Transcript Chapter 13 Curve Sketching

INTRODUCTORY MATHEMATICAL ANALYSIS
For Business, Economics, and the Life and Social Sciences
Chapter 13
Curve Sketching
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
Chapter Objectives
• To find critical values, to locate relative maxima
and relative minima of a curve.
• To find extreme values on a closed interval.
• To test a function for concavity and inflection
points.
• To locate relative extrema by applying the
second-derivative test.
• To sketch the graphs of functions having
asymptotes.
• To model situations involving maximizing or
minimizing a quantity.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
Chapter Outline
13.1) Relative Extrema
13.2) Absolute Extrema on a Closed Interval
13.3) Concavity
13.4) The Second-Derivative Test
13.5) Asymptotes
13.6) Applied Maxima and Minima
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.1 Relative Extrema
Increasing or Decreasing Nature of a Function
• Increasing f(x) if x1 < x2 and f(x1) < f(x2).
• Decreasing f(x) if x1 < x2 and f(x1) > f(x2).
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.1 Relative Extrema
Extrema
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.1 Relative Extrema
RULE 1 - Criteria for Increasing or Decreasing Function
• f is increasing on (a, b) when f’(x) > 0
• f is decreasing on (a, b) when f’(x) < 0
RULE 2 - A Necessary Condition for Relative Extrema
f ' a   0

relativeextremum


or


at a

 f ' a  does not exist

implies
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.1 Relative Extrema
RULE 3 - Criteria for Relative Extrema
1. If f’(x) changes from +ve to –ve, then f has a
relative maximum at a.
2. If f’(x) changes from -ve to +ve, then f has a
relative minimum at a.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.1 Relative Extrema
First-Derivative Test for Relative Extrema
1. Find f’(x).
2. Determine all critical values of f.
3. For each critical value a at which f is continuous,
determine whether f’(x) changes sign as x
increases through a.
4. For critical values a at which f is not continuous,
analyze the situation by using the definitions of
extrema directly.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.1 Relative Extrema
Example 1 - First-Derivative Test
4
y  f x   x 
for x  1 ,
x 1
If
use the first-derivative test
to find where relative extrema occur.
Solution:
x 2  2x  3 x  3x  1
STEP 1 - f ' x  

for x  1
2
2
x  1
x  1
STEP 2 - Setting f’(x) = 0 gives x = −3, 1.
STEP 3 - Conclude that at−3, there is a relative
maximum.
STEP 4 – There are no critical values at which f is not
continuous.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.1 Relative Extrema
Example 3 - Finding Relative Extrema
Test y  f x   x 2e x for relative extrema.
Solution: By product rule, f ' x   xex x  2
Relative maximum when x = −2
Relative minimum when x = 0.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.2 Absolute Extrema on a Closed Interval
Extreme-Value Theorem
• If a function is continuous on a closed interval, then
the function has a maximum value and a minimum
value on that interval.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.2 Absolute Extrema on a Closed Interval
Procedure to Find Absolute Extrema for a
Function f That Is Continuous on [a, b]
1. Find the critical values of f .
2. Evaluate f(x) at the endpoints a and b and at the
critical values in (a, b).
3. The maximum value of f is the greatest value
found in step 2. The minimum value is the least
value found in step 2.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.2 Absolute Extrema on a Closed Interval
Example 1 - Finding Extreme Values on a Closed Interval
Find absolute extrema for f x   x 2  4x  5 over the
closed interval [1, 4].
Solution:
Step 1: f ' x   2x  4x  2x  2
Step 2: f 1  2
f 4  5 values of f at endpoints
f 2  1 values of f at critical value 2 in 1, 4
Step 3: max is f 4  5 and min is f 2  1
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.3 Concavity
• Cases where curves concave upward:
• Cases where curves concave downward:
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.3 Concavity
• f is said to be concave up on (a, b) if f is
increasing on (a, b).
• f is said to be concave down on (a, b) if f is
decreasing on (a, b).
• f has an inflection point at a if it is continuous at a
and f changes concavity at a.
Criteria for Concavity
• If f’’(x) > 0, f is concave up on (a, b).
• If f”(x) < 0, f is concave down on (a, b).
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.3 Concavity
Example 1 - Testing for Concavity
Determine where the given function is concave up
and where it is concave down.
a. y  f x   x  1  1
3
Solution:
2
Applying the rule, y '  3x  1
y ' '  6x  1
Concave up when 6(x − 1) > 0 as x > 1.
Concave down when 6(x − 1) < 0 as x < 1.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.3 Concavity
Example 1 - Testing for Concavity
b. y  x 2
Solution:
Applying the rule, y '  2x
y''  2
As y’’ is always positive, y = x2 is always concave up.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.3 Concavity
Example 3 - A Change in Concavity with No Inflection Point
Discuss concavity and find all inflection points for
f(x) = 1/x.
Solution: f ' x    x 2 for x  0
f ' ' x   2x 3 for x  0
x > 0 f”(x) > 0 and x < 0  f”(x) < 0.
f is concave up on (0,∞) and
concave down on (−∞, 0)
f is not continuous at 0  no inflection point
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.4 The Second-Derivative Test
• The test is used to test certain critical values for
relative extrema.
Suppose f’(a) = 0.
• If f’’(a) < 0, then f has a relative maximum at a.
• If f’’(a) > 0, then f has a relative minimum at a.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.4 The Second-Derivative Test
Example 1 - Second-Derivative Test
Test the following for relative maxima and minima.
Use the second-derivative test, if possible.
a. y  18x  32 x 3
Solution: y '  23  x 3  x 
y ' '  4 x
When y '  0, w ehave x  3
When x  3, y ' '  43  12  0
When x  3, y ' '  4 3  12  0
Relative minimum when x = −3.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.4 The Second-Derivative Test
Example 1 - Second-Derivative Test
b. y  6x 4  8x 3  1
Solution: y '  24 x 3  24 x 2  24 x 2 x  1
y ' '  72x 2  48 x
When y '  0, w ehave x  0, 1
When x  0, y ' '  0
When x  1, y ' '  0
No maximum or minimum
exists when x = 0.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.5 Asymptotes
Vertical Asymptotes
• The line x = a is a vertical asymptote if at least
one of the following is true:
lim f x   
x a
lim f x   
x a
Vertical-Asymptote Rule for Rational Functions
• P and Q are polynomial functions and the quotient
is in lowest terms.
P x 
f x  
Q x 
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.5 Asymptotes
Example 1 - Finding Vertical Asymptotes
Determine vertical asymptotes for the graph of
x 2  4x
f x   2
x  4x  3
Solution: Since f is a rational function,
x 2  4x
f x  
x  3x  1
Denominator is 0 when x is 3 or 1.
The lines x = 3 and x = 1
are vertical asymptotes.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.5 Asymptotes
Horizontal and Oblique Asymptotes
• The line y = b is a horizontal asymptote if at least
one of the following is true:
lim f x   b or lim f x   b
x 
x  
Nonvertical asymptote
• The line y = mx +b is a nonvertical asymptote if
at least one of the following is true:
limf x   mx  b   0 or lim f x   mx  b   0
x 
2011 Pearson Education, Inc.
x  
Chapter 13: Curve Sketching
13.5 Asymptotes
Example 3 - Finding an Oblique Asymptote
Find the oblique asymptote for the graph of the
rational function
10 x 2  9 x  5
y  f x  
5x  2
Solution: f x   10 x  9 x  5  2x  1  3
5x  2
5x  2
3
lim f x   2 x  1  lim
0
x  
x   5 x  2
2
y = 2x + 1 is an oblique asymptote.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.5 Asymptotes
Example 5 - Finding Horizontal and Vertical Asymptotes
Find horizontal and vertical asymptotes for the graph
y  f x   e x  1
Solution: Testing for horizontal asymptotes,
 
lim e  1  lim e
lim e x  1  
x 
x
x 
x 
x
 lim 1  0  1  1
x 
The line y = −1 is a horizontal asymptote.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.5 Asymptotes
Example 7 - Curve Sketching
Sketch the graph of
Solution:
Intercepts
Symmetry
4x
y 2
.
x 1
(0, 0) is the only intercept.
There is only symmetry about the origin.
Asymptotes Denominator  0  No vertical asymptote
4x
Since lim 2
0
x  x  1
y = 0 is the only non-vertical asymptote
  
Max and Min For y '  4 1 x 12 x , relative maximum is (1, 2).
x
Concavity
For y ' ' 

1
8x x  3 x  3

2
x
2


1
3

, inflection points are
(-√ 3, -√3), (0, 0), (√3, √3).
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.5 Asymptotes
Example 7 - Curve Sketching
Solution: Graph
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
• Use absolute maxima and minima to explain the
endpoints of the domain of the function.
Example 1 - Minimizing the Cost of a Fence
A manufacturer plans to fence in a 10,800-ft2 rectangular
storage area adjacent to a building by using the building as
one side of the enclosed area. The fencing parallel to the
building faces a highway and will cost $3 per foot installed,
whereas the fencing for the other two sides costs $2 per foot
installed. Find the amount of each type of fence so that the
total cost of the fence will be a minimum.
What is the minimum cost?
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
Example 1 - Minimizing the Cost of a Fence
Solution:
Cost function is C  3x  2y  2y  C  3x  4y
10800
Storage area is xy  10,800  y 
x
Analyzing the equations, Cx   3x  4 10800   3x  43200

dC
43200
0 3
dx
x2
x  120 since x  0
Only critical value is
120.

2
d
C 86400
and

2
dx
x3
d 2C
When x  120, 2  0
dx
x =120 gives a relative minimum.
43200
Thus, C 120  3 x 
 720
120
2011 Pearson Education, Inc.
x
x
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
Example 3 - Minimizing Average Cost
A manufacturer’s total-cost function is given by
q2
c  c q  
 3q  400
4
where c is the total cost of producing q units. At what
level of output will average cost per unit be a
minimum? What is this minimum?
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
Example 3 - Minimizing Average Cost
Solution:
Average-cost function is q 2
 3q  400
c
q
400
4
c  c q   
 3
q
q
4
q
To find critical values, we set
dc
q 2  1600
0
 q  40 since q  0
2
dq
4q
2
d c 800
 3
2
dq
q is positive when q = 40, which is the only relative
extremum.
40
400
3
 23
The minimum average cost is c 40  
4
40
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
Example 5 - Economic Lot Size
A company annually produces and sells 10,000 units of a
product. Sales are uniformly distributed throughout the
year. The company wishes to determine the number of
units to be manufactured in each production run in order
to minimize total annual setup costs and carrying costs.
The same number of units is produced in each run. This
number is referred to as the economic lot size or
economic order quantity. The production cost of each unit
is $20, and carrying costs (insurance, interest, storage,
etc.) are estimated to be 10% of the value of the average
inventory. Setup costs per production run are $40. Find
the economic lot size.
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
Example 5 - Economic Lot Size
Solution:
Let q be the number of units in a production run.
Total of the annual carrying costs and setup is
 10000 
40000
q
  q 
C  0.120   40
q
2
 q 
Setting
dC
400000 q 2  400000
 1

2
dq
q
q2
2
dC/dq = 0, we get dC  0  q  400000
dq
q2
q  400000  632.5
Since q > 0, there is an absolute minimum at q = 632.5.
Number of production runs = 10,000/632.5  15.8
16 lots  Economic size = 625 units
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
Example 7 - Maximizing the Number of Recipients of
Health-Care Benefits
An article in a sociology journal stated that if a
particular health-care program for the elderly were
initiated, then t years after its start, n thousand elderly
people would receive direct benefits, where
t3
n   6t 2  32t 0  t  12
3
For what value of t does the maximum number
receive benefits?
2011 Pearson Education, Inc.
Chapter 13: Curve Sketching
13.6 Applied Maxima and Minima
Example 7 - Maximizing the Number of Recipients of Health-Care Benefits
Solution: Setting dn/dt = 0, we have
t3
n   6t 2  32t 0  t  12
3
Absolute maximum value of n must occur at t = 0, 4,
dn
8, or 12:
 0  t 2  12t  32
dt
t  4 or t  8
160
128
n 0   0, n 4  
, n 8  
, n 12  96
3
3
Absolute maximum occurs when t = 12.
2011 Pearson Education, Inc.