Chapter 4 practice set for math 261 The actual exam is different.
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Find the absolute extreme values of the function on the interval. π 7π 1) f(θ) = sin θ + ,0≤θ≤ 2 4 Find the absolute extreme values of the function on the interval. π π 2) f(x) = tan x, - ≤ x ≤ 3 6 3) f(x) = x2/3 , -1 ≤ x ≤ 27
1)
2)
3)
Determine all critical points for the function. 2x 4) f(x) = x+8
4)
5) f(x) = x2 + 4x + 4
5)
6) f(x) = 20x3 - 3x5
6)
7) y = 3x2 - 96 x
7)
Find the extreme values of the function and where they occur. 8) y = x3 - 3x2 + 7x - 10 9) y = x2 + 2x - 3
10) y =
8) 9)
1 2 x -1
10)
Find the derivative at each critical point and determine the local extreme values. 11) y = x2/3(x2 - 9); x ≥ 0
11)
12) y = 6 - 2x, x ≤ 1 x + 3, x > 1
12)
13) y = x2 25 - x
13)
1
Provide an appropriate response. 14) Let f(x) = (x - 1)2/3
14)
(a) Does f′(1) exist? (b) Show that the only local extreme value of f occurs at x = 1. (c) Does the result of (b) contradict the Extreme Value Theorem? (d) Repeat parts (a) and (b) for f(x) = (x - c)2/3. Give reasons for your answers. Find the extreme values of the function and where they occur. x+1 15) y = 2 x + 2x + 2
15)
Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval. Explain the reason for your answer. 16) g(x) = x3/4, 0,1 16) Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval. 17) s(t) = t(1 - t), -1,5 17)
Find the value or values of c that satisfy the equation
f(b) - f(a) = f′(c) in the conclusion of the Mean Value Theorem for b-a
the function and interval. 18) f(x) = x2 + 2x + 2, [-2, 1]
18)
Show that the function has exactly one zero in the given interval. 4 19) f(x) = x3 + + 3, (-∞, 0). x2
19)
Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval. 20) g(x) = x3/4, 0,2 20) Find all possible functions with the given derivative. 21) y′ = 5x2 - 3x
21)
Find the absolute extreme values of the function on the interval. 22) f(x) = 2x - 3, -2 ≤ x ≤ 3
22)
Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval. 23) f(x) = x1/3, 23) -1,2 Show that the function has exactly one zero in the given interval. 1 24) r(θ) = 5 cot θ + + 2, (0, π). θ2
2
24)
Find the function with the given derivative whose graph passes through the point P. 25) r′(θ) = 4 + sec2 θ, P(π, 0) 26) r′(t) = sec2 t - 4, P(0, 0)
25) 26)
Identify the function's local and absolute extreme values, if any, saying where they occur. 27) f(x) = x2 + 18x + 162
27)
Identify the function's extreme values in the given domain, and say where they are assumed. Tell which of the extreme values, if any, are absolute. 28) f(x) = (x + 4)2 , -∞ < x ≤ 0 28) Find the function with the given derivative whose graph passes through the point P. 1 29) g′(x) = + 2x , P(-5, 5) x2 Find all possible functions with the given derivative. 30) y′ = csc2 3θ
29)
30)
Find the extrema of the function on the given interval, and say where they occur. π 31) sin 4x, 0 ≤ x ≤ 2 Find all possible functions with the given derivative. 3 32) y′ = 4t t
31)
32)
33) y′ = 7x2 + 1
33)
Find the function with the given derivative whose graph passes through the point P. 1 34) g′(x) = + 2x , P(-2, 2) x2
34)
Identify the function's extreme values in the given domain, and say where they are assumed. Tell which of the extreme values, if any, are absolute. 35) f(x) = 36 - x2 , -6 ≤ x < 6 35)
3
Graph the equation. Include the coordinates of any local and absolute extreme points and inflection points. 36) y = 8x2 + 48x 36) 200
y
100
-10
-5
5
10
x
-100
-200
Use the graph of the function f(x) to locate the local extrema and identify the intervals where the function is concave up and concave down. 37) 37) 10
y
5
-10
-5
5
10 x
-5
-10
Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. 38) f′(x) = (5 - x)(8 - x) 38)
4
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Solve the problem. 39) Find the table that matches the given graph.
39)
y
a
b
c
x
A)
B) x f′(x) a does not exist b 0 c 1
x f′(x) a 0 b 0 c -1
C)
D) x f′(x) a does not exist b 0 c -1
x f′(x) a does not exist b does not exist c -1
5
40) Find the graph that matches the given table. x -1 1 3
40)
f′(x) 0 does not exist 0
A)
B) y 6
6
5
5
4
4
3
3
2
2
1
1
-6 -5 -4 -3 -2 -1 -1
1
2
3
4
5
y
-6 -5 -4 -3 -2 -1 -1
6 x
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
C)
1
2
3
4
5
6 x
1
2
3
4
5
6 x
D) 6
y
6
5
5
4
4
3
3
2
2
1
1
-6 -5 -4 -3 -2 -1 -1
1
2
3
4
5
6 x
-6 -5 -4 -3 -2 -1 -1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
6
y
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Find the location of the indicated absolute extremum for the function. 41) Minimum
41)
f(x) 5 4 3 2 1 -5
-4
-3
-2
-1
1
2
3
4
5
x
-1 -2 -3 -4 -5
42) Maximum
42) h(x) 5 4 3 2 1
-5
-4
-3
-2
-1
1
2
3
4
5 x
-1 -2 -3 -4 -5
7
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Determine from the graph whether the function has any absolute extreme values on the interval [a, b]. 43)
43)
A) No absolute extrema. B) Absolute minimum and absolute maximum. C) Absolute maximum only. D) Absolute minimum only. 44)
44)
A) Absolute minimum only. B) No absolute extrema. C) Absolute minimum and absolute maximum. D) Absolute maximum only. SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. 45) f′(x) = (1 - x)(8 - x) 45)
8
Graph the equation. Include the coordinates of any local and absolute extreme points and inflection points. 24x 46) y = 46) 2 x +9 y
x
Solve the problem. 47) If the price charged for a candy bar is p(x) cents, then x thousand candy bars will be sold in x a certain city, where p(x) = 142 . How many candy bars must be sold to maximize 20
47)
revenue? Find an antiderivative of the given function. 28 48) x8
48)
1 9 x
49)
49)
x-6 +
50)
7 cos 5x
50)
Graph the equation. Include the coordinates of any local and absolute extreme points and inflection points. 51) y = 2x3 - 3x2 - 12x 51) y
x
9
Graph the rational function. x2 - 8x + 15 52) y = x-2 40
52)
y
30 20 10 -40 -30 -20 -10 -10
10
20
30
40 x
-20 -30 -40
53) y =
x-4
53)
x2 - 7x + 12 y 8 4
-8
-4
4
8
x
-4 -8
Solve the problem. 54) A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 64 ft3 . What dimensions yield the minimum surface area? Round to the nearest tenth, if necessary.
10
54)
55) You are planning to close off a corner of the first quadrant with a line segment 21 units long running from (x, 0) to (0, y). Show that the area of the triangle enclosed by the segment is largest when x = y.
55)
56) A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only one-fifth as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.
56)
57) A rectangular field is to be enclosed on four sides with a fence. Fencing costs $8 per foot for two opposite sides, and $5 per foot for the other two sides. Find the dimensions of the field of area 760 ft2 that would be the cheapest to enclose.
57)
58) Find the optimum number of batches (to the nearest whole number) of an item that should be produced annually (in order to minimize cost) if 150,000 units are to be made, it costs $1 to store a unit for one year, and it costs $360 to set up the factory to produce each batch.
58)
59) Suppose that c(x) = 4x3 - 22x2 + 6849x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.
59)
Use Newton's method to estimate the requested solution of the equation. Start with given value of x 0 and then give x 2 as the estimated solution. 60) 3x2 + 2x - 1 = 0; x0 = 1; Find the right-hand solution.
11
60)
61) -x2 + 4x -1 = 0; x0 = 0; Find the left-hand solution.
61)
Solve the problem. 62) A private shipping company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?
62)
Graph the equation. Include the coordinates of any local and absolute extreme points and inflection points. 63) y = x1/3(x2 - 63) 63) y
x
Find an antiderivative of the given function. 9 64) x6/5 5
64)
65) 27x2 - 8x - 8
65)
66) 7 x + 5
66)
12
67) - 2 csc2
x 3
68) cos πx + 6 sin
67)
x 6
68)
Solve the initial value problem. dy 69) = 2x-3/4, y(1) = 10 dx
69)
70)
dy 1 = + 3, y(1) = -1 dx 2 x
70)
71)
ds π = cos t - sin t, s =5 dt 2
71)
72)
d2 r 4 dr = 2, r(1) = 5 = ; 2 dt t3 dt t=1
72)
73)
d3 y = 5; y ′′ (0) = 1, y ′(0) = 7, y(0) = 5 dx3
73)
13
Answer Key Testname: 261CH4P
1) absolute maximum is 1 at θ = 0; absolute minimum is -1 at θ = π 3 π π 2) absolute maximum is at x = ; absolute minimum is - 3 at x = 3 6 3 3) absolute maximum is 9 at x = 27; absolute minimum is 0 at x = 01 4) x = -8 5) x = -2 6) x = 0, x = -2, and x = 2 7) x = 0 and x = 4 8) None 9) The minimum is -4 at x = -1. 10) Local maximum at (0, -1). 11) Critical Pt. derivative Extremum Value x=0 Undefined local max 0 x = 1.5 0 minimum -8.845 12) Critical Pt. derivative Extremum Value x=1 undefined minimum 4 13) Critical Pt. derivative Extremum Value 0 0 min x=0 0 undefined min x = 25 x = 20
0
local max 400 5 2 14) (a) No, since f′(x) = (x - 1)-1/3, which is undefined at x = 1. 3 (b) The derivative is defined and nonzero for all x ≠ 1. (c) No, f(x) need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form [a, b] would have both a maximum value and a minimum value on the interval. (d) The answers are the same as (a) and (b) with 1 replaced by c. 1 1 15) The maximum is at x = 0; the minimum is - at x = -2. 2 2 16) Yes 17) No 1 18) 2 19) The function f(x) is continuous on the open interval (-∞, 0). Also, f(x) approaches -∞ as x approaches -∞, and f(x) approaches ∞ as x approaches 0 from the left. Since f(x) is continuous and changes sign along the interval, it must have at least one root on the interval. The first derivative of f(x) is f′(x) = 3x2 -
8 , which is everywhere positive on (-∞, 0). Thus, f(x) has a single root on (x3
∞, 0). 20) Yes 14
Answer Key Testname: 261CH4P
21)
5 3 3 2 x - x +C 3 2
22) absolute maximum is 3 at x = 3; absolute minimum is - 7 at x = -2 23) No 24) The function r(θ) is continuous on the open interval (0, π). Also, r(θ) approaches ∞ as θ approaches 0 from the right, and r(θ) approaches -∞ as θ approaches π from the left. Since r(θ) is continuous and changes sign along the interval, it must have at least one root on the interval. The first derivative of r(θ) is r′(θ) = -5 csc2 θ -
2 , which is everywhere negative on (0, π). Thus, r(θ) has a single root θ3
on (0, π). 25) r(θ) = 4θ + tan θ - 4π 26) r(t) = tan t - 4t 27) absolute minimum: 9 at x = -9 28) local maximum: 16 at x = 0; local and absolute absolute minimum: 0 at x = -4 1 101 29) g(x) = - + x2 x 5 30) -
1 cot 3θ + C 3
31) local maxima: 1 at x =
π π and 0 at x = ; 8 2
local minima: 0 at x = 0 and -1 at x =
3π 8
32) 2t2 - 6 t + C 7 33) x3 + x + C 3 34) g(x) = -
1 5 + x2 x 2
35) local and absolute minimum: 0 at x = -6; local and absolute maximum: 6 at x = 0 36) absolute minimum: (-3,-72) no inflection points 200
y
100
-10
-5
5
10
x
-100
-200
37) Local minimum at x = 1; local maximum at x = -1; concave up on (0, ∞); concave down on (-∞, 0) 38) Decreasing on (5, 8); increasing on (-∞, 5) ∪ (8, ∞) 15
Answer Key Testname: 261CH4P
39) C 40) A 41) x = -2 42) x = 1 43) B 44) B 45) Decreasing on (1, 8); increasing on (-∞, 1) ∪ (8, ∞) 46) local minimum: (-3, -4) local maximum: (3, 4) inflection points: (0, 0), (-3 3, -6 3), (3 3, 6 3) y 6 4 2
-4
-2
2
x
4
-2 -4 -6
47) 1420 thousand candy bars 4 48) x7 49) 50)
1 5x5
+
2 1/2 x 9
7 sin 5x 5
51) local minimum: (2, -20) local maximum: (-1, 7) 1 13 inflection point: , 2 2 y 24
12
-8
-4
4
8
x
-12
-24
16
Answer Key Testname: 261CH4P
52) 40
y
30 20 10 -40 -30 -20 -10 -10
10
20
30
40 x
8
x
-20 -30 -40
53) y 8 4
-8
-4
4 -4 -8
54) 5 ft × 5 ft × 2.5 ft 55) If x , y represent the legs of the triangle, then x2 + y2 = 212 . Solving for y, y = 441 - x2 A(x) = xy = x 441 - x2 x2 A'(x) = + 2 441 - x2 Solving A'(x) = 0, x = ±
441 - x2 2 21 2 2
Substitute and solve for y: ( 56)
21 2 2 21 2 ) + y2 = 441 ; y = ∴ x = y. 2 2
width 20 = height 10 + 4π
57) 21.8 ft @ $8 by 34.9 ft @ $5 58) 14 batches 59) There is not a production level that will minimize average cost. 60) 0.35 61) 0.23 62) 18 in. × 18 in. × 36 in.
17
Answer Key Testname: 261CH4P
63) local minimum: 3, -54
3
3
3
local maximum: -3, 54 3 inflection point: (0, 0) 400
y
300 200 100 -20
-10
10
20 x
-100 -200 -300 -400
64)
9 11/5 x 11
65) 9x3 - 4x2 - 8x 14 3/2 66) x + 5x 3 67) 6 cot 68)
x 3
1 x sin πx - 36 cos π 6
69) y = 8x1/4 + 2 70) y = x + 3x - 5 71) s = sin t + cos t + 4 2 72) r = + 4t - 1 t 73) y =
5 3 1 2 x + x + 7x + 5 6 2
18
