CