Practice Problems
**Note this list of problems is by no means complete and to focus solely on these problems would be
unwise.**
1.
By hand find the derivative for each of the following. You do not need to simplify.
a.
fx =
5
3x 8
+
22
x2
+ x +
f ′ x = − 403 x −9 − 44x −3 +
b.
fx =
′
f x =
c.
fx =
5
x +3x−3
1
5
x
−4
5
x− 2 −
1
x−
1
9
17
8
1
1
+3 − x 5 +3x−3
240x 5 +4−8x −2
2
34 x +
40x 6 +4x+8x −1
11
REWRITE
2
=
x
+
5
x3
10
f x = 11 3x + 5x − 2 + 2x −1
1
4
′
d.
27
8
x −8 + 22x −2 + x 2 + 3x − 8
x 5 +3x−3
40x 6 +4x+8x −1
=
40x 6 +4x+8x −1
x9
1
2
REWRITE
40x 6 +4x+ 8x
REWRITE 5
=
3
3
8
3
3
4
3x 4 + 5x − 2 + 2x −1
1
x− 4 −
3
3
15
2
11
x − 2 − 2x −2
5
fx = x 5 + 8 7 2x 7 + 6x 3 − 12x 9
f ′ x = 2x 7 + 6x 3 − 12x 9 7x 5 + 8 6 5x 4 
+ x 5 + 8 7 92x 7 + 6x 3 − 12x 8 14x 6 + 18x 2 − 12
e.
fx = 8 12 12x 10 + 9x 6
f ′ x =
f.
fx =
REWRITE
=
1
812x 10 + 9x 6  12
− 11
8
12x 10 + 9x 6  12 120x 9
12
REWRITE
3
=
39 − x −2
9−x 2
−3
+ 54x 5 
f ′ x = −69 − x −1
g.
fx =
x 4 +7x 3 −6x+3
x 10 −4x 7 +2x−4
x 3 −x 2 +x+1
x 3 − x 2 + x + 1
f x =
2
7
7
2
5
x 3 + 7x 2 + 3 2 3x 2 + 14x
f ′′ x = 3x 2 + 14x
4.
x 3 −x 2 +x+1
Find the second derivative of fx = x 3 + 7x 2 + 3 2 .
f ′ x =
3.
10x 9 − 28x 6 + 2
−x 4 + 7x 3 − 6x + 3x 10 − 4x 7 + 2x − 43x 2 − 2x + 1
′
2.
x 10 − 4x 7 + 2x − 44x 3 + 21x 2 − 6 + x 4 + 7x 3 − 6x + 3
35
4
3
x 3 + 7x 2 + 3 2 3x 2 + 14x +
7
2
5
x 3 + 7x 2 + 3 2 6x + 14
REWRITE
4 −1
Find f 3 x if fx = 3x4
=
x
3
4 −2
′
f x = − 3 x
f ′′ x = 83 x −3
f 3 x = −8x −4
Recall that a function fx is continuous at x = a if all of the following conditions are satisfied:
i. fa is defined.
ii. lim
fx exists
x→a
iii. lim
fx = fa
x→a
On each set of axes below, sketch the graph of a function that meets the given criteria.
a. fx is not continuous at x = 2 because it does not satisfy condition (i).
b. gx is not continuous at x = −1 because it does not satisfy condition (ii) but it does satisfy
condition (i).
1
c.
d.
hx is not continuous at x = 4 because it does not satisfy condition (iii) but it does satisfy
conditions (i) & (ii).
kx is continuous on −∞, ∞ but k ′ −2 does not exist.
y
-6
-4
-2
y
6
4
2
2
2
-2
4
6
-6
x
-6
-4
-2
-2
-6
-6
y
4
2
2
4
6
6
x
6
4
2
4
b) gx is not continuous at x = −1
6
-2
2
-2
-4
-6
x
-4
-2
2
-2
-4
-4
-6
-6
4
6
x
d) kx is continuous, k ′ −2 DNE
c) hx is not continuous at x = 4
e.
-4
-4
a) fx is not continuous at x = 2
y
6
4
Consider the graph of y = fx given below.
y
15
10
5
-12
-10
-8
-6
lim fx = 5
x→2 −
lim fx = 10
x→2 +
lim fx DNE
x→2
f2 = 10
f.
-4
-2
2
4
6
8
lim fx = 5
f −3 = 12
lim fx = 5
f ′ −3 = 0
lim fx = 5
f−10 = 10
f−8 DNE
f ′ −10 = − 52
x→ −8 −
x→ −8 +
x→ −8
10
x
(5 points) Find the given interval. Write your answer in interval notation. Write "NONE" if
appropriate.
i. The intervals wheref ′ x > 0.
−8, −3 ∪ 2, ∞
ii. The intervals where f ′ x < 0.
−∞, −8 ∪ −3, 2
2
iii. The intervals where f ′′ x > 0.
5.
NONE
iv. The intervals where f ′′ x < 0.
−8, 2
Consider the function
fx = 52 + 23
x
x
a.
Write down the difference quotient for fx in unsimplified form.
5
x+h 2
+
2
x+h 3
−
5
x2
+
2
x3
h
b.
Find the simplified difference quotient for fx.
fx+h−fx
= − 3 1 3 5h 2 x + 2h 2 + 15hx 2 + 6hx + 10x 3 + 6x 2 
h
x h+x
c.
By hand, find the limit as h approaches zero of the simplified0 difference quotient.
lim − 3 1 3 5h 2 x + 2h 2 + 15hx 2 + 6hx + 10x 3 + 6x 2 
h→0
=−
=−
x h+x
1
x 3 0+x 3
1
10x 3
x6
50 2 x + 20 2 + 150x 2 + 60x + 10x 3 + 6x 2
+ 6x 2  = − 10
−
x3
6
x4
What is the name of the function you found in part b)?
The derivative of fx
Consider the function
fx = 9x 4 + 119x 3 + 366x 2 − 765x + 1254
d.
6.
a.
b.
c.
d.
e.
f.
Find all the critical numbers of the function.
f ′ x = 0, Solution is: − 173 , −5, 34
How can you be sure you found all of the critical numbers?
Because f is a polynomial, there are no values of x for which the derivative does not exits.
Find the intervals over which the function is increasing and decreasing. Give your answer in
interval notation.
f ′ x > 0, Solution is: − 173 , −5 ∪ 34 , ∞
f ′ x < 0, Solution is: −5, 34 ∪ −∞, − 173
Increasing on − 173 , −5 ∪ 34 , ∞
Decreasing on −5, 34 ∪ −∞, − 173
Find the relative extreme values for f, if any. Be sure to use calculus to justify your answers.
f ′′ − 173  = 154 > 0 ⌣ min
f ′′ −5 = − 138 < 0 ⌢ max
f ′′  34  = 5313
> 0 ⌣ min
4
17
134 141
f− 3  = 27 = 4968. 185 185 185 19
f−5 = 4979
f 34  = 240256429 = 939. 175 781 25
Rel min values: 4968. 185 and 939. 176
Rel max value: 4979
Find the absolute extreme values for f, if any.
Absolute min is 939. 176
No absolute max.
Find the intervals over which the function is concave up and concave down. Give your answer
in interval notation.
3
7.
f ′′ x > 0, Solution is: −∞, − 361 5377 − 119
∪ 361 5377 − 119
,∞
36
36
− 361 5377 − 119
= − 5. 342 445 280 868 74
36
1
119
5377 − 36 = − 1. 268 665 830 242 37
36
′′
f x < 0, Solution is: − 361 5377 − 119
, 361 5377 − 119
36
36
1
119
Concave up on −∞, − 361 5377 − 119
∪
5377
−
,∞
36
36
36
OR −∞, −5. 343 ∪ −1. 269, ∞
Concave down on − 361 5377 − 119
, 361 5377 − 119
36
36
OR −5. 343, −1. 268 665 830 242 37
g. Find the point(s) of inflection. Be sure to give both the x and the y coordinates. Explain what
each inflection point tells you about f.
f ′′ x = 0, Solution is: 361 5377 − 119
, − 361 5377 − 119
36
36
Changes concavity.
f 361 5377 − 119
= 2593. 935 740 857 85
36
1
119
f− 36 5377 − 36  = 4973. 481 868 881 52
Inflection points: −1. 269, 2593. 936 and −5. 343, 4973. 482
At −5. 343, 4973. 482, f changes from concave up to concave down. The rate of change of f is
changing from an increasing rate to a decreasing rate. At −1. 269, 2593. 936, f changes from
concave down to concave up. The rate of change of f is changing from a decreasing rate to
an increasing rate. At
h. What is the absolute maximum and minimum value of the function on the closed interval
−4, 4?
f−4 = 4858
f4 = 13 970
Absolute min is 939. 176
Absolute max is 13970
Consider a function fx with the following characteristics:
′
● f x > 0 on the interval −∞, −4 ∪ 3, 8 ∪ 16, ∞
′
● f x < 0 on the interval −4, 3 ∪ 8, 16
′
′
● f −4 is undefined, f 8 is undefined
′
′
● f 3 = 0, f 16 = 0
● f−4 = 6, f3 = −4, f8 is undefined, f16 = 2
′′
● f 26 = 0
′′
● f x > 0 on the interval −∞, −4 ∪ −4, 8 ∪ 8, 26
′′
● f x < 0 on the interval 26, ∞
These characteristics can be summarized in "sign charts" as given below.
DNE
2
8, 16 16 16, ∞
x
−∞, −4
−4
−4, 3
3
3, 8
8
f x
+
DNE
−
0
+
DNE
′
6
fx
a.
−4
6
fx
−
0
DNE
11
8, 26 26 26, ∞
x
−∞, −4
−4
−4, 8
8
f ′′ x
+
DNE
+
DNE
+
0
+
−
Sketch a possible graph of y = fx on the axes below.
4
y
20
10
-20
-10
10
20
30
40
x
-10
-20
What are the critical numbers for f?
−4, 3, 16
c. What are the relative maximum and relative minimum values of f?
Rel min: −4, 2
Rel max: 6
d. What are the inflection points for f?
26, 11
Consider the function
fx = 50x 4 − 8x 3 + 0. 149x 2 + 12
b.
8.
a.
b.
Find the equation of the tangent line at x = 0. 6.
f ′ 0. 6 = 34. 738 8
f0. 6 = 16. 805 64
y − f0. 6 = f ′ 0. 6x − 0. 6, Solution is: 34. 738 8x − 4. 037 64
Answer: y = 34. 738 8x − 4. 037 64
Sketch the graph of y = fx and the tangent line at x = 0. 6 on the same set of axes.
y
100
80
60
40
20
-2
9.
-1
1
2
x
Find the absolute maximum and minimum values of the function
fx = 20x 5 + 95x 4 − 428x 3 − 1494x 2 + 2160x
5
y 40000
30000
20000
10000
-10
-8
-6
-4
-2
2
4
x
on the interval −1, 5.
f ′ x = 0, Solution is: 3, − 125 , 35 , −5
f3 = − 5967
f 35  = 424625737 = 679. 579 2
f−1 = − 3151
f5 = 41 825
Abs max 41825
Abs min −5967
10. Find the absolute minimum and maximum values (if they exist) of the function gx = 4x +
1
x
on 0, ∞.
g ′ x = 0, Solution is: − 12 , 12
g ′′  12  = 16 ⌣ min
g 12 = 4
Abs min 4
No abs max
11. The marketing department of Camcon has determined that the weekly demand for their digital cameras
is given by
p = dx = −0. 03x + 300
where p denotes the camera’s unit price (in dollars) and x denotes the quantity demanded.
a. Find the revenue function Rx.
Rx = x−0. 03x + 300
b. Find the actual revenue gained on the 2001st camera.
R2001 − R2000 = 179. 97
$179. 97
c. Find the marginal revenue when x = 2000. Explain why this is so close to your answer in part
b).
R ′ 2000 = 180. 0
$180. 00
When 2000 cameras are demanded, the rate of change of the revenue is $180. 00 per
camera. This means that the revenue gained on the next camera is approximately $180. 00.
d. Suppose that the marketing department has determined that the weekly cost incurred for
producing each camera is $86. 00. The weekly fixed costs incurred is $5600. 00. Find a
function,Cx, which represents weekly total cost.
Cx = 86x + 5600
e. Find the weekly profit function, Px.
Px = Rx − Cx = − 0. 03x 2 + 214x − 5600
f. How many cameras should be produced in order to maximize profit? Be sure to use calculus
6
g.
h.
i.
to justify your answer.
P ′ x = 0, Solution is: 3566. 666 666 666 67
P ′′ 3566. 67 = − 0. 06 ⌢ max
3567 cameras
What price should Camcom charge for each camera in order to maximize profit?
d3567 = 192. 99
$192. 99
How many cameras should be produced in order to break-even?
Px = 0, Solution is: 7107. 068 401 561 23, 26. 264 931 772 101 8
7107 or 26 cameras
Sketch a graph of the profit, revenue and cost functions on the same set of axes below which
will help the executives at Camcom to analyze their business practices. Label at least three
points on each axis. Indicate which graph corresponds to which function. Show the
executives where their maximum profit and break-even points occur.
8e+5
break-even
6e+5
max profit
4e+5
2e+5
0
break-even
0
2000
4000
6000
break-even 8000
x
Profit - solid, Revenue - dashed, Cost - circles
P3567 = 376033. 33
R ′ x = 0, Solution is: 5000. 0
R5000 = 750000. 0
R7107 = 616816. 53
12. Phillip, the proprietor of a vineyard, estimates that the first 10, 000 bottles of wine produced this season
will fetch a profit of $5 per bottle. However, the profit from each bottle beyond 10, 000 drops by $0. 0002
for each additional bottle sold. Assuming at least 10, 000 bottles of wine are produced and sold, how
many bottles should be sold in order to maximize profit? What will be the maximum profit? Be sure to
justify your answer using calculus.
7
x
# bottles
profit
0
10000
5
1
10001
4. 999 8
2
10002
4. 999 6
3
10003
4. 999 4
10
10010
4. 998
20
10020
4. 996
⋮
# bottles = 10000 + x
price = 5 − 0. 0002x
Px = 10000 + x5 − 0. 0002x
P ′ x = 0, Solution is: 7500. 0
P ′′ 7500 = − 0. 000 4 ⌢ max
Should sell 17500 bottles.
P7500 = 61250. 0
Max profit is $61, 250. 00
13. A poster is to have an area of 180 in 2 with 1-inch margins at the bottom and sides and a 2-inch margin
at the top. What dimensions will give the largest printed area? What will be the largest area? Be sure to
justify your answer using calculus.
2”
1”
1”
1”
Let x = width, y = height.
xy = 180, Solution is:
 180
x  if x ≠ 0
∅
if x = 0
y = 180
x
Printed area is x − 2y − 3
Ax = x − 2 180
x − 3
′
A x = 0, Solution is: x = −10. 954 451 150 103 3, x = 10. 954 451 150 103 3
A ′′ 10. 955 = − 0. 547 640 238 269 669 ⌢ max
180
Dimensions: width 10. 955, height 10.955
= 16. 431
A10. 955 = 120. 273 293 016 887
Max area 120. 273 in 2
14. A rectangular storage container with an open top is to have a volume of 10 m 3 . The length of its base is
twice the width. Material for the base costs $10 per square meter and material for the sides costs $6 per
square meter. Find the dimensions of the container that will minimize the cost of materials. What will be
8
the minimum cost? Be sure to justify your answer using calculus.
Let l = length of base, w = width of base, h = height
l = 2w
Volume is lwh = 2wwh = 10, Solution is:
h=
5
w2
if w ≠ 0
∅
if w = 0
5
w2
Cost for base is 10lw
Cost for front and back is 26lh
Cost for two sides is 26wh
Total cost is 10lw + 12lh + 12wh
Cw = 102ww + 122w w52 + 12w
5
w2
w = −0. 825 481 812 223 657 − 1. 429 776 439 495 4i,
C ′ w = 0, Solution is:
w = −0. 825 481 812 223 657 + 1. 429 776 439 495 4i,
w = 1. 650 963 624 447 31
C ′′ 1. 651 = 119. 994 712 331 738 ⌣ min
w = 1. 651
5
h=
2 = 1. 834 323 195 273 61
1.651
l = 21. 651 = 3. 302
Dimensions: l = 3. 302, w = 1. 651, h = 1. 834
C1. 651 = 163. 540 853 434 282
Minimum cost is $163. 55.
15. Postal regulations specify that a parcel sent by parcel post may have a combined length, depth and
height of no more than 108 in. In other words, the length plus the depth plus the height must be no
more than 108 in. Suppose a rectangular package with equal depth and height has a combined length,
depth and height of exactly 108 in.
Find the dimensions of the box that will yield the maximum volume. What will be the maximum volume?
l + h + h = 108, Solution is: 108 − 2h
l = 108 − 2h
Volume is ldh = lh 2
Vh = 108 − 2hh 2
V ′ h = 0, Solution is: 36, 0
V ′′ 36 = − 216 ⌢ max
108 − 236 = 36
V36 = 46 656
Dimensions: l = 36, d = 36, h = 36
9
Max volume 46656 in 3
16. A ship carrying math students has the misfortune to be wrecked on Calculus Island. The population of
the island after time t in years is given by
3
2
Pt = 3. 625t − 0. 399 2t3 + 366t + 1684
0. 001547t + 4. 34
a.
b.
How many students were initially shipwrecked? How many students were on the island after
1 year? 3 years? 10 years?
P0 = 388. 018 433 179 724
P1 = 472. 924 927 450 975
P3 = 656. 420 317 912 697
P10 = 1516. 745 371 156 79
Initial: 388 people
1 year: 473 people
3 years: 656 people
10 years: 1517 people
What is the limiting population to the island? That is, what will the population be in the long
run?
lim Pt = 2343. 244 990 303 81
t→∞
c.
Limiting population: 2343 people
At what rate is the population increasing after time t years?
P ′ t = 0.001 5471 t 3 +4. 34 10. 875t 2 − 0. 798 4t + 366 −
t2
3
2
0. 004 641
2 3. 625t − 0. 399 2t + 366t + 1684
3
0.001 547 t +4. 34
d.
How fast is the population increasing initially? after 1 year? 3 years? 10 years?
e.
P ′ 0 = 84. 331 797 235 023
P ′ 1 = 86. 117 173 305 206 7
P ′ 3 = 99. 060 854 130 170 8
P ′ 10 = 125. 971 542 932 926
Initially: 84 people per year
1 year: 86 people per year
3 years: 99 people per year
10 years: 126 people per year
How fast will the population be increasing in the long run?
lim P ′ t = 0. 0
t→∞
f.
g.
Population will not be increasing in the long run.
Over what interval is the population increasing? Explain why this is the same as solving
P ′ t > 0.
P ′ t > 0, Solution is: 1798. 205 401 396 06, ∞ ∪ −∞, −14. 103 702 503 783 6
∪ −14. 103 702 503 783 6, 36. 473 414 686 683 7
Increasing: 0, 36. 473  ∪ 1798. 205, ∞
When the rate of change is positive, the function is increasing.
Over what interval is the population decreasing? Explain why this is the same as solving
P ′ t < 0
P ′ t < 0, Solution is: 36. 473 414 686 683 7, 1798. 205 401 396 06
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h.
i.
Decreasing: 36. 473 , 1798. 205 
When the rate of change is negative, the population is decreasing
Over what interval is the rate of population increase increasing? Explain why this is the same
as solving P ′′ t > 0.
P ′′ t > 0, Solution is:
−∞, −14. 103 702 503 783 6 ∪ 4. 416 105 070 232 83 × 10 −2 , 8. 048 486 353 068 45
∪ 48. 884 277 379 678 6, 2703. 323 832 059 2
Increasing: 0. 0442, 8. 048 5 ∪ 48. 884 , 2703. 324
When the rate of change of the population increase is positive, then the population growth
rate is increasing.
Over what interval is the rate of population increase decreasing? Explain why this is the
same as solving P ′′ t < 0.
P ′′ t < 0, Solution is: −14. 103 702 503 783 6, 4. 416 105 070 232 83 × 10 −2 
∪ 2703. 323 832 059 2, ∞ ∪ 8. 048 486 353 068 45, 48. 884 277 379 678 6
Decreasing: 0, 0. 0442 ∪ 8. 048 55, 48. 884  ∪ 2703. 324, ∞
When the rate of change of the population increase is negative, then the population growth
rate is decreasing.
10. The concentration of a certain drug in an organ at any time t (in hours) is given by
gt = 0. 08 + 0. 121 − t −1.02 
where gt is measured in grams/cubic centimeter (g/cm 3 ).
a. What is the initial concentration of the drug in the organ?
g0 undefined. Unknown.
b. What is the concentration of the drug in the organ after 2 hours?
g2 = 0. 140 826 037 730 398
0. 141 g/cm 3
c. What will be the concentration of the drug in the organ in the long run?
lim gt = 0. 2
t→∞
d.
0. 2 g/cm 3
How fast is the concentration of the drug in the organ changing after 2 hours? Be sure to
include the units.
g ′ 2 = 3. 017 872 075 749 68 × 10 −2
Increasing by 0. 03018 g/cm 3 per hour.
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