Chabot Mathematics
§3.1 Relative
Extrema
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
1
Bruce Mayer, PE
[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
Chabot College Mathematics
2
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
§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
Chabot College Mathematics
3
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
Chabot College Mathematics
4
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Inc & Dec Values Graphically
MTH15 • Bruce Mayer, PE
7
DEcreasing
6
5
4
y = f(x)
3
2
INcreasing
1
0
-1
-2
-3
-4
XY f cnGraph6x6BlueGreenBkGndTemplate1306.m
0
1
Chabot College Mathematics
5
2
3
4
5
x
6
7
8
9
10
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Inc & Dec with Derivative
 If for every c on f ' (c)  df
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]
Chabot College Mathematics
6
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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]
Chabot College Mathematics
7
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
Chabot College Mathematics
8
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
x
Example  Inc & Dec Profit
 For this Situation 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 = x× 10 - x  10 x  x 3 / 2
Chabot College Mathematics
9
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Inc & Dec Profit
 And now cost as a function of x:
Cx  variable cost  fixed cost  1x  50
 Then the Profit is the Revenue minus
the Costs:      
P x  R x C x
= (10x - x3/2 ) - ( x+ 50)
= 9x- x3/2 - 50.
Chabot College Mathematics
10
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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).
Profit Curve
Chabot College Mathematics
11
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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- x3/2 - 50
exact intervals
Profit Curve
Chabot College Mathematics
12
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
Chabot College Mathematics
13
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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]
Chabot College Mathematics
14
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
Chabot College Mathematics
15
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Peaks & Valleys
 Extrema is precise math terminology for
MTH15 • Bruce Mayer, PE
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
Chabot College Mathematics
16
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
Chabot College Mathematics
17
20
30
x
40
50
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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 OR
x c
HORIZONTAL
slope
at c
Chabot College Mathematics
18
df
dx

x c
VERTICAL
slope
at c
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
Chabot College Mathematics
19
0.2
0.25
0
0
0.5
1
1.5
2
x
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
2.5
3
20
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
21
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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])
14Mar17 CatchUp
Discuss the
next 7 slides,
slides 23→29
Chabot College Mathematics
22
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Direction Diagrams (Slope Chart)
 Use “Direction Diagrams” to determine
for Critical Points: MAX, MIN, Neither
 MAX
form
| −−−−−−
x  C.P.
−−−−−− | ++++++
x  C.P.
++++++
 MIN
form
 “Neither” forms
Direction of Graph
df/dx Sign
Direction of Graph
df/dx Sign
Direction of Graph
| ++++++
| −−−−−−
++++++
df/dx Sign
x  C.P.
x  C.P.
−−−−−−
Chabot College Mathematics
23
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
40
2
f x   4 x  2
x
30
20
y = f(x)
10
0
-10
-20
-30
-40
-10
Chabot College Mathematics
24
XY f cnGraph6x6BlueGreenBkGndTemplate1306.m
-8
-6
-4
-2
0
x
2
4
6
8
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
 0 OR
dx
df
 UnDefined
dx
Think Division by Zero
Chabot College Mathematics
25
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
0  f ' ( x)  4 x  2  
4 x  2 x 2
dx 
x  dx
0 = 4 + 2 × -2x-3
-4 = -4x-3
1
1 x  3
x
x 1
3
 Now need to consider critical points due
to the derivative being undefined
Chabot College Mathematics
26
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx

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
27
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
f ' (0.5)  4  4(0.5) 3  28 → f is DEcreasing
 Similarly for x>1, say 2:
f ' (2)  4  4(2) 3  3.5 → f is INcreasing
Chabot College Mathematics
28
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
Chabot College Mathematics
29
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
 The average temperature, in degrees
Fahrenheit, in an ice cave t hours after
midnight is T t   10t  1 Note When 𝑡 2 − 𝑡 + 1 = 0,
modeled by:
t 2  t  1 then 𝑇 𝑡 is Undefined
 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?
Chabot College Mathematics
30
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
31
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
 Taking dT/dt:
2
d 
10t  1 
dT d
2
T t   2

 10t  1 t  t  1


dt 
t  t  1
dt dt

 Using the Quotient Rule
d
d 2

t  t  1 10t  1  10t  1  t  t  1
dT
dt
dt

2
2
dt
t  t  1
2



dT t 2  t  1 10  10t  12t  1

2
2
dt
t  t 1
Chabot College Mathematics
32

Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx

Example  Evaluating Temperature
 Expanding and Simplifying
2
2
2
dT 10t  10t  10  20t  8t  1  10t  2t  11


2
2
2
2
dt
t  t  1
t  t  1
 When dT/dt → ∞
2
2
dT
 10t  2t  11
2

   t  t  1  0
2
dt
t 2  t  1
• 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

33
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
No Point → No Critical Point
 Notice the last Statement
however,(t2−t +1)2 is zero exactly when t2−t + 1 is
zero, so it results in NO critical values
 ReCall that since the original
10t  1
T t   2
Function is Undefined at
t  t 1
2
𝑡 − 𝑡 − 1 = 0 then any point where
𝑡 2 − 𝑡 − 1 = 0 can NOT be a Critical Pt.
 From the previous slide 𝑑𝑇 𝑑𝑡 → ∞
when 𝑡 2 − 𝑡 − 1 2 = 0, an Undefined Pt
 Thus No CritPoint if 𝑡 2 − 𝑡 − 1 = 0
Chabot College Mathematics
34
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
No Point → No Critical Point
 Solving 𝑡 2 − 𝑡 + 1 = 0 yields Undefined Points
for the ORIGINAL Function, which then can
NOT produce critical points at these values
Undefined For
10𝑇+1
𝑡 2 −𝑡+−1
 Orginal Fcn
→
𝑡 2 − 𝑡 + 1 = 0 → NO
valid pts at these values
 Thus NO Critical Points
Exist at these Point as
denom = 𝑡 2 − 𝑡 + 1 2 = 0
Complex Solns!
←Same→
Chabot College Mathematics
35
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
 When dT/dt = 0 Numerator = 0

2
2
 10t 2  2t  11 2
2
2

t

t

1

0

t

t

1


10
t
 2t  11
0 

2
2
0
t  t 1








 Thus Find:  10t 2  2t  11  0
 Using the quadratic
formula (or a
computer algebra
system such as
MuPAD), find
Chabot College Mathematics
36
Critical
Points
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
37
See Previous Slide
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
dT
dt
 10 2   2(2)  11
2

t  2
 2
2

 (2)  1
2
25

 0.51
49
 The negative Slope indicates that T is
Decreasing left of −1.15
Chabot College Mathematics
38
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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 to the Right of −1.15
Chabot College Mathematics
39
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
 The Direction
|
|
Diagram
 1.15
0.954
 Now we test in the second region using
2
t = 1: dT
 101  2(1)  11
dt

t 1
1
2

 (1)  1
2
 1
 Again the negative Slope indicates that
T is DEcreasing to the Left of 0.954
Chabot College Mathematics
40
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
Coldest
Warmest
 The Completed Slope
|
|
Direction-Diagram:
 1.15
0.954
 Can 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
41
•
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
 Graphically
Relative Max
(Stove OFF)
10t  1
T t   2
t  t 1
Relative Min
(Stove On)
Chabot College Mathematics
42
Note that there are NO
Vertical Asymtotes
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
MuPAD Plot Code
Chabot College Mathematics
43
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
44
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
All Done for Today
Critical
(Mach)
Number
Ernst Mach
Chabot College Mathematics
45
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
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
Chabot College Mathematics
50
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
51
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
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
P3.1-40 Hand Sketch
Chabot College Mathematics
54
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
P3.1-40 MuPAD Graph
Chabot College Mathematics
55
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
WhiteBd Graphic
for P3.1-52
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
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
Chabot College Mathematics
63
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
64
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
P3.1-60 MuPAD
Chabot College Mathematics
65
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Chabot College Mathematics
66
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx
Example  Evaluating Temperature
 Find where 𝑑𝑓 𝑑𝑡 → ∞ when
2
2
  t  t  1 0  t 2  t  1  0
 Solve by Quadratic Eqn as 𝑡 2 − 𝑡 + 1
does not Factor.
 Using MuPAD
For Quadratic
Equation
Chabot College Mathematics
67
Bruce Mayer, PE
[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx