[email protected] MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical

[email protected] • MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§3.1 RelativeExtrema



Review §

Any QUESTIONS About• §2.6 → Implicit Differentiation

Any QUESTIONS About HomeWork• §2.6 →

HW-12

2.6

http://www.google.com/url?sa=i&rct=j&q=implicit+differentiation&source=images&cd=&cad=rja&docid=-pM1sO_6siWT_M&tbnid=8eZm4590N40-_M:&ved=0CAUQjRw&url=http://education-portal.com/academy/lesson/how-to-find-derivatives-of-implicit-functions.html&ei=S4PcUaWlJ6fsiQKfjICgBg&bvm=bv.48705608,d.cGE&psig=AFQjCNHNGLcOF21zRz3w35wB7qUPfxJvFQ&ust=1373492313941109



§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:

• INcreasing is Moving UP from Left→Right

A function f is DEcreasing if whenever a<b, then:

• DEcreasing is Moving DOWN from Left→Right

)()( bfaf



Inc & Dec Values Graphically

0 1 2 3 4 5 6 7 8 9 10-4

-3

-2

-1

0

1

2

3

4

5

6

7

x

y =

f(x)

MTH15 • Bruce Mayer, PE

XYf cnGraph6x6BlueGreenBkGndTemplate1306.m

INcreasing

DEcreasing



Inc & Dec with Derivative

If for every c on the interval [a,b]• That is, the Slope is POSITIVE

Then f is INcreasing on [a,b]

If for every c on the interval [a,b]• That is, the Slope is NEGATIVE

Then f is DEcreasing on [a,b]

0)(' cxdx

dfcf

0)(' cxdx

dfcf



Example Inc & Dec

The function, y = f(x),is decreasing on [−2,3] and increasing on [3,8]

43 2 xxfy



Example Inc & Dec Profit

The default list price of 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




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:

xpxxR 2/310 xx




And now cost as a function of x:

Then the Profit is the Revenue minus the Costs:

cost fixedcost variable xC 501 x

xCxRxP




Now we turn to determining the intervals of increase and decrease.

The graph of the profit function is shown next on the interval [0,100] (where the price and quantity demanded are both non-negative).




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 exact intervals




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 thederivative, or Slope,function:

509' 2/3 xxdx

dxP

2/1

2

39 x




On Increasing intervals the Slope is POSTIVE or NonNegative so in this case need

SolvingThisInEquality:

The profit function is DEcreasing on the interval [36,100]

02

39 2/1 x

dx

dP

02

39 2/1 x

3

292/1 x 36x

92

3 2/1 x



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



Peaks & Valleys

Extrema is precise math terminology for Both of• The TOP of a

Hill; that is, a PEAK

• The Bottom of a Trough, That is a VALLEY

0 10 20 30 40 50-3

-2

-1

0

1

2

3

4

x

y =

f(x)


PEAKPEAK

VALLEY

VALLEY



Rel&

Ab

s Max&

Min

0 10 20 30 40 50-3

-2

-1

0

1

2

3

4

x

y =

f(x)


RelativeMax

AbsoluteMax

RelativeMin

AbsoluteMin



Critical Points

Let c be a value in the domain of f Then c is a Critical Point If, and only if

cxcx dx

df

dx

dfOR0

HORIZONTAL slopeat c

VERTICAL slopeat c



Critical Points GeoMetrically Horizontal Vertical

0 0.5 1 1.5 2 2.5 30

2

4

6

8

10

12

14

16

18

20

x

y =

f(x)

MTH15 • Critical-Pt

0.05 0.1 0.15 0.2 0.250.6

0.7

0.8

0.9

1

1.1

1.2

1.3

x

y=f(

x)

MTH15 • Zero Critical-Pt

(0.1695, 1.2597)



MA

TL

AB

Co

de

% Bruce Mayer, PE% MTH-15 • 07Jul13% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m%clear; clc;% The Limitsxmin = 0; xmax = 0.27; ymin =0; ymax = 1.3;% The FUNCTIONx = 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 Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenplot(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 onplot(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])hold off



MA

TL

AB

Co

de

% Bruce Mayer, PE% MTH-15 • 23Jun13% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m% ref:%% The Limitsxmin = 0; xmax = 3; ymin = 0; ymax = 20;% The FUNCTIONx = linspace(xmin,1.99,1000); y = -1./(x-2);% % The ZERO Lineszxh = [xmin xmax]; zyh = [0 0]; zxv = [0 0]; zyv = [ymin ymax];%% the 6x6 Plotaxes; set(gca,'FontSize',12);whitebg([0.8 1 1]); % Chg Plot BackGround to Blue-Greenplot(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 onplot([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:

2

24

xxxf

-10 -8 -6 -4 -2 0 2 4 6 8 10-40

-30

-20

-10

0

10

20

30

40

x

y =

f(x)


XYf cnGraph6x6BlueGreenBkGndTemplate1306.m




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

UnDefinedOR0 dx

df

dx

df

Think Division by Zero




For theZeroCriticalPoint

Now need to consider critical points due to the derivative being undefined

22

242

4)('0

xx

dx

d

xx

dx

dxf

1x




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




Use Direction Diagram to Classify the Critical Point at x = 1

Calculating the derivative/slope at a test point to the left of 1 (e.g. x = 0.5) find

Similarly for x>1, say 2:28)5.0(44)5.0(' 3 f → f is DEcreasing

5.3)2(44)2(' 3 f → f is INcreasing




From our Direction Diagram it appears that f has a relative minimum at x = 1.

