# Applications §4.4 Exp &amp; Log Chabot Mathematics Bruce Mayer, PE

```Chabot Mathematics
&sect;4.4 Exp &amp; Log
Applications
Bruce Mayer, PE
BMayer@ChabotCollege.edu
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Review &sect;
4.3
• &sect;4.3 → Exp &amp; Log
Derivatives
 Any QUESTIONS
• &sect;4.3 → HW-20
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
&sect;4.4 Learning Goals
 Use exponential
and logarithmic
derivatives in
curve sketching
 Examine
applications
involving
exponential
models
Chabot College Mathematics
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Summary of Log Rules
 Solving Logarithmic Equations Often
Requires the Use of Logarithms Laws
 For any
log a ( MN )  log a M  log a N ;
positive
numbers
log a M p  p log a M ;
M, N, and
M
a with a ≠ 1, log a
 log a M  log a N ;
N
p a whole
k
number
log a  k .
a
Chabot College Mathematics
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Typical Log-Confusion
 Beware that Logs do NOT behave
Algebraically. In General:
log a ( MN )  (log a M )(log a N ),
M log a M
log a

,
N log a N
log a ( M  N )  log a M  log a N ,
log a ( M  N )  log a M  log a N .
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Exponent↔Logarithm Duality
 Some Important Implications of the
Properties of Logs &amp; Exponents
a a  xu
ln x  v log x  w
log a x  z
x
e
ln x
e
xe
v
v
Chabot College Mathematics
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u
10
log x
 10
x  10
w
w
a
a
log a x
xa
z
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
z
Alternative Graph: Swap x &amp; y
MTH15 • y=2.3x &amp; x = 2.3y
• Note that y =
and y = logux are
Mirror images
ux
Chabot College Mathematics
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5
4
3
2
y = 2.3x, y
 It will be helpful in
later work to be
able to graph an
equation in which
the x and y in
y = ax are
interchanged
6
1
y  2.3 x
0
-1
x  2.3 y
x  log 2.3 y
-2
-3
yx
-4
-5
-6
-6
Bruce May er, PE • 18Jul13
-5
-4
-3
-2
-1
0
1
x, x = 2.3
y  2.3 x
2
3
4
5
6
y
 log 2.3 y  log 2.3 2.3 x
log 2.3 y  x  2.3 y  x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 18Jul13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
% ref:
%
% The Limits
xmin = -6; xmax = 6;
ymin = -6; ymax = 6;
% The FUNCTION
x = linspace(xmin,xmax,1000); x1=x; y1=2.3.^x; x2=y1; y2=x; x3=x;
y3=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(x1,y1, x2,y2, 'LineWidth', 4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x, x = 2.3^y'),
ylabel('\fontsize{14}y = 2.3^x, y '),...
title(['\fontsize{16}MTH15 • y=2.3^x &amp; x = 2.3^y
',]),...
annotation('textbox',[.51 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
18Jul13','FontSize',7)
hold on
plot(x3,y3, '--m', zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)
set(gca,'XTick',[xmin:1:xmax]); set(gca,'YTick',[ymin:1:ymax])
Recall: Better Graphing GamePlan
1. Find THE y-Intercept, if Any
a. Set x = 0, find y
b. Only TWO Functions do NOT have a
y-intercepts
–
–
Of the form 1/x
x = const; x ≠ 0
2. Find x-Intercept(s), if Any
a. Set y = 0, find x
b. Many functions do NOT have x-intercepts
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Better Graphing GamePlan
3. Find VERTICAL (↨) Asymptotes, If Any
a. Exist ONLY when fcn has a denom
b. Set Denom = 0, solve for x
–
These Values of x are the Vertical Asymptote
(VA) Locations
4. Find HORIZONTAL (↔) Asymptotes
(HA), If Any
a. HA’s Exist ONLY if the fcn has a finite
limit-value when x→+∞, or when x→−∞
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Better Graphing GamePlan
f x 
b. Find y-value for: yHA  xlim
 
–
These Values of y are the HA Locations
5. Find the Extrema (Max/Min) Locations
a. Set dy/dx = 0, solve for xE
b. Find the corresponding yE = f(xE)
c. Determine by 2nd Derivative, or
ConCavity, then test whether (xE, yE) is a
Max or a Min
–
See Table on Next Slide
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Better Graphing GamePlan
– Determine Max/Min By Concavity
𝒅𝟐 𝒚
Sign
Concavity
Max or Min
POSitive
NEGative
Neither (Zero)
Up ↑
Down ↓
No Information
Min
Max
Flat Spot
𝒅𝒙𝟐 𝒙𝑬
6. Find the Inflection Pt Locations
a. Set d2y/dx2 = 0, solve for xi
b. Find the corresponding yi = f(xi)
c. Determine by 3rd Derivative test The
Inflection form: ↑-↓ or ↓-↑
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Better Graphing GamePlan
7. Find the Inflection Pt Locations
a. Set d2y/dx2 = 0, solve for xi
b. Find the corresponding yi = f(xi)
c. Determine by 3rd Derivative test The
Inflection form: ↑-↓ or ↓- ↑
–
𝒅𝟑 𝒚
𝒅𝒙𝟑 𝒙𝒊
Determine Inflection form by 3rd Derivative
Sign
POSitive
NEGative
Neither (Zero)
Chabot College Mathematics
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ConCavity Change
Inflection Form
Down-to-Up
Up-to-Down ↓
No Information
↓-↑
↑-↓
↑-↑ OR ↓-↓
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Better Graphing GamePlan
8. Sign Charts for Max/Min and ↑-↓/↓-↑
a. To Find the “Flat Spot” behavior for
dy/dx = 0, when d2y/dx2 exists, but
[d2y/dx2]xE = 0 use the Direction-Diagram
Slope
df/dx Sign
Critical (Break)
Points
Chabot College Mathematics
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−−−−−−
++++++
a
Max
−−−−−−
b
NO
Max/Min
++++++
c
Min
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
x
Better Graphing GamePlan
9. Sign Charts for Max/Min and ↑-↓/↓-↑
a. To Find the ↑-↑ or ↓-↓ behavior for
d2y/dx2 = 0, when d3y/dx3 exists, but
[d3y/dx3]xi = 0 use the Dome-Diagram
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
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−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
x
Example  Exp Inoculation
 In a researcher’s model, inoculating x
individuals to a virus suggests kPeople
will become
x
infected as I x  a  (0.90)  b

• Where a &amp; b are Constants
 Find
a. If there are 5000 thousand susceptible
individuals in the population, then find the
values of constants a and b.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Exp Inoculation
b. How many individuals become infected
when 2000 are inoculated?

 SOLUTION a. I x  a  (0.90)  b
 5000 susceptible individuals could imply
that the point (0,5) should be on the
graph of the function (no individuals
inoculated means all get sick). It also
means that if everyone is inoculated,
nobody should get sick. In other words,
(5,0) is on the graph.
Chabot College Mathematics
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x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Exp Inoculation
 Using (x,I) = (0,5)
I  x   a  (0.90)  b
x
5  a  (0.90) 0  b
5  a 1  b
b  a 5
 Now Use (5,0)
0  a  (0.90) 5  b
b  a  (0.90)
Chabot College Mathematics
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5
 But From Before
b  a 5
 Substituting
a  5  a  (0.90) 5
a  a  (0.90) 5  5
a  1  0.5905  5
a  5 0.4095
a  12.21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Exp Inoculation
 But From Before
b  a 5
b  12.21  5  7.21
 Thus ans a)
I  x   12.21(0.90) x  7.21
 Doing the algebra
x
2  7.21  12.21(0.90)
2  7.21
 (0.90) x  0.7543
12.21


ln (0.90) x  0.7543
 SOLUTION b)
x ln0.90  ln 0.7543
 Using 𝐼 𝑥 above to
find 𝑥 when 𝐼 𝑥 =2k
x  ln 0.7543 ln0.90
2  12.21(0.90) x  7.21
Chabot College Mathematics
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x   0.2820  0.1054
x  2.676
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
5
 A version of the
f x  
x
“Logistic Function” →
1 e
 Determine where the fcn is increasing &amp;
decreasing and where its graph is
concave Up &amp; concave Down.
 Sketch the graph of the function. Show
as many key features as possible
• high and low points, points of inﬂection,
vertical/horizontal asymptotes, intercepts,
cusps, vertical tangents
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 SOLUTION:
 Finding intervals of increase and
decrease (along with any relative
extrema) can be accomplished using
the derivative.
 First, rewrite the function in f (x) = 5(1+ e- x )-1,
a form avoids the quotient rule
df
 x 2
 Then
 5  11  e   e  x   1
dx
Chabot College Mathematics
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

 5e  x (1  e  x ) 2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 Note that df/dx is always positive (each
factor is always positive), so the original
function is increasing on its entire
domain.
• This also implies that the function has NO
relative extrema.
 Now find intervals on which the function
is concave up or concave down.
• This requires the use of the second
derivative.
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 Taking the Second Derivative


d2 f
d
x
 x 2

5
e
(
1

e
)
2
dx
dx
d&eacute;
d &eacute; -x &ugrave;
-x
- x -2 &ugrave;
= 5e &times; &euml;(1+ e ) &ucirc; + &euml;5e &ucirc; &times; (1+ e- x )-2
dx
dx
 5e
x
 x 3
x
x
 x 2




  2 (1  e )  e    1 5e  (1  e )
=10e-2 x &times; (1+ e-x )-3 - 5e- x &times; (1+ e-x )-2
= 5e-2 x (1+ e-x )-3 &eacute;&euml;2 - e x &times; (1+ e-x )&ugrave;&ucirc;
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 Concavity changes at Inflection-Points
when the 2nd Derivative equals Zero
0 = 5e-2 x (1+ e- x )-3 &eacute;&euml;1- e x &ugrave;&ucirc;
 Because the first two factors are always
NonZero, the equation reduces to
0  1 ex
 ex  1  x  0
 Now chk the sign of the 2nd derivative
on either side of 0, at x = −1 &amp; x = 1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 The
Sign
Tests d 2 f
d2 f
dx 2
dx


 5e 2( 1) (1  e ( 1) ) 3 1  e 1 &raquo; 0.454
x  1


 5e 2(1) (1  e (1) ) 3 1  e1 &raquo; -0.454
2
x  1
 The Sign Chart (Dome-Diagram
ConCavity
Form
d2f/dx2 Sign
++++++
1
Inflection
0
−−−−−−
1
Critical Point
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
x
Example  Logistic Curve
 The 2nd Derivative function is
• Concave UP for all real no.s less than 0
• Concave DOWN for all real no.s greater
than 0.
 Because the graph changes concavity
at x = 0, an inflection point exists at his
location.
 Next investigate asymptotes.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 Because the function has no errors
(Div-by-Zero) in its domain, conclude
that there are NO vertical asymptotes
 Letting x→&plusmn;∞ reveals TWO horizontal
Asymptotes
5
5
lim

0

x
x   1  e
1 
5
5
lim

5

x
x   1  e
1 0
• Thus Have Horizontal Asymptotes at
–y = 0
–y = 5
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
5
 Check for
f 0  
 2.5
0
1 e
y-intercept at x = 0
• Have y-intercept at (0, 2.5)
 Check for
5
f (x) =
= 0,
-x
x-intercept at y = 0
1+ e
x
5
1

e


x

0

5

0
1

e
 5  0 ???
1  e  x
 1
• This CONTRADICTION (5=0) means that
there is NO soln to the eqn, and thus NO
x-intercepts exist

Chabot College Mathematics
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 Finally, to find any cusps or vertical
tangents, look for those values of x
where the derivative function is
undefined. Recall df/dx
df
5
x
 x 2
 5e (1  e )  x
dx
e (1  e  x ) 2
 UnDefinition Occurs when x
e (1+ e- x )2 = 0
the Divisor Equals Zero, or:
• But Since 𝑒 𝑥 and 𝑒 −𝑥 are ALWAYS Positive
this eqn has NO Solutions, so the fcn has
no Cusps or Vertical Tangents
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Logistic Curve
 Graphically
Horizontal
Asymptotes
5
f (x) =
1+ e- x
Inflection Point
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Marginal Inoculation
 Consider the inoculation function from
the Previous Example
I x   12.21 (0.90)  7.21
x
 Use marginal/incremental analysis to
estimate the change in the number of
infected individuals when increasing the
number of inoculated person from 1000
to 1010
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Marginal Inoculation
 SOLUTION:
 ReCall Marginal analysis is the process
of using the derivative to predict change
in a function in the short run. Recall
that for a function f(x), value a, and
small number
df
f a  x   f a  
 x
∆x; to Whit:
dx a
 In this case with
x in kPeople, I 1  0.01  I 1  dI  0.01
dx 1
estimate:
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Marginal Inoculation
 Calc 𝐼 1 : I x   12.21 (0.90) x  7.21
I 1  12.21 1  7.21  3.779
1
 Next
d
d
x
I  x   12.21  (0.90)  7.21
𝑑𝐼
dx
Find : dx
𝑑𝑥
 Now Let
0.9x = eu
Chabot College Mathematics
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dI d
 12.21eu   7.21
dx dx
dI
d
du
u


12.21e  7.72
dx du
dx
dI
u du
 12.21e 
dx
dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Marginal Inoculation
 Now
du/dx


 
0.9  e  ln 0.9  ln e
x ln 0.9   u ln e  u  u  x ln 0.9 
du d
  x ln 0.9   1  ln 0.9 
dx dx
x
u
x
u
 BackSub eu = 0.9x &amp; du/dx = ln(0.9)
dI
u du
 12.21e 
dx
dx
Chabot College Mathematics
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
dI
x
 12.21 0.9  ln 0.9
dx
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Marginal Inoculation
 Find 𝑑𝐼 𝑑𝑥 dI
1.01
 12.21 0.9  ln 0.9  1.157
at x = 1.01 dx 1.01
 By marginal analysis the Estimated
value at an inoculation level of 1010
dI
I 1.01  I 1 
 0.01
dx 1
 3.779  1.1578 0.01  3.767
 The estimated number of infected is
3,767 using marginal analysis.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
WhiteBoard Work
 Problems From &sect;4.4
• P36 → Marginal Analysis
• Special Prob → Sketch Log Fcn
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
All Done for Today
Finding
Pwr Fcn by
Log-Log
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu
–
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
ConCavity Sign Chart
ConCavity
Form
d2f/dx2 Sign
++++++
Critical (Break)
Points
Chabot College Mathematics
39
−−−−−−
a
Inflection
−−−−−−
b
NO
Inflection
++++++
c
Inflection
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
x
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
P4.4-36 Graph
MTH15 • P4.4-36
1600
Bruce May er, PE • 19JUl13
y = 1000e-x/50(x-125)
1400
1200
1000
800
600
400
200
0
0
50
Chabot College Mathematics
50
100
150
200
250
x
300
350
400
450
500
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 19Jul13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
% The Limits
xmin = 0; xmax = 500;
ymin = 0; ymax = 1600;
% The FUNCTION
x = linspace(xmin,xmax,1000); y = 1000*exp(-x/50).*(x-125);
%
% 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 =
1000e^-^x^/^5^0(x-125)'),...
title(['\fontsize{16}MTH15 • P4.4-36',]),...
annotation('textbox',[.67 .81 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'Bruce Mayer, PE •
19JUl13','FontSize',7)
hold on
set(gca,'XTick',[xmin:50:xmax]); set(gca,'YTick',[ymin:200:ymax])
Example  Graph y  4 ln x  x
2
 Use Graphing GamePlane
1. Find y-intercept if it exists
2. Find any x-intercept(s)
3. Use Denom→0 to Check for Vertical
Asymptote(s)
4. Use Denom→∞ to Check for Horizontal
Asymptote(s)
5. Find max/min pts by dy/dx = 0
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Graph y  4 ln x  x
2
 Use Graphing GamePlane
6. Find Inflection Points by
[d2y/(dx)2] = 0
7. Check form of inflection points using 3rd
Derivative Test [d3y/(dx)3]InflPts
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
4
Bruce May er, PE • 18Jul13
3.5
3
2.5
2
1.5
1
0.5
0
0
2
4
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6
8
10
12
14
16
18
20
Bruce Mayer, PE
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 18Jul13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
% The Limits
xmin = 0; xmax = 20;
ymin = 0; ymax = 4;
% The ZERO Lines
zxh = [xmin xmax]; zyh = [0 0]; zxv = [.05 .05]; zyv =
[ymin ymax];
%
% the 6x6 Plot
axes; set(gca,'FontSize',12);
plot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth',
2),axis([xmin xmax ymin ymax]),...
grid, annotation('textbox',[.68 .82 .0 .1],
'FitBoxToText', 'on', 'EdgeColor', 'none', 'String',
'Bruce Mayer, PE • 18Jul13','FontSize',7)
set(gca,'XTick',[xmin:2:xmax]);
set(gca,'YTick',[ymin:0.5:ymax])
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
MTH15 • Sketch ln
3
2.5
y = 4ln2x/x
2
1.5
1
0.5
0
BMay er • 18Jul13
0
2
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4
6
8
x
10
12
14
16
Bruce Mayer, PE
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
MATLAB Code
% Bruce Mayer, PE
% MTH-15 • 18Jul13
% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m
%
% The Limits
xmin = 0; xmax =16;
ymin = 0; ymax = 3;
% The FUNCTION
x = linspace(xmin,xmax,1000); y = 4*(log(x).^2)./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,y, 'LineWidth', 4),axis([xmin xmax ymin ymax]),...
grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y =
4ln^2x/x'),...
title(['\fontsize{16}MTH15 • Sketch ln',]),...
annotation('textbox',[.75 .05 .0 .1], 'FitBoxToText', 'on',
'EdgeColor', 'none', 'String', 'BMayer • 18Jul13','FontSize',7)
hold on
set(gca,'XTick',[xmin:2:xmax]); set(gca,'YTick',[ymin:.5:ymax])
Example  Exp Inoculation
I(0) = a &times;(0.90)0 - b = 5
 Using Pts (0,5) &amp;
(5,0) in the Model I (5)  a  (0.90)5  b  0
 Simplifying, we can
solve the first
equation for a and
then substitute into
the second equation.
Chabot College Mathematics
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&igrave;a - b = 5
&iacute;
&icirc;0.59049a - b = 0
&igrave;a = 5 + b
&iacute;
&icirc;0.59049a - b = 0
0.59049 ( 5+ b) - b = 0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Exp Inoculation
 Running 2.95245+ 0.59049b - b = 0
the
-2.95245
b=
&raquo; 7.21.
Numbers
-0.40951
 Now Back
a-b = 5
SubStitute a - 7.21= 5
to find a:
a  12.21
 Sub the Values of a &amp; b into Model:
I x   12.21 (0.90) x  7.21.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-21_sec_4-4_EXP-n-LOG_Applications.pptx
Example  Exp Inoculation
 Then the target
12.21&times; (0.90)x - 7.21 = 2
level of infection is
9.21
x
(0.90) =
2000 People, which
12.21
x
translatesto solving
ln(0.90) = ln ( 0.7543)
the equation I(x) = 2
x &times; ln(0.90) = ln ( 0.7543)
 State: when 2,676
individuals are inoculated, x = ln ( 0.7543)
ln(0.90)
only 2000 will get sick
• This suggests that even Partial x  2.676
inoculation reduces disease transmission
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Bruce Mayer, PE
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```