bmayer@chabotcollege.edu mth15_lec-14_sec_3-2_concavity_inflection_.pptx 1 bruce mayer, pe chabot...

BMayer@ChabotCollege.edu • MTH15_Lec-14_sec_3-2_Concavity_Inflection_.pptx1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

§3.2 Concavity

& Inflection

Review §

Any QUESTIONS About• §3.1 → Relative Extrema

Any QUESTIONS About HomeWork• §3.1 →

§3.2 Learning Goals

Introduce Concavity (a.k.a. Curvature) Use the sign of the second derivative to

find intervals of concavity Locate and examine

inflection points Apply the second

derivatives test for relative extrema

ConCavity Described

Concavity quantifies the Slope-Value Trend (Sign & Magnitude) of a fcn when moving Left→Right on the fcn Graph

1 2 3 4-5

Position, x

MTH15 • BLUE

1 2 3 4-5

Position, x

MTH15 • RED

m≈−1.4

m≈−4.4

m≈−1.4

% Bruce Mayer, PE% MTH-15 •11Jul133% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m% % The datablue =[2.2 0 -1.4 -4.4]red = [-4.4 -1.4 0 2.2]%% the 6x6 Plotaxes; set(gca,'FontSize',12);subplot(1,2,1)bar(blue, 'b'), grid, xlabel('\fontsize{14}Position, x'), ylabel('\fontsize{14}m = df/dx'),... title(['\fontsize{16}MTH15 • BLUE',]), axis([0 5 -5,3])subplot(1,2,2)bar(red, 'r'), grid, xlabel('\fontsize{14}Position, x'), axis([0 5 -5,3]),... title(['\fontsize{16}MTH15 • RED',])set(get(gco,'BaseLine'),'LineWidth',4,'LineStyle',':')

ConCavity Defined

A differentiable function f on a < x < b is said to be:

… concave DOWN (↓) if df/dx is DEcreasing on the interval

…concave up if df/dx is INcreasing on the interval.

Example Graphical Concavity

Consider the function f given in the graph and defined on the interval (−4,4).

Approximate all intervals on which the function is INcreasing, DEcreasing, concave up, or concave down

SOLUTION Because we have NO equation for the

function, we need to use our best judgment: • around where the

graph changes directions (increasing/decreasing)

• where the derivative of the graph changes directions (concave up or down).

To determine where the function is INcreasing, we look for the graph to “Rise to the Right (RR)”

Rising

Similarly, the function is DEcreasing where the graph “Falls to the Right (FR)”:

Falling

Conclude that f is increasing on the interval (0,4) and decreasing on the interval (−4,0)

Now ExamineConcavity.

Falling to Rt Rising to Rt

A function is concave UP wherever its derivative is INcreasing. Visually, we look for where the graph is“curved upward”, or “Bowl-Shaped”Similarly, A function is concave DOWN wherever its derivative is DEcreasing. Visually, we look for where the graph is “curved downward”, or “Dome-Shaped”

The graph is “curved UPward” for values of x near zero, and might guess the curvature to be positive between −1 & 1

f is ConCave UP

The graph is “curved DOWNward” for values of x on the outer edges of the domain.

f is ConCave DOWN f is ConCave DOWN

Thus the function is concave UP approximately on the interval (−1,1) and concave DOWN on the intervals (−4, −1) & (1,4)

f is ConCave UPf is ConCave DOWN f is ConCave DOWN

Inflection Point Defined

A function has an inflection point at x=a if f is continuous and the CONCAVITY of f CHANGES at Pt-a

-2 -1 0 1 2 3 4 5 6 7 8 9-50

MTH15 • Inflection Point

ConCave DOWN

ConCave UP

InflectionPoint

% Bruce Mayer, PE% MTH-15 • 10Jul13% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m%% The Limitsxmin = -2; xmax = 9; ymin =-50; ymax = 50;% The FUNCTIONx = linspace(xmin,xmax,1000); y =(x-4).^3/4 + (x+5).^2/7;yOf4 = (4-4).^3/4 + (4+5).^2/7 % % 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', 5),axis([xmin xmax ymin ymax]),... grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = f(x)'),... title(['\fontsize{16}MTH15 • Inflection Point',])hold onplot(4, yOf4, 'd r', 'MarkerSize', 9,'MarkerFaceColor', 'r', 'LineWidth', 2)plot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)set(gca,'XTick',[xmin:1:xmax]); set(gca,'YTick',[ymin:10:ymax])hold off

Example Inflection Graphically

The function shown above has TWO inflection points.

change from concave

down to up

change from concave

up to down

2nd Derivative Test

Consider a function for Which is Defined on some interval containing a critical Point (Recall that ) Then:• If , then is Concave UP at so is a

Relative MIN• If , then is Concave DOWN at so is

a Relative MAX

Example Apply 2nd Deriv Test

Use the 2nd Derivative Test to Find and classify all critical points for the Function

SOLUTION Find the

critical points by solving:

0' xfdx

By Zero-Products:

Also need to check for values of x that make the derivative undefined.• ReCall the

1st Derivative: • Thus df/dx is UNdefined for x = −1, But the

ORIGINAL function is ALSO Undefined at the this value– Thus there is NO Critical Point at x = −1

