Transcript W A T K I N S - J O H N S O N C O M P A N Y Semiconductor

Chabot Mathematics
§3.1 Relative
Extrema
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
1
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[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Review §
2.6
 Any QUESTIONS About
• §2.6 → Implicit Differentiation
 Any QUESTIONS
About
HomeWork
• §2.6 →
HW-12
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§3.1 Learning Goals
 Discuss increasing and decreasing
functions
 Define critical points and relative
extrema
 Use the first
derivative test to
study relative
extrema and
sketch graphs
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Increasing & Decreasing Values
 A function f is INcreasing if whenever
a<b, then:
f (a) < f (b)
• INcreasing is Moving UP from Left→Right
 A function f is DEcreasing if whenever
a<b, then:
f ( a )  f (b )
• DEcreasing is Moving DOWN from Left→Right
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Inc & Dec Values Graphically
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DEcreasing
6
5
4
y = f(x)
3
2
INcreasing
1
0
-1
-2
-3
-4
XY f cnGraph6x6BlueGreenBkGndTemplate1306.m
0
1
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2
3
4
5
x
6
7
8
9
10
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Inc & Dec with Derivative
df
 If for every c on
f ' (c ) 
the interval [a,b]
dx
0
xc
• That is, the Slope is POSITIVE
Then f is INcreasing on [a,b]
 If for every c on f ' ( c )  df
dx
the interval [a,b]
0
xc
• That is, the Slope is NEGATIVE
Then f is DEcreasing on [a,b]
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Example  Inc & Dec
y  f x   x  3  4
2
 The function, y = f(x),is decreasing on
[−2,3] and increasing on [3,8]
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Example  Inc & Dec Profit
 The default list price of
p(x)
=10
a small bookstore’s
paperbacks Follows this Formula
• Where
– x ≡ The Estimated Sales Volume in No. Books
– p ≡ The Book Selling-Price in $/book
 The bookstore buys paperbacks for $1
each, and has daily overhead of $50
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x
Example  Inc & Dec Profit
 For this Situation Find:
• Find the profit as a function of x
• intervals of increase and decrease for the
Profit Function
 SOLUTION
 Profit is the difference of revenue and
cost, so first determine the revenue as a
function of x:
R x   x  p x 
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(
= x × 10 - x
)
 10 x  x
3/2
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Example  Inc & Dec Profit
 And now cost as a function of x:
C  x   variable
cost  fixed cost  1 x  50
 Then the Profit is the Revenue minus
the Costs: P  x   R  x   C  x 
= (10x - x 3/2 ) - ( x + 50)
= 9x - x 3/2 - 50.
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Example  Inc & Dec Profit
 Now we turn to determining the intervals
of increase and decrease.
3/2
 The graph of the profit
y = 9x - x - 50
function is shown
next on the interval
[0,100] (where the
price and quantity
demanded are
both non-negative).
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Example  Inc & Dec Profit
 From the Plot Observe that The profit
function appears to be increasing until
some sales level below 40, and then
decreasing thereafter.
 Although a graph is
informative, we turn
to calculus to
determine the
y = 9x - x 3/2 - 50
exact intervals
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Example  Inc & Dec Profit
 We know that if the derivative of a
function is POSITIVE on an open
interval, the function is INCREASING on
that interval. Similarly, if the derivative
is negative, the function is decreasing
 So first compute the P '  x   d 9 x  x 3 / 2  50 
dx
derivative, or Slope,
3 1/ 2
function:
9 x
2
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Example  Inc & Dec Profit
 On Increasing intervals the Slope is
POSTIVE or NonNegative dP
3 1/ 2
9 x 0
so in this case need
dx
2
 Solving
3 1/ 2
3 1/ 2
9 x 0
 x
 9
This
2
2
InEquality:
2
1/ 2
x  36
x  9
 The profit
3
function is
DEcreasing on the interval [36,100]
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Relative Extrema (Max & Min)
 A relative maximum of a function f is
located at a value M such that
f(x) ≤ f(M) for all values of x on an
interval a<M<b
 A relative minimum of a function f is
located at a value m such that
f(x) ≥ f(m) for all values of x on an
interval a<m<b
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Peaks & Valleys
 Extrema is precise math terminology for
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Both of
4
• The Bottom
of a Trough,
That is a
VALLEY
3
2
y = f(x)
• The TOP of a
Hill; that is,
a PEAK
PEAK
PEAK
VALLEY
1
0
-1
-2
-3
0
VALLEY
10
20
30
x
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40
50
MTH15 • Bruce Mayer, PE
4
3
y = f(x)
2
Rel&Abs Max& Min
Absolute
Max
Relative
Max
Relative
Min
1
0
-1
-2
-3
0
Absolute
Min
10
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20
30
x
40
50
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Critical Points
 Let c be a value in the domain of f
 Then c is a Critical Point If, and only if
df
dx
0
xc
HORIZONTAL
slope
at c
Chabot College Mathematics
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OR
df
dx
 
xc
VERTICAL
slope
at c
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Critical Points GeoMetrically
 Horizontal
 Vertical
MTH15 • Zero Critical-Pt
MTH15 •  Critical-Pt
1.3
20
1
0.9
16
14
y = f(x)
1.1
y=f(x)
18
(0.1695, 1.2597)
1.2
12
10
8
6
0.8
4
0.7
2
0.6
0.05
0.1
0.15
x
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0.2
0.25
0
0
0.5
1
1.5
2
x
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2.5
3
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MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 07Jul13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
clear; clc;
% The Limits
xmin = 0; xmax = 0.27;
ymin =0; ymax = 1.3;
% The FUNCTION
x = linspace(xmin,xmax,1000); y1 = x.*(12-10*x-100*x.^2);
%
% The Max Condition
[yHi,I] = max(y1); xHi = x(I);
y2 = yHi*ones(1,length(x));
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
plot(x,y1, 'LineWidth', 5),axis([.05 xmax .6 ymax]),...
grid, xlabel('\fontsize{14}x'),
ylabel('\fontsize{14}y=f(x)'),...
title(['\fontsize{16}MTH15 • Zero Critical-Pt',]),...
annotation('textbox',[.15 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', ' ','FontSize',7)
hold on
plot(x,y2, '-- m', xHi,yHi, 'd r', 'MarkerSize',
10,'MarkerFaceColor', 'r', 'LineWidth', 2)
set(gca,'XTick',[xmin:.05:xmax]); set(gca,'YTick',[ymin:.1:ymax])
Bruce Mayer, PE
Chabot
holdCollege
off Mathematics
Chabot College Mathematics
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MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 23Jun13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
% ref:
%
% The Limits
xmin = 0; xmax = 3;
ymin = 0; ymax = 20;
% The FUNCTION
x = linspace(xmin,1.99,1000); y = -1./(x-2);
%
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Green
plot(x,y, 'LineWidth', 4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y =
f(x)'),...
title(['\fontsize{16}MTH15 • \infty Critical-Pt',]),...
annotation('textbox',[.51 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', ' ','FontSize',7)
hold on
plot([2 2], [ymin,ymax], '--m', 'LineWidth', 3)
set(gca,'XTick',[xmin:0.5:xmax]); set(gca,'YTick',[ymin:2:ymax])
Example  Critical Numbers
 Find all critical numbers and classify
them as a relative maximum, relative
minimum, or neither for The Function:
MTH15 • Bruce Mayer, PE
f x   4 x 
40
2
x
30
2
20
y = f(x)
10
0
-10
-20
-30
-40
-10
Chabot College Mathematics
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XY f cnGraph6x6BlueGreenBkGndTemplate1306.m
-8
-6
-4
-2
0
x
2
4
6
8
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10
Example  Critical Numbers
 SOLUTION
 Relative extrema can only take place at
critical points (but not necessarily all
critical points end up being extrema!)
 Thus we need to find the critical points
of f. In other words, values of x so that
df
dx
0
OR
df
 UnDefined
dx
Think Division by Zero
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Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Critical Numbers
 For
the
Zero
Critical
Point

d 
2 
d
2
0  f '( x) 
4
x


4
x

2
x
2
dx 
x  dx
0 = 4 + 2 × -2x -3
-4 = -4x -3
-3
1= x
x 1
 Now need to consider critical points due
to the derivative being undefined
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
Example  Critical Numbers
 The Derivative Fcn, f’ = 4 − 4/x3 is
undefined when x = 0.
 However, it is very important to note
that 0 cannot be the location of a critical
point, because f is also undefined at 0
 In other words, no critical point of a
function can exist at c if no point on f
exists at c
Chabot College Mathematics
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Example  Critical Numbers
 Use Direction Diagram to Classify the
Critical Point at x = 1
|
1
 Calculating the derivative/slope at a test
point to the left of 1 (e.g. x = 0.5) find
3
f ' ( 0 . 5 )  4  4 ( 0 . 5 )   28 → f is DEcreasing
 Similarly for x>1, say 2:
f ' (2)  4  4(2)
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3
  3 .5
→ f is INcreasing
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Example  Critical Numbers
 From our Direction Diagram it appears
that f has a relative minimum at x = 1.
 A graph of the function corroborates this
assessment.
f (x) = 4x +
2
x2
Relative Minimum
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Bruce Mayer, PE
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Example  Evaluating Temperature
 The average temperature, in degrees
Fahrenheit, in an ice cave t hours after
10 t  1
midnight is
T t   2
modeled by:
t  t 1
 Use the Model to Answer Questions:
• At what times was the temperature
INcreasing? DEcreasing?
• The cave occupants light a camp stove in
order to raise the temperature. At what
times is the stove turned on and then off?
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Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
 SOLUTION:
 The Temperature “Changes Direction”
before & after a Max or Min (Extrema)
• Thus need to find the Critical Points which
give the Location of relative Extrema
• To find critical points of T, determine
values of t such that one these occurs
– dT/dt = 0 or
– dT/dt → ±∞ (undefined)
Chabot College Mathematics
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Example  Evaluating Temperature
 Taking dT/dt:
d 
10 t  1 
dT
d
2




T
t



10
t

1
t

t

1
2
dt 
t  t  1 
dt
dt

 Using the Quotient Rule
dT
t
2

 t 1 

d
dt
dt
dT

t

2
t
2
 t 1

dt
t
2
 t 1
2
 t  1   10  10 t  1  2 t  1 
dt
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10 t  1   10 t  1  
d
t
2
 t  1
2
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

2
Example  Evaluating Temperature
 Expanding and Simplifying
dT
10 t  10 t  10  20 t  8 t  1
2

2
t
dt
2
 t 1

2
 10 t  2 t  11
2

t
2
 t 1

2
 When dT/dt → ∞
dT
 10 t  2 t  11
2
  
dt
t
2
 t  1
2
   t  t  1   0
2
2
• The denominator being zero causes the
derivative to be undefined
– however,(t2−t +1)2 is zero exactly when t2−t + 1
is zero, so it results in NO critical values
Chabot College Mathematics

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Example  Evaluating Temperature
 When dT/dt = 0
2

 10 t  2 t  11  2
 t  t 1
0 

2
2
t  t 1




2

 0 t  t 1
2


 Thus Find:  10 t 2  2 t  11  0
 Using the quadratic
formula (or a
computer algebra
system such as
MuPAD), find
Chabot College Mathematics
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
2
  10 t  2 t  11
2
Bruce Mayer, PE
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Example  Evaluating Temperature
 For dT/dt = 0 find: t ≈ −1.15 or t ≈ 0.954
 Because T is always continuous (check
that the DeNom fcn, (t2−t +1)2 has no
real solutions) these are the only two
values at which T can change direction
 Thus Construct a Direction Diagram
with Two BreakPoints:
• t ≈ −1.15
• t ≈ +0.954
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Evaluating Temperature
 The Direction
|
|
Diagram
 1 . 15
0 . 954
 We test the derivative function in each
of the three regions to determine if T is
increasing or decreasing. Testing t = −2
2
dT
 10   2   2 (  2 )  11
25


2
2
dt t   2
49
 2   (  2 )  1
 The negative Slope indicates that T is
DEcreasing
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Evaluating Temperature
 The Direction
|
|
Diagram
 1 . 15
0 . 954
 Now we test in the second region using
2
t = 0: dT
 10 0   2 ( 0 )  11

dt
t0
0 
2
 (0)  1

2
  11
 The positive Slope indicates that T is
INcreasing
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Bruce Mayer, PE
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Example  Evaluating Temperature
 The Direction
|
|
Diagram
 1 . 15
0 . 954
 Now we test in the second region using
t = 1: dT
2
 10 1   2 (1)  11

dt
t 1
1
2
 (1)  1

2
 1
 Again the negative Slope indicates that
T is DEcreasing
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Bruce Mayer, PE
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Example  Evaluating Temperature
 The Completed Slope
|
|
Direction-Diagram:
 1 . 15
0 . 954
 We conclude that the function is
increasing on the approximate interval
(−1.15, 0.954) and decreasing on the
intervals (−∞, −1.15) & (0.954, +∞)
• It appears that the stove was lit around
10:51pm (1.15 hours before midnight) and
turned off around 12:57am (0.95 hours
after midnight), since these are the relative
extrema of the graph.
Chabot College Mathematics
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•
Bruce Mayer, PE
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Example  Evaluating Temperature
 Graphically
Relative Max
(Stove OFF)
Relative Min
(Stove On)
Chabot College Mathematics
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T t  
10 t  1
t  t 1
2
Bruce Mayer, PE
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MuPAD Plot Code
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WhiteBoard Work
 Problems From §3.1
• P40 → Use Calculus to Sketch Graph
• Similar to P52 →
Sketch df/dx for
f(x) Graph at right
• P60 → Machine
Tool Depreciation
Chabot College Mathematics
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All Done for Today
Critical
(Mach)
Number
Ernst Mach
Chabot College Mathematics
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
42
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
43
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
44
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
45
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
46
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
47
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
48
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
49
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
P3.1-40 Hand Sketch
Chabot College Mathematics
50
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
P3.1-40 MuPAD Graph
Chabot College Mathematics
51
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
WhiteBd Graphic
for P3.1-52
Chabot College Mathematics
52
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
53
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
54
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
55
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
56
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
57
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
58
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
59
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
60
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
P3.1-60 MuPAD
Chabot College Mathematics
61
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
62
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx