Introductory maths analysis chapter 13 official

INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 13Chapter 13
Curve SketchingCurve Sketching

INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics

9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

• 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.
Chapter 13: Curve Sketching
Chapter ObjectivesChapter Objectives

Relative Extrema
Absolute Extrema on a Closed Interval
Concavity
The Second-Derivative Test
Asymptotes
Applied Maxima and Minima
13.1)
13.2)
13.3)
Chapter OutlineChapter Outline
13.4)
13.5)
13.6)

13.1 Relative Extrema13.1 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).

13.1 Relative Extrema
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
( )
( )



 =
⇒



existnotdoes'
or
0'
at
extremumrelative
af
af
a
implies

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.

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.

Example 1 - First-Derivative Test
If , use the first-derivative test
to find where relative extrema occur.
Solution:
STEP 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
( ) 1for
1
4
≠
+
+== x
x
xxfy
( )
( )
( )( )
( )
1for
1
13
1
32
' 22
2
−≠
+
−+
=
+
−+
= x
x
xx
x
xx
xf

Example 3 - Finding Relative Extrema
Test for relative extrema.
Solution: By product rule,
Relative maximum when x = −2
Relative minimum when x = 0.
( ) x
exxfy 2
==
( ) ( )2' += xxexf x

13.2 Absolute Extrema on a Closed Interval13.2 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.

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.

13.2 Absolute Extrema on a Closed Interval
Example 1 - Finding Extreme Values on a Closed Interval
Find absolute extrema for over the
closed interval [1, 4].
Solution:
Step 1:
Step 2:
Step 3:
( ) 542
+−= xxxf
( ) ( )2242' −=−= xxxxf
( )
( ) endpointsatofvalues54
21
ff
f
=
=
( ) ( )41,in2valuecriticalatofvalues12 ff =
( ) ( ) 12isminand54ismax == ff

13.3 Concavity13.3 Concavity
• Cases where curves concave upward:
• Cases where curves concave downward:

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).

13.3 Concavity
Example 1 - Testing for Concavity
Determine where the given function is concave up
and where it is concave down.
Solution:
Applying the rule,
Concave up when 6(x − 1) > 0 as x > 1.
Concave down when 6(x − 1) < 0 as x < 1.
( ) ( ) 11a.
3
+−== xxfy
( )
( )16''
13'
2
−=
−=
xy
xy

13.3 Concavity
Example 1 - Testing for Concavity
Solution:
Applying the rule,
As y’’ is always positive, y = x2
is always concave up.
2
b. xy =
2''
2'
=
=
y
xy

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:
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
( )
( ) 0for2''
0for'
3
2
≠=
≠−=
−
−
xxxf
xxxf

13.4 The Second-Derivative Test13.4 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.

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.
Solution:
Relative minimum when x = −3.
3
3
2
18. xxya −=
( )( )
xy
xxy
4''
332'
−=
−+=
3havewe,0'When ±== xy
( )
( ) 01234'',3When
01234'',3When
>=−−=−=
<−=−=+=
yx
yx

13.4 The Second-Derivative Test
Example 1 - Second-Derivative Test
Solution:
No maximum or minimum
exists when x = 0.
186. 34
+−= xxyb
( )
xxy
xxxxy
4872''
1242424'
2
223
−=
−=−=
1,0havewe,0'When == xy
0'',1When
0'',0When
>=
==
yx
yx

13.5 Asymptotes13.5 Asymptotes
Vertical Asymptotes
• The line x = a is a vertical asymptote if at least
one of the following is true:
Vertical-Asymptote Rule for Rational Functions
• P and Q are polynomial functions and the quotient
is in lowest terms.
( )
( ) ±∞=
±∞=
−
+
→
→
xf
xf
ax
ax
lim
lim
( ) ( )
( )xQ
xP
xf =

13.5 Asymptotes
Example 1 - Finding Vertical Asymptotes
Determine vertical asymptotes for the graph of
Solution: Since f is a rational function,
Denominator is 0 when x is 3 or 1.
The lines x = 3 and x = 1
are vertical asymptotes.
( )
34
4
2
2
+−
−
=
xx
xx
xf
( )
( )( )13
42
−−
−
=
xx
xx
xf

13.5 Asymptotes
Horizontal and Oblique Asymptotes
• The line y = b is a horizontal asymptote if at least
one of the following is true:
Nonvertical asymptote
• The line y = mx +b is a nonvertical asymptote if
at least one of the following is true:
( ) ( ) bxfbxf
xx
==
−∞→∞→
limorlim
( ) ( )( ) ( ) ( )( ) 0limor0lim =+−=+−
−∞→∞→
bmxxfbmxxf
xx

13.5 Asymptotes
Example 3 - Finding an Oblique Asymptote
Find the oblique asymptote for the graph of the
rational function
Solution:
y = 2x + 1 is an oblique asymptote.
( )
25
5910 2
+
++
==
x
xx
xfy
( )
25
3
12
25
5910 2
+
++=
+
++
=
x
x
x
xx
xf
( )( ) 0
25
3
lim12lim =
+
=+−
±∞→±∞→ x
xxf
xx

13.5 Asymptotes
Example 5 - Finding Horizontal and Vertical Asymptotes
Find horizontal and vertical asymptotes for the graph
Solution: Testing for horizontal asymptotes,
The line y = −1 is a horizontal asymptote.
( ) 1−== x
exfy
( )
( ) 1101limlim1lim
1lim
−=−=−=−
∞=−
−∞→−∞→−∞→
∞→
x
x
x
x
x
x
x
ee
e

13.5 Asymptotes
Example 7 - Curve Sketching
Sketch the graph of .
Solution:
Intercepts (0, 0) is the only intercept.
Symmetry There is only symmetry about the origin.
Asymptotes Denominator ≠ 0  No vertical asymptote
Since
y = 0 is the only non-vertical asymptote
Max and Min For , relative maximum is (1, 2).
Concavity For , inflection points are
(-√ 3, -√3), (0, 0), (√3, √3).
1
4
2
+
=
x
x
y
0
1
4
lim 2
=
+∞→ x
x
x
( )( )
( )22
1
114
'
+
−+
=
x
xx
y
( )( )
( )32
1
338
''
+
−+
=
x
xxx
y

13.5 Asymptotes
Example 7 - Curve Sketching
Solution: Graph

13.6 Applied Maxima and Minima13.6 Applied Maxima and Minima
Example 1 - Minimizing the Cost of a Fence
• Use absolute maxima and minima to explain the
endpoints of the domain of the function.
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?

13.6 Applied Maxima and Minima
Example 1 - Minimizing the Cost of a Fence
Solution:
Cost function is
Storage area is
Analyzing the equations,
Thus,
yxCyyxC 43223 +=⇒++=
x
yxy
10800
800,10 =⇒=
( )
x
x
x
xxC
43200
3
10800
43 +=





+=
0since120
43200
30 2
>=
−==
xx
xdx
dC
0,120When
86400
2
2
32
2
>=
=
dx
Cd
x
xdx
Cd
and
Only critical value is
120.
x =120 gives a relative minimum.
( ) 720
120
43200
3120 =+= xC

Example 3 - Minimizing Average Cost
A manufacturer’s total-cost function is given by
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?
( ) 4003
4
2
++== q
q
qcc

Example 3 - Minimizing Average Cost
Solution:
Average-cost function is
To find critical values, we set
is positive when q = 40, which is the only relative
extremum.
The minimum average cost is
( )
q
q
q
q
q
q
c
qcc
400
3
4
4003
4
2
++=
++
===
0since40
4
1600
0 2
2
>=⇒
−
== qq
q
q
dq
cd
32
2
800
qdq
cd
=
( ) 23
40
400
3
4
40
40 =++=c

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.

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
Setting dC/dq = 0, we get
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
( )
2
2
2
400000400000
1
4000010000
40
2
201.0
q
q
qdq
dC
q
q
q
q
C
−
=−=
+=





+





=
5.632400000
400000
0 2
2
≈=
−
==
q
q
q
dq
dC

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
For what value of t does the maximum number
receive benefits?
120326
3
2
3
≤≤+−= ttt
t
n

Example 7 - Maximizing the Number of Recipients of Health-Care Benefits
Solution: Setting dn/dt = 0, we have
Absolute maximum value of n must occur at t = 0, 4,
8, or 12:
Absolute maximum occurs when t = 12.
120326
3
2
3
≤≤+−= ttt
t
n
8or4
32120 2
==
+−==
tt
tt
dt
dn
( ) ( ) ( ) ( ) 9612,
3
128
8,
3
160
4,00 ==== nnnn

Introductory maths analysis chapter 13 official

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Introductory maths analysis chapter 13 official