College Mathematics: Learning Worksheets Chapter 11
Name ________________________________ Date ______________ Class ____________
Goal: To find limits of functions that are in the form zero over zero or infinity over infinity
In Problems 1 and 2, use L’Hôpital’s rule to find the limit.
3
27
x
−
22
33(3)27
x
2.
4
5
317
lim 625
x
3
4
12 12
lim lim 0
x
Section 11-3 L’Hôpital’s Rule
Theorems: L’Hôpital’s Rule for 0/0 Indeterminate forms
1. For c, a real number, if lim ( ) 0
xc
fx
and lim ( ) 0,
xc
gx
then
f
xfx
2. L’Hôpital’s rule for the indeterminate form 0/0 are also valid if the limit of f
and the limit of g are both infinite.
324
3. Explain why L’Hôpital’s rule does not apply to the following problem. If the limit
exists, find it by other means.
3
2
4
235
lim
35
x
x
x
→
−
+
L’Hôpital’s rule does not apply because when the value is substituted into the
function, the result is not 0/0.
4. Use L’Hôpital’s rule to find the following limit and then verify that limit
by a second method.
2
235
lim
xx
−−
2102(7)1041
lim 22 2(7)2123
x
5.
2
1
lim
x
x
x
If the value –1 is substituted into the function, the result will be 0/0. Therefore, we
use L’Hôpital’s rule.
322
76 3 73(1)710 10
xx x
College Mathematics: Learning Worksheets Chapter 11
6.
4
3
2
63
lim 26
x
xx
xx
The value can be substituted into the function and not result in 0/0. Therefore, use
simple substitution.
7.
3
4
745
lim 967
x
xx
xx
If the value is substituted into the function, the result will be /. Therefore, we
use L’Hôpital’s rule.
43
2
967 366
42
lim 108
42
lim 216
0
xx
x
x
xx x
x
x
x
8. 2
3
26
lim
524
x
x
xx
→−
+
−−
If the value –3 is substituted into the function, the result will be 0/0. Therefore, we
use L’Hôpital’s rule.
26 2 2 2
x
+
33
3
524
2
lim 8
22
38 11
xx
x
xx
xx
x
→− →−
→−
+−
−−
=
−
==−
−−
College Mathematics: Learning Worksheets Chapter 11
326
College Mathematics: Learning Worksheets Chapter 11
327
Name ________________________________ Date ______________ Class ____________
Goal: To graph functions using the final version of the graphing strategy.
In Problems 1–3, summarize the pertinent information obtained by applying the graphing
strategy and sketch the graph.
1. 4
() 4
x
fx x
Step 1: Analyze the function.
b) The x intercepts are found by setting the function (or the numerator) equal to zero
and the y intercept is found by setting the value of x equal to zero.
c) The vertical asymptote will be at the point of discontinuity or 4.x
The horizontal asymptote is found by taking the limit of the function as it goes to
Section 11-4 Curve-Sketching Techniques
Procedure: Graphing Strategy (Final Version)
Step 1. Analyze ().
f
x
Step 2. Analyze ‘( ).
f
xFind the partition numbers for, and the critical values of,
f
Step 3. Analyze ”( ).
f
xFind the partition numbers for ”( ).
f
x Determine the
Step 4. Sketch the graph of the function.
College Mathematics: Learning Worksheets Chapter 11
328
Step 2: Analyze the first derivative.
2
8
‘( ) (4)
fx
x
There is only one partition number, which is the value that makes the denominator
zero, or 4.x Pick a test number on each side of the partition number and find the
value of the first derivative at these test numbers:
Test number: –5 Test number: –3
8
8
Step 3: Analyze the second derivative
2
‘( ) 8( 4)
fx x
College Mathematics: Learning Worksheets Chapter 11
Step 4: Sketch the graph
2. 4
() 3
x
fx x
Step 1: Analyze the function.
b) The x intercepts are found by setting the function (or the numerator) equal to zero
and the y intercept is found by setting the value of x equal to zero.
c) The vertical asymptote will be at the point of discontinuity or 3.x
The horizontal asymptote is found by taking the limit of the function as it goes to
infinity. Since the exponent in the numerator is the same as the denominator, the
horizontal asymptote is 4
14.y
Step 2: Analyze the first derivative.
4
() 3
x
fx x
College Mathematics: Learning Worksheets Chapter 11
Test number: 2 Test number: 4
2
12
‘(2) (2 3)
f
2
12
‘(4) (4 3)
f
Step 3: Analyze the second derivative
2
12
‘( ) (3)
fx
x
There is only one partition number; the value that makes the denominator zero is
3.x Pick a test number in each interval. Find the value of the second derivative at
these test numbers:
Test number: 2 Test number: 4
24
24
College Mathematics: Learning Worksheets Chapter 11
331
Step 4: Sketch the graph
3. 2
3
()
x
fx e
Step 1: Analyze the function.
b) There are no x intercepts and the y intercept is found by setting the value of x
equal to zero.
3(0) 0
c) There is no vertical asymptote because the domain is all real numbers.
Step 2: Analyze the first derivative.
3
x
There is only one partition number; and also a value that makes the function zero or
0.x Pick a test number on each side of the critical value and find the value of the first
derivative at these test numbers:
Test number: –1 Test number: 1
2
3( 1)
‘( 1) 6( 1)
fe
2
3(1)
‘(1) 6(1)
fe
College Mathematics: Learning Worksheets Chapter 11
Therefore, the function is increasing on the interval ( , 0)
and decreasing on the
interval (0, ), and has a local maximum at 0.x
Step 3: Analyze the second derivative
2
3
() 6
x
fx xe
There are two partition numbers. Set the second derivative equal to zero to find the
partition numbers 0.41.x Pick a test number in each interval of the partition number and
find the value of the second derivative at these test numbers:
Test number: –1 Test number: 0
2
3( 1) 2
‘‘( 1) 6 (6( 1) 1)
fe
2
3(0) 2
‘‘(0) 6 (6(0) 1)
fe
Test number: 1
2
3(1) 2
‘‘(1) 6 (6(1) 1)
fe
Therefore, the function is concave upward on the intervals ( , 0.41) and
Step 4
: Sketch the graph
College Mathematics: Learning Worksheets Chapter 11
Name ________________________________ Date ______________ Class ____________
Goal: To find absolute maxima or minima on open and closed intervals
Section 11-5 Absolute Maxima and Minima
Definition: Absolute Maxima and Minima
If () ()
f
cfxfor all x in the domain of f, then ()
f
cis called the absolute maximum.
If () ()
f
cfxfor all x in the domain of f, then ()
f
cis called the absolute minimum.
Theorems:
1. A function f that is continuous on a closed interval [a, b] has both an absolute
maximum and an absolute minimum on that interval.
3. Second Derivative Test
Let f be continuous on an interval I with only one critical value c in I.
f
f
Procedure: Finding absolute extrema on closed intervals
1. Check to make certain that f is continuous over [a, b].
3. Evaluate f at the endpoints a and b and at the critical values found in step 2.
5. The absolute minimum is the smallest value found in step 3.
f
1. Find the absolute maximum and the absolute minimum for the given function on
the given interval.
2
() 7 14 5
f
xx x=−+ [2,5]−
First, find the critical value(s) for the function.
2
() 7 14 5
f
xx x
=−+
Calculate the function value at the endpoints and the critical value.
2
( 2) 7( 2) 14( 2) 5
f
−=− − −+
2
(1) 7(1) 14(1) 5
f
=−+
2
(5) 7(5) 14(5) 5
f
=−+
Therefore, the absolute maximum is at the point (5, 110) and the absolute minimum is
at the point (1, –2).
2. 34
() 5 6
f
xxx
First find the critical value(s) for the function.
34
f
() 5 6
fx x x
2
30
x
58 0
x
College Mathematics: Learning Worksheets Chapter 11
The second-derivative test fails at 0 and shows that there is a local maximum at 5
8.
Calculate the function value at the critical values.
34
(0) 5(0) 6(0)
f
34
55 5
88 8
() 5() 6()
f
Therefore, the local maximum is at the point 5 625
8 2048
( , ) and the local minimum is at the
3.
2
2
5
()
2
x
fx
x
−
=
+
First find the critical value(s) for the function.
2
2
5
()
2
x
fx
x
−
=
+
To determine if this is a maximum or a minimum, we find the second derivative and
College Mathematics: Learning Worksheets Chapter 11
42
42 3
422
14
‘( )
44
( 4 4)(14) (14 )(4 8 )
”( )
x
fx
xx
x
xxxx
fx
=
++
++ − +
=
Therefore, (0,–2.5) is a local minimum point because the second derivative is
4. Absolute minimum value on (0, )
for 7
() 6 2 .fx x
x
First find the critical value(s) for the function.
1
2
() 6 2 7
f
xxx
2
72
x
337
5. Absolute minimum value on [0, )for 2
() 10 8 26.fx x x
First find the critical value(s) for the function.
2
() 10 8 26
fx x x
There is only one critical value. To determine if this is a minimum, we will find the
second derivative and use the second-derivative test.
‘( ) 20 8
f
xx
338
6. Absolute maximum value on (0, )for 64
() 3 18 .
f
xx x
First find the critical value(s) for the function.
64
53
() 3 18
‘( ) 18 72
f
xx x
f
xx x
0
x
2
4
2
x
x
Only
2x is in the interval (0, ). To determine if the critical value 2xis a
maximum, we find the second derivative and use the second-derivative test.
53
42
‘( ) 18 72
”( ) 90 216
fx x x
fx x x
College Mathematics: Learning Worksheets Chapter 11
339
Name ________________________________ Date ______________ Class ____________
Goal: To solve application problems that involve maximum and minimum values
Solve the following problems using the four-step procedure outlined above.
1. Find two numbers whose sum is 50 and whose product is maximum.
Step 1: Maximize Axy subject to 50xy+=
Step 2: 50
50
xy
x
y
+=
=−
()
() (50 )
Ay xy
Ay yy
=
=−
Section 11-6 Optimization
Procedure: Strategy for Solving Optimization Problems
1. Introduce variables, look for relationships among the variables, and construct
f
2. Find the critical values of ().
f
x
3. Use the procedure developed in Section 11-5 to find the absolute maximum
f
4. Use the solution to the mathematical model to answer all the questions asked
in the problem.
College Mathematics: Learning Worksheets Chapter 11
340
Step 3: Use the second-derivative test to determine whether the critical value
produces a maximum value.
Step 4: The numbers 25 and 25 have a sum 50 and gives the maximum product
2. A fence is to be built to enclose a rectangular area of 900 square feet. The fence
along three sides is to be made of material that costs $14 per foot. The material for the fourth
side costs $42 per foot. Find the dimensions of the rectangle that will allow for the most
economical fence to be built.
Step 1: Minimize 28 56Cxy subject to 900xy
Step 2:
900
900
xy
xy
1
( ) 28 56
( ) 28(900 ) 56
Cy x y
Cy y y
College Mathematics: Learning Worksheets Chapter 11
341
Step 3: Use the second-derivative test to determine whether the critical value
produces a maximum value.
2
‘( ) 25, 200 56
Cy y
Step 4: A rectangular field that has a width of 21.21 feet (at a cost of $14.00 per
3. A commercial apple grower estimates from past records that if 50 trees are planted
per acre, then each tree will yield an average of 70 pounds of apples per season. If, for each
additional tree planted per acre (up to 30), the average yield is reduced by 1 pound, how
many trees should be planted per acre to obtain the maximum yield per acre? What is the
maximum yield?
Step 1: Maximize ( ) (50 )(70 )Px x x=+ − subject to 30x
Step 2:
() (50 )(70 )
Px x x
=+ −
College Mathematics: Learning Worksheets Chapter 11
342
Step 4: The variable x is the number of additional trees that should be planted.
Therefore, 10 additional trees should be planted per acre for a total of 60 trees per acre. The
4. A 400-room hotel in New York City is filled to capacity every night at $100 a room.
For each $2 increase in rent, 4 fewer rooms are rented. If each rented room costs $12 to
service per day, how much should the management charge for each room to maximize gross
profit? What is the maximum gross profit?
Step 1: Maximize () () ()Px Rx Cx subject to 100x
Step 2:
( ) (400 4 )(100 2 ) 12(400 4 )
Px x x x
Step 4: The variable x is the number of $2 increases that the management
should add to the price of the room. Therefore, management should increase the price of the