2OR020 xxxx

Thus the only critical points are at −2 & 0 Now use the second derivative test to

determine whether each is a MAXimum or MINimum (or if the test is InConclusive):

1122221

Before expanding the BiNomials, note that the numerator and denominator can be simplified by removing a common factor of (x+1) from all terms:

1122221

222211

Now expand BiNomials:

Now Check Value of f’’’(0) & f’’’(−2)

422222

fd3)1(

The 2nd Derivative is NEGATIVE at x = −2• Thus the orginal fcn is ConCave

DOWN at x = −2, and aRelative MAX exists at this Pt

Conversely, 2nd Derivative is POSITIVE at x = 0• Thus the orginal fcn is ConCave UP at x = 0

and a Relative MIN exists at this Pt

Confirm by Plot → Note the relative

MINimum at 0, relative MAXimumat −2, and a vertical asymptote where the function is undefined at x=−1 (although the vertical line is not part of the graph of the function)

ConCavity Sign Chart A form of the df/dx (Slope) Sign Chart

(Direction-Diagram) Analysis Can be Applied to d2f/dx2 (ConCavity)

Call the ConCavity Sign-Charts “Dome-Diagrams” for INFLECTION Analysis

−−−−−−++++++ −−−−−− ++++++

ConCavityForm

d2f/dx2 Sign

Critical (Break)Points Inflection NO

InflectionInflection

Example Dome-Diagram

Find All Inflection Points for • Notes on this (and all other) PolyNomial

Function exists for ALL x

Use the ENGR25 Computer Algebra System, MuPAD, to find • Derivatives• Critical Points

153 45 xxxfy

Example Dome-Diagram The Derivatives

The Critical Points

The ConCavity Values Between Break Pts• At x = −1

• At x = ½

MyPAD Code

Draw Dome-Diagram

The ConCavity Does NOT change at 0, but it DOES at 1• Since Inflection requires Change, the

only Inflection-Pt occurs at x = 1

−−−−−−−−−−−− ++++++

ConCavityForm

d2f/dx2 Sign

Critical (Break)Points NO

TheFcnPlotShowingInflectionPoint at(1,y(1))= (1,−3)

-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-15

x5 - 5

x4 - 1

MTH15 • Dome-Diagram

(1,−3)

% Bruce Mayer, PE% MTH-15 • 11Jul13% XYfcnGraph6x6BlueGreenBkGndTemplate1306.m%% The Limitsxmin = -1.5; xmax = 2.5; ymin =-15; ymax = 15;% The FUNCTIONx = linspace(xmin,xmax,1000); y =3*x.^5 - 5*x.^4 - 1;% % 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', 5),axis([xmin xmax ymin ymax]),... grid, xlabel('\fontsize{14}x'), ylabel('\fontsize{14}y = f(x) = 3x^5 - 5x^4 - 1'),... title(['\fontsize{16}MTH15 • Dome-Diagram',])hold onplot(1,-3, 'd r', 'MarkerSize', 10,'MarkerFaceColor', 'r', 'LineWidth', 2)plot(zxv,zyv, 'k', zxh,zyh, 'k', 'LineWidth', 2)set(gca,'XTick',[xmin:0.5:xmax]); set(gca,'YTick',[ymin:5:ymax])hold off

Example Population Growth

A population model finds that the number of people, P, living in a city, in kPeople, t years after the beginning of 2010 will be:

Questions • In what year will the population be

decreasing most rapidly? • What will be the population at that time?

105109 23 ttttP

SOLUTION: “Decreasing most rapidly” is a phrase

that requires some examination. “Decreasing” suggests a negative derivative.

“Decreasing most rapidly” means a value for which the negative derivative is as negative as possible. In other words, where the derivative is a MIN

Need to find relative minima of functions (derivative functions are no exception) where the rate of change is equal to 0.

“Rate of change in the population derivative, set equal to zero” TRANSLATES mathematically to

The only time at which the second derivative of P is equal to zero is the beginning of 2013.• Need to verify that the derivative is, in fact,

negative at that point:

10183' 2 tttPdt

10)3(18)3(33' 2

171054273'3

Thus the function is decreasing most rapidly at the inflection point at the beginning of 2013:

The Model Predicts 2013 Population:

Peoplek 81105)3(10)3(9)3(3 23 P

WhiteBoard Work

Problems From §3.2• P45 → Sketch Graph using General

Description• P66 → Spreading a Rumor

All Done for Today

RememgeringConCavity:

cUP & frOWN

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

Appendix

srsrsr 22

ConCavity Sign Chart

−−−−−−++++++ −−−−−− ++++++

ConCavityForm

d2f/dx2 Sign

Critical (Break)Points Inflection NO

Max/Min Sign Chart

−−−−−−++++++ −−−−−− ++++++

df/dx Sign

Critical (Break)Points Max NO

Max/MinMin

bmayer@chabotcollege.edu mth15_lec-14_sec_3-2_concavity_inflection_.pptx 1 bruce mayer, pe chabot...

chabot mathematics

concavity inflection

intervals of concavity

fcn graph

differentiable function

right f

relative extrema

mth15 red

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