Introductory maths analysis chapter 13 official

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 13 Chapter 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 ExtremaIncreasing 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

existnot does '

or

0'

at

extremum relative

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

continuous.

1 for 1

4

x

xxxfy

1 for 1

13

1

32'

22

2

xx

xx

x

xxxf




Example 3 - Finding Relative Extrema

Test for relative extrema.

Solution: By product rule,Relative maximum when x = −2 Relative minimum when x = 0.

xexxfy 2

2' xxexf x



13.2 Absolute Extrema on a Closed Interval13.2 Absolute Extrema on a Closed IntervalExtreme-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

endpoints at of values 54

21

ff

f

4 1, in 2 value critical at of values 12 ff

12 is min and 54 is max 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.

11 a. 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

0 for 2''

0 for '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.

33218 . xxya

xy

xxy

4''

332'

3 have we,0' When xy

01234'' ,3 When

01234'' ,3 When

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 ,0 have we,0' When xy

0'' ,1 When

0'' ,0 When

yx

yx



13.5 Asymptotes13.5 AsymptotesVertical 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

xPxf



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

42

2

xx

xxxf

13

42

xx

xxxf



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:

bxfbxfxx

lim or lim

0lim or 0lim

bmxxfbmxxfxx



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

xxxfy

25

312

25

5910 2

x

xx

xxxf

025

3lim12lim

xxxf

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 xexfy

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

42

x

xy

01

4lim

2

x

xx

22 1

114'

x

xxy

32 1

338''

x

xxxy



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

xyxy

10800800,10

x

xx

xxC43200

310800

43

0 since 120

4320030

2

xxxdx

dC

0 ,120 When

86400

2

2

32

2

dx

Cdx

xdx

Cdand

Only critical value is 120. x =120 gives a relative minimum.

720120

432003120 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?

40034

2

qq

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

qq

q

cqcc

4003

4

40034

2

0 since 404

16000

2

2

qqq

q

dq

cd

32

2 800

qdq

cd

2340

4003

4

4040 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

4000004000001

400001000040

2201.0

q

q

qdq

dC

qq

q

qC

5.632400000

4000000

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?

120 3263

23

tttt

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.

120 3263

23

tttt

n

8 or 4

32120 2

tt

ttdt

dn

9612 ,3

1288 ,

3

1604 ,00 nnnn

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