Yvette Granillo, Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A trailer rental agency rents 10 trailers per day at a rate of $8 per day. It discovers that for each $2 increase in rate, one fewer trailer is rented. Determine the rate, rmax , which maximizes the rental income. 1. rmax = $14 correct 2. rmax = $16 2. rebate = $100 correct 3. none of these 3. rebate = $95 4. rmax = $18 5. rmax = $12 Explanation: Let r be the daily rate of renting a trailer. Then the number of trailers rented daily is (r - 8) 10 - . 2 Hence the daily income, I(r), of the agency is given by I(r) = r 10 - (r - 8) 2 . 4. rebate = $110 5. rebate = $105 6. none of these is the rate that maximizes income. keywords: Stewart5e, 002 (part 1 of 1) 10 points
1
Circuit City has been selling 100 television sets a week at $600 each. A market survey indicates that for each $20 rebate offered to a buyer, the number of sets sold will increase by 5 per week. How large a rebate should Circuit City offer a buyer in order to maximize its revenue? 1. rebate = $115
Explanation: Let $20x be the rebate offered to a buyer. Then the price of a TV will be $(600 - 20x) and the number of sets sold at this price will be 100 + 5x. The revenue with this rebate is thus R(x) = (600 - 20x)(100 + 5x) = 100(30 - x)(20 + x) = 100(600 + 10x - x2 ) . But then R (x) = 100(10 - 2x) , while
The income will be maximized, therefore, at the critical points of I(r). Now after differentiation, I (r) = 10 - (r - 8) 2 - r . 2
Thus the critical points of I(r) occur at 10 - (r - 8) 2 - r = 0, i.e., r = 14 . 2
Consequently, rmax = $14
R (x) = -100 2 < 0 .
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell Consequently, the Revenue is maximized at x0 = 5, in which case the rebate = $100 .
2
and so the only critical point of c(x) for x > 0 is the positive solution of c (x) = 0, i.e., replacement time = 3200 = 8 years . 50
keywords: Stewart5e, 003 (part 1 of 1) 10 points A high speed office copier has an initial price of $3200. A service contract cost $260 for the first year and increases by $100 per year thereafter. It can be shown that over x years the total cost of the copier is given by C(x) = 3200 + 210x + 50x2 . When is the average cost per year smallest? (This is often referred to as the replacement time for a piece of equipment.) 1. replacement time = 9 years 2. replacement time = 7 years
keywords: Stewart5e, 004 (part 1 of 2) 10 points The marketing department of a company has determined that the cost (in dollars) of producing x units of a new product is given by C(x) = 300 + 180x, while the demand equation for that product will be given by p = 300 - 3 x. 100
(i) What will be the marginal profit when x = 50, assuming all units produced can be sold? 1. P (50) = $122.00
3. replacement time = 8 years correct 2. P (50) = $132.00 4. replacement time = 11 years 5. replacement time = 10 years Explanation: The average cost C(x) 3200 + 210x + 50x2 c(x) = = x x 3200 = + 210 + 50x . x In practical terms it represents the total cost (initial cost plus maintenance costs) of the computer averaged out over an x year period. The marginal average cost, therefore, is given by 3200 c (t) = - 2 + 50 , x 3. P (50) = $117.00 correct 4. P (50) = $137.00 5. P (50) = $127.00 Explanation: The revenue R(x) from the sale of x units will be R(x) = xp(x) = 300x - 3 2 x , 100
so the profit on these units will be P (x) = R(x) - C(x) = 120x - 3 2 x - 300. 100
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell The marginal profit will thus be given by 3 P (x) = 120 - x. 50 At x = 50, therefore, P (50) = 117.00. 005 (part 2 of 2) 10 points (ii) At what production level will the marginal profit be zero? 1. x = 2250 units 2. x = 1750 units 3. x = 2750 units 4. x = 2000 units correct 5. x = 2500 units Explanation: The marginal profit will be zero when P (x) = 120 - i.e., when x = 2000 units. F3 (x) = 3 x = 0, 50 3. F (x) = 2x3 + 7x2 + 8x 4. F (x) = 6x3 - 14x2 + 8x + C 5. F (x) = 2x3 - 7x2 + 8x Explanation: Since
3
d r x = rxr-1 , dx the most general anti-derivative of f is the function F (x) = 6 x3 3 - 14 x2 2 + 8x + C
with C an arbitrary constant. Consequently, F (x) = 2x3 - 7x2 + 8x + C . keywords: Stewart5e, antiderivatives, polynomial 007 (part 1 of 1) 10 points Consider the following functions: F1 (x) = sin2 x , F2 (x) = cos 2x , 4 cos2 x . 2
Which are anti-derivatives of f when keywords: Stewart5e, 006 (part 1 of 1) 10 points Find the most general antiderivative, F , of the function f (x) = 6x2 - 14x + 8 . 1. F (x) = 2x3 + 7x2 + 8x + C 2. F (x) = 2x3 - 7x2 + 8x + C correct f (x) = sin x cos x ? 1. none of them correct 2. F3 only 3. F2 only 4. F2 and F3 only 5. F1 and F2 only
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell 6. F1 and F3 only 7. all of them 8. F1 only Explanation: By trig identities, cos 2x = 2 cos2 x - 1 = 1 - 2 sin2 x , while sin 2x = 2 sin x cos x . But d d sin x = cos x, cos x = - sin x . dx dx Consequently, by the Chain Rule, A. False. B. False. C. False. keywords: Stewart5e, antiderivatives, differentiation of trigonometric functions, trigonometric formula 008 (part 1 of 1) 10 points The acceleration of a particle moving along the x-axis at time t is given by a(t) = 6t - 2. If the velocity is 25 when t = 3 and the position is 10 when t = 1, then find the position x(t). 1. x(t) = 36t - 4t - 77t + 55 2. x(t) = 9t2 + 1 3. x(t) = 3t2 - 2t + 4 4. x(t) = t - t + 4t + 6 correct Explanation: 5. x(t) = t3 - t2 + 9t - 20
3 2 3 2
4
v(3) = 25 = 25 = 27 - 6 + C 4=C Thus v(t) = s = 3t2 - 2t + 4 1 1 s = 3 t3 - 2 t2 + 4t + C1 3 2 s(1) = 10 = 10 = 1 - 1 + 4 + C1 6 = C1 3 Thus s(t) = t - t2 + 4t + 6. 009 (part 1 of 1) 10 points Stewart Section 4.10, Example 2, page 301 Find all functions g such that g (x) = 1. g(x) = 2 x rect 2. g(x) = 2 x x2 + 4x + 5 + C 3. g(x) = x x2 + 4x + 5 + C 1 2 4 x + x-5 +C 5 3 x2 + 4x + 5 . x 1 2 4 x + x + 5 + C cor5 3
4. g(x) = 2 x
5. g(x) = 2 x x2 + 4x - 5 + C 6. g(x) = x 1 2 4 x + x+5 +C 5 3
Explanation: After division g (x) = x3/2 + 4x1/2 + 5x-1/2 , so we can now find an antiderivative of each term separately. But d dx axr r = axr-1
a(t) = v = 6t - 2 1 v = 6 t2 - 2t + C 2
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell for all a and all r = 0. Thus 2 5/2 8 3/2 x + x + 10x1/2 5 3 = 2 x 1 2 4 x + x+5 5 3 keywords: Stewart5e, trigonometric functions
5
antiderivatives,
011 (part 1 of 1) 10 points Find the value of f (0) when f (t) = 2(3t + 4) and f (1) = 3, 1. f (0) = 5 correct 2. f (0) = 8 3. f (0) = 7 4. f (0) = 9 5. f (0) = 6 Explanation: The most general anti-derivative of f has the form f (t) = 3t2 + 8t + C where C is an arbitrary constant. f (1) = 3, then f (1) = 3 + 8 + C = 3 , i.e., C = -8. From this it follows that f (t) = 3t2 + 8t - 8 , and the most general anti-derivative of the latter is f (t) = t3 + 4t2 - 8t + D , where D is an arbitrary constant. f (1) = 2, then f (1) = 1 + 4 - 8 + = D 2 , i.e., D = 5. Consequently, f (t) = t3 + 4t2 - 8t + 5 . But if But if f (1) = 2 .
is an antiderivative of g . Consequently, g(x) = 2 x 1 2 4 x + x+5 +C 5 3
with C an arbitrary constant. keywords: Stewart5e, antiderivatives, power functions 010 (part 1 of 1) 10 points Find the most general function f such that f (x) = 12 cos 2x . 1. f (x) = 3 sin x + Cx + D 2. f (x) = 3 cos x + Cx + D 3. f (x) = 2 sin 2x + Cx + D 4. f (x) = -2 sin x + Cx2 + D 5. f (x) = -2 cos 2x + Cx2 + D 6. f (x) = -3 cos 2x + Cx + D correct Explanation: When f (x) = 12 cos 2x then f (x) = 6 sin 2x + C with C an arbitrary contant. Consequently, the most general function f is f (x) = - 3 cos 2x + Cx + D with D also an arbitrary constant.
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell At x = 0, therefore, 013 (part 1 of 1) 10 points f (0) = 5 . If the graph of f is keywords: Stewart5e, 012 (part 1 of 1) 10 points A car braked with a constant deceleration of 35 ft/s2 , producing skid marks measuring 80 ft before coming to a stop. How fast was the car traveling when the brakes were first applied? 1. 74.83 ft/s correct 2. 63.895 ft/s 3. 37.415 ft/s 1. 4. 85.295 ft/s 5. 96.335 ft/s Explanation: a(t) = -35 v(t) = -35 t + v0 where v0 is the car's speed in (ft/s) when the brakes were applied. -35 t + v0 = 0 The car stops when 1 v0 t= 35 Now 1 s(t) = (-35)t2 + v0 t 2 = -17.5t2 + v0 t The car travels 80 ft in the time that it takes to stop, so 1 s v0 = 80 35 80 = -17.5 1 v0 35
2
6
which one of the following contains only graphs of anti-derivatives of f ? cor-
rect 2.
3.
+ v0 v0 2
1 1 v0 = v0 2 35 70 = 70 80 = 5600 v0 = 74.83 ft/s
keywords: Stewart5e,
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell
7
4.
(i) an anti-derivative of f will have a local extremum at the x-intercepts of f . This eliminates two more figures since they contains graphs whose local extrema occur at points other than the x-intercepts of f . (ii) An anti-derivative of f is increasing on interval where the graph of f lies above the x-axis, and decreasing where the graph of f lies below the x-axis. Consequently, of the two remaining figures only
5.
6.
consists entirely of graphs of anti-derivatives of f .
Explanation: If F1 and F2 are anti-derivatives of f then independently of x; this means that for any two anti-derivatives of f the graph of one is just a vertical translation of the graph of the other. In general, no horizontal translation of the graph of an anti-derivative can be the graph of an anti-derivative, nor can a horizontal and vertical translation be the graph of an anti-derivative. This rules out two sets of graphs. Now in each of the the remaining four figures the dotted and dashed graphs consist of vertical translations of the graph whose linestyle is a continuous line. To decide which of these figures consists of anti-derivatives of f , therefore, we have to look more carefully at the actual graphs. But calculus ensures that F1 (x) - F2 (x) = constant
keywords: Stewart5e, antiderivative, graph, geometric interpretation 014 (part 1 of 1) 10 points
Which of the following functions will fail to have an inverse?
5 4 4 1.3 2 2 1 0 -1 -4 -2 2 4 -2 -2 -3 -4 -4 -5 -5 -4 -3 -2 -1 0 1 2 3 4 5
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell
5 4 2.3 2 1 0 -1 -4 -2 -3 -4 -5 -5 -4 5 4 3.3 2 1 0 -1 -4 -2 -3 -4 -5 -5 -4 5 4 4.3 2 1 0 -1 -4 -2 -3 -4 -5 -5 -4 5 4 5.3 2 1 0 -1 -4 -2 -3 -4 -5 -5 -4
8
4 2 -2 -2 -4 2 4
-3 -2 -1 0 1 2 3 4 5
5 4 4 3 2 2 1 0 -1 -4 -2 2 4 -2 -2 -3 -4 -4 -5 -5 -4 -3 -2 -1 0 1 2 3 4 5
4 2 -2 -2 -4 2 4
keywords: Stewart5e, 015 (part 1 of 1) 10 points Determine the inverse function, f -1 , of f when f (x) = 9x , (x 0) . 1. f -1 (x) = 1 2 x , 9 (x 0) (x 0)
-3 -2 -1 0 1 2 3 4 5
4 2 -2 -2 -4 2 4
correct
1 2. f -1 (x) = - x2 , 9
3. f -1 (x) does not exist 4. f -1 (x) = 5. f -1 (x) = 1 2 x , 9 1 2 x 9 (x 0) correct
-3 -2 -1 0 1 2 3 4 5
4 2 -2 -2 -4 2 4
Explanation: The function f (x) = 9x is a one-to-one function for x 0; hence the inverse f -1 (x) exists. Now the solution to the equation 9x = y , y > 0
1 is x = 9 y 2 . But the original function f (x) can take only non-negative values, i.e., the range of f is [0, ); this means that the domain of f -1 is [0, ). Consequently,
-3 -2 -1 0 1 2 3 4 5
Explanation: If a function f has an inverse, then it must be a one-to-one function; equivalently, the graph of f must pass the horizontal line test. The only graph failing to pass this test is
f -1 (x) =
1 2 x , 9
(x 0) .
keywords: Stewart5e,
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell keywords: Stewart5e, 016 (part 1 of 1) 10 points
1 Find the value of g ( 2 ) when g is the inverse of the function f (x) = sin x, - x . 2 2
9
017 (part 1 of 1) 10 points
Consider the graph of the function y = f (x).
1. g 2. g 3. g 4. g
1 2 1 2 1 2 1 2
= - 2 = 2 correct
1 = - 2 2 = 2 2 1
8 6 4 2 -6 -4 -2 -2 -4 -6 -8 2 4 6
5. None of these 6. g 7. g 1 2 1 2 = -2 2 = 2 2
Which of the following choices is the graph of the inverse function f -1 (x)?
10 9 8 7 1.6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
Explanation: Since f (x) = sin x, we see that f (x) = cos x, Consequently, g Thus g 1 2 = 1 1 f (f -1 ( 2 )) = 1 = f () 4 1
1 2
x , 2 2 4 1 = . 2
8 6 4 2 -8 -6 -4 -2 -2 -4 -6 -8 2 4 6 8
f
1 2
= f -1
1 2
= . 4
=
2.
-9-8-7-6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 10 -10 9 8 8 7 2.6 6 5 4 4 3 2 2 1 0 -1 2 4 6 8 -2 -8 -6 -4 -2 -2 -3 -4 -4 -5 -6 -6 -7 -8 -8 -9 -10 -10 -9-8-7-6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell
10 9 8 7 3.6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
10
8 6 4 2 -8 -6 -4 -2 -2 -4 -6 -8 2 4 6 8
correct
8 6 4 2 -8 -6 -4 -2 -2 -4 -6 -8 keywords: Stewart5e, 018 (part 1 of 1) 10 points (11) for Find f
-1
2 4 6 8
-10 -9-8-7-6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 10 9 8 7 4.6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
8 6 4 2 -8 -6 -4 -2 -2 -4 -6 -8 2 4 6 8
f (x) = x3 + 2x - 1 1. 2. 3. 4. 1 correct 14 1 34 1 31 1 16 1 19
-10 -9-8-7-6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10 10 9 8 7 5.6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10
8 6 4 2 -8 -6 -4 -2 -2 -4 -6 -8 2 4 6 8
5.
6. None of these 1 9 1 8. 18 1 9. 15 Explanation: 7. f (x) = x3 + 2 x - 1 f (x) = 3x2 + 2
-10 -9-8-7-6-5-4-3-2-10 1 2 3 4 5 6 7 8 9 10
Explanation: The graph of the inverse function of f -1 (x) is obtained by reflecting the graph of f over the line y = x:
f (2) = 11, so f -1 (11) = 2, and f -1 (11) = 1 f [f -1 (11)]
Granillo, Yvette Homework 12 Due: Nov 17 2005, 3:00 am Inst: Edward Odell = 1 f (2) 1 = 14
11
019 (part 1 of 1) 10 points Use f (x) to determine whether f (x) = x4 - 2x2 4
has an inverse on (-, ). 1. f has inverse 2. f does not have inverse correct Explanation: The function will have an inverse if f is increasing everywhere on (-, ) or decreasing everywhere, so we need to look at the sign of f (x). Now f (x) = x3 - 4x = x(x2 - 4) , and so has sign chart - + - +
- -2 0 2 Thus f is decreasing on (-, -2] and on [0, 2], but increasing on [-2, 0] and on [2, ). Consequently, f does not have an inverse .