A graph of the function corroborates this assessment.

Relative Minimum



Example Evaluating Temperature

The average temperature, in degrees Fahrenheit, in an ice cave t hours after midnight is modeled by:

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?

1

1102

tt

ttT




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)

http://www.google.com/url?sa=i&rct=j&q=ice+cave&source=images&cd=&cad=rja&docid=OkbyGnHaaD3QuM&tbnid=8qqHFcDAeV3NIM:&ved=0CAUQjRw&url=http://geology.about.com/od/glaciers_ice/ig/glacier-pictures/ice-cave.htm&ei=zHvdUeCROquajAL7hoCgCQ&bvm=bv.48705608,d.cGE&psig=AFQjCNGF7rI2HeW1SvwwRyM98DJox4j5EA&ust=1373555973461217




Taking dT/dt:

Using the Quotient Rule

22

22

1

11101101

tt

ttdtd

ttdtd

tt

dt

dT

22

2

1

12110101

tt

tttt

dt

dT

222

11101

110

tttdt

d

dt

dT

tt

ttT

dt

d

http://www.google.com/url?sa=i&rct=j&q=quotient+rule+for+derivatives&source=images&cd=&cad=rja&docid=dLTfZKytHoMY_M&tbnid=YuLyJgq80aZK3M:&ved=0CAUQjRw&url=http://www.karlscalculus.org/calc4_3.html&ei=l37dUb_zIozQigKW7oEo&bvm=bv.48705608,d.cGE&psig=AFQjCNEyG8UdY1GmSRLzorHzZ1bETMX64A&ust=1373556736094156




Expanding and Simplifying

When dT/dt → ∞

• 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

22

2

22

22

1

11210

1

1820101010

tt

tt

tt

tttt

dt

dT

011

11210 2222

2

tt

tt

tt

dt

dT




When dT/dt = 0

Thus Find: Using the quadratic

formula (or a computer algebra system such as MuPAD), find

112101011

112100 22222

22

2

tttttt

tt

tt

011210 2 tt




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




The DirectionDiagram

We test the derivative function in each of the three regions to determine if T is increasing or decreasing. Testing t = −2

The negative Slope indicates that T is DEcreasing

954.015.1

| |

49

25

1)2(2

11)2(221022

2

2

tdt

dT





Now we test in the second region using t = 0:

The positive Slope indicates that T is INcreasing

954.015.1

| |

11

1)0(0

11)0(201022

2

0

tdt

dT





Now we test in the second region using t = 1:

Again the negative Slope indicates that T is DEcreasing

954.015.1

| |

1

1)1(1

11)1(211022

2

1

tdt

dT




The Completed SlopeDirection-Diagram:

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.

954.015.1

| |




GraphicallyRelative Max (Stove OFF)

Relative Min (Stove On)

1

1102

tt

ttT



MuPAD Plot Code



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 → MachineTool Depreciation

http://www.google.com/url?sa=i&rct=j&q=WJ-1000&source=images&cd=&cad=rja&docid=jIuN-fb_VuigZM&tbnid=ZvgsuEnByu7IbM:&ved=0CAUQjRw&url=http://www.prweb.com/releases/AvizaChapter11/WJAPCVD/prweb2527634.htm&ei=8qDdUbyYHOSKiAKl1IGYCQ&bvm=bv.48705608,d.cGE&psig=AFQjCNGHG7uiJSSOCz7KLcYHZgJ6Or2gWA&ust=1373565485609747



All Done for Today

Critical(Mach)Number

Ernst Mach

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Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix srsrsr 22

















P3.1-40 Hand Sketch



P3.1-40 MuPAD Graph



Wh

iteBd

Grap

hic

for P

3.1-52



















P3.1-60 MuPAD



http://www.google.com/url?sa=i&rct=j&q=derivative+power+rule&source=images&cd=&cad=rja&docid=NjbfKj_JZBSzwM&tbnid=jJkxz7X095T-NM:&ved=0CAUQjRw&url=http://mathworld.wolfram.com/Derivative.html&ei=TfPWUe_nA6akiQKZ64DIDw&bvm=bv.48705608,d.cGE&psig=AFQjCNGw-AM0Kc2tARyUGs-zPM7iK-qH3g&ust=1373127702659906

http://www.google.com/url?sa=i&rct=j&q=related+rates+calculus&source=images&cd=&cad=rja&docid=wSGtOAIrRxIoIM&tbnid=l4Q1f5xFVNRraM:&ved=0CAUQjRw&url=http://www.cantonschools.org/~lforastiere/ap%20calculus?Templates=Printable&ei=2TLcUY-gOa2UjALp4IHgDw&bvm=bv.48705608,d.cGE&psig=AFQjCNH42r6LfHpjrPyJXK2DJrvdwHWZIQ&ust=1373471717102765

http://www.google.com/url?sa=i&rct=j&q=related+rates+calculus&source=images&cd=&cad=rja&docid=nuuUeFKHM5pqjM&tbnid=x7raEhpCGJ8edM:&ved=0CAUQjRw&url=http://designatedderiver.wikispaces.com/Games+and+fun+stuff&ei=IzPcUYPxBaOLiwLjs4GYAQ&bvm=bv.48705608,d.cGE&psig=AFQjCNH42r6LfHpjrPyJXK2DJrvdwHWZIQ&ust=1373471717102765

Documents

[email protected] MTH15_Lec-12_sec_3-1_Rel_Extrema_.pptx 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical