + 1 cot A) = csc cot - sec B) = - csc cot + sec C) = cos - csc D) = csc cot - csc 9) Find the indicated derivative. 30) Find if = 6 sin. A) = - 1 cos + 6 sin B) = 1 cos - 6 sin C) = 6 cos - 1 sin D) = - 6 sin 30) Solve the problem. 31) Find the tangent to = cot at = π 4. 31) A) = - + π B) = + 1 C) = - π + 1 D) = - + π + 1 Write the function in the form = f(u) and u = g(). Then find d/d as a function of. 3) = (-3 + 7)5 A) = u5; u = ; d d = -15(-3 + 7) 4 C) = 5u + 7; u = 5; d d = B) = u5; u = ; d d = 5(-3 + 7) 4 D) = u5; u = ; d d = -3(-3 + 7) 5 3) 6

7 Suppose that the functions f and g and their derivatives with respect to have the following values at the given values of. Find the derivative with respect to of the given combination at the given value of. f() g() f () g () 33) ) g(), = 3 A) 1 4 B) C) 7 4 D) ) f() g() f () g () ) f(g()), = 4 A) -3 B) 4 C) 8 D) -8 Find the derivative of the function. cos 5 35) h() = 1 + sin A) -5 cos4 (1 + sin )5 C) - 4 sin cos cos sin B) -5 sin 4 cos D) 5 cos sin 35) Find d/dt. 36) = cos( 8t + 11) 1 A) - 8t + 11 sin( 8t + 11) B) 4 sin( 8t + 11) 8t + 11 C) -sin( 8t + 11) D) -sin 4 8t ) Find d for the given function. d 37) = 5 sin( + 7) A) -10 sin( + 7) B) -0 sin( + 7) C) 10 cos( + 7) D) -0 cos( + 7) 37) Solve the problem. 38) The position of a particle moving along a coordinate line is s = 5 + 4t, with s in meters and t in seconds. Find the particle's velocit at t = 1 sec. 38) A) 1 6 m/sec B) 4 3 m/sec C) m/sec D) 3 m/sec 7

8 Use implicit differentiation to find d/d. 39) = 8 39) A) B) + + C) D) ) cos + 6 = 6 40) A) sin 65 B) sin 65 - sin C) sin 65 D) sin 65 + sin At the given point, find the slope of the curve or the line that is tangent to the curve, as requested. 41) = + 1, tangent at (0, 1) A) = - 3 B) = C) = + 1 D) = ) Use implicit differentiation to find d/d and d/d. 4) - + = A) d d = ; d d = + 1 ( + 1) C) d d = ; d d = - ( + 1) B) d d = ; d d = + ( + 1) D) d d = ; d d = - ( + 1) 4) Solve the problem. 43) A piece of land is shaped like a right triangle. Two people start at the right angle of the triangle at the same time, and walk at the same speed along different legs of the triangle. If the area formed b the positions of the two people and their starting point (the right angle) is changing at 4 m/s, then how fast are the people moving when the are 5 m from the right angle? (Round our answer to two decimal places.) A) 0.80 m/s B) 0.40 m/s C) 1.60 m/s D) 6.5 m/s 43) Solve the problem. Round our answer, if appropriate. 44) As the zoom lens in a camera moves in and out, the size of the rectangular image changes. Assume that the current image is 8 cm 5 cm. Find the rate at which the area of the image is changing (da/df) if the length of the image is changing at 0.6 cm/s and the width of the image is changing at 0. cm/s. A) 11.6 cm/sec B) 9. cm/sec C) 5.8 cm/sec D) 4.6 cm/sec 44) 45) Water is being drained from a container which has the shape of an inverted right circular cone. The container has a radius of 5.00 inches at the top and a height of 6.00 inches. At the instant when the water in the container is 4.00 inches deep, the surface level is falling at a rate of 0.7 in./sec. Find the rate at which water is being drained from the container. A) 7.7 in.3/s B) 4.4 in.3/s C).0 in.3s D) 3.3 in.3/s 45) 46) The radius of a right circular clinder is increasing at the rate of 8 in./sec, while the height is decreasing at the rate of 10 in./sec. At what rate is the volume of the clinder changing when the radius is 6 in. and the height is 18 in.? A) 504 in.3/sec B) 1368π in.3/sec C) -8 in.3/sec D) 504π in.3/sec 46) 8

9 Determine all critical points for the function. 47) f() = A) = 0 and = 1 B) = -1 and = 1 C) = 0 and = D) = 0 47) Find the absolute etreme values of the function on the interval. 48) g() = , 5 6 A) absolute maimum is 1 4 B) absolute maimum is 41 4 C) absolute maimum is 5 4 D) absolute maimum is at = ; absolute minimum is 0 at 6 and 0 at = 5 11 at = ; absolute minimum is 0 at 6 and 0 at = 5 13 at = ; absolute minimum is 0 at 6 and 0 at = 5 13 at = ; absolute minimum is 0 at 6 and 0 at = 5 48) Find the etreme values of the function and where the occur. 49) f() = A) None B) Local maimum at (0, 0). C) Local maimum at (, -14), local minimum at (-, 18). D) Local maimum at (-, 18), local minimum at (, -14). 49) 1 50) f() = - 1 A) Local maimum at (1, 0), local minimum at (-1, 0). B) Local maimum at (-1, 0), local minimum at (1,0). C) Local maimum at (0, -1). D) None 50) 9

10 Solve the problem. 51) Select an appropriate graph of a twice-differentiable function = f() that passes through the points (-,1), - following sign patterns. 6 3,5 9,(0,0), 6 3,5 9 and (,1), and whose first two derivatives have the 51) : : A) B) C) D) Find the largest open interval where the function is changing as requested. 5) Decreasing f() = 3-4 A) 3 3, B) - 3 3, 3 3 C) -, D) -, ) 53) Increasing = ( - 9) A) (3, ) B) (-3, 3) C) (-, 0) D) (-3, 0) 53) 10

11 Use the graph of the function f() to locate the local etrema and identif the intervals where the function is concave up and concave down. 54) 54) A) Local minimum at = 1; local maimum at = -1; concave up on (-, ) B) Local minimum at = 1; local maimum at = -1; concave down on (0, ); concave up on (-, 0) C) Local minimum at = 1; local maimum at = -1; concave down on (-, ) D) Local minimum at = 1; local maimum at = -1; concave up on (0, ); concave down on (-, 0) 11

12 Solve the problem. 55) Using the following properties of a twice-differentiable function = f(), select a possible graph of f. 55) Derivatives < > 0, < 0-11 = 0, < 0 - < < 0 < 0, < < 0, = 0 0 < < < 0, > 0-1 = 0, > 0 > > 0, > 0 A) B) C) D) For the given epression, find '' and sketch the general shape of the graph of = f(). 56) = (4 - ) 56) 1

13 A) B) C) D) 13

14 Solve the problem. 57) 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 10 in. What dimensions will give a bo with a square end the largest possible volume? 57) A) 0 in. 0 in. 40 in. B) 40 in. 40 in. 40 in. C) 0 in. 0 in. 100 in. D) 0 in. 40 in. 40 in. 58) A rectangular sheet of perimeter 33 cm and dimensions cm b cm is to be rolled into a clinder as shown in part (a) of the figure. What values of and give the largest volume? 58) A) = 13 cm; = 7 cm B) = 1 cm; = 9 cm C) = 11 cm; = 11 cm D) = 10 cm; = 13 cm 14

15 59) At noon, ship A was 1 nautical miles due north of ship B. Ship A was sailing south at 1 knots (nautical miles per hour; a nautical mile is 000 ards) and continued to do so all da. Ship B was sailing east at 6 knots and continued to do so all da. The visibilit was 5 nautical miles. Did the ships ever sight each other? A) Yes. The were within 3 nautical miles of each other. B) No. The closest the ever got to each other was 6.4 nautical miles. C) No. The closest the ever got to each other was 5.4 nautical miles. D) Yes. The were within 4 nautical miles of each other. 59) 60) The 9 ft wall shown here stands 30 feet from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall. 60) 9' wall 30' A) 53.3 ft B) 51.3 ft C) 39 ft D) 5.3 ft 61) Find the optimum number of batches (to the nearest whole number) of an item that should be produced annuall (in order to minimize cost) if 300,000 units are to be made, it costs $ to store a unit for one ear, and it costs $440 to set up the factor to produce each batch. A) 6 batches B) 0 batches C) 8 batches D) 18 batches 61) 6) Suppose c() = ,000 is the cost of manufacturing items. Find a production level that will minimize the average cost of making items. A) 1 items B) 11 items C) 10 items D) 13 items 6) Find the linearization L() of f() at = a. 63) f() = , a = 0 A) L() = 3-8 B) L() = C) L() = D) L() = ) 15

16 Epress the relationship between a small change in and the corresponding change in in the form d = f () d. 64) f() = A) d = d B) d = (18 + 6) d C) d = d D) d = 18 d 64) Find the value or values of c that satisf the equation the function and interval. 65) f() = + 3 +, [1, ] f(b) - f(a) b - a = f (c) in the conclusion of the Mean Value Theorem for 65) A) 1, B) - 3, 3 C) 3 D) 0, 3 66) f() = + 3, [, ] 66) A) 0, 4 B) -4, 4 C), D) 4

17 Answer Ke Testname: AP CALC SEM 1 REVIEW 1) A ) B 3) D 4) B 5) B 6) C 7) C 8) A 9) C 10) D 11) A 1) B 13) A 14) D 15) Let f() = and let 0 = 0. f() = 11 and f(3) = -46. Since f is continuous on [, 3] and since 0 = 0 is between f() and f(3), b the Intermediate Value Theorem, there eists a c in the interval (, 3) with the propert that f(c) = 0. Such a c is a solution to the equation = 0. ) A 17) A 18) B 19) C 0) C 1) B ) B 3) B 4) D 5) D 6) C 7) C 8) A 9) B 30) B 31) D 3) A 33) C 34) A 35) A 36) B 37) B 38) D 39) C 40) D 41) B 4) D 43) A 44) D 45) B 46) B 47) C 17

18 Answer Ke Testname: AP CALC SEM 1 REVIEW 48) A 49) D 50) C 51) D 5) B 53) A 54) D 55) C 56) C 57) A 58) C 59) C 60) D 61) A 6) B 63) D 64) B 65) C 66) D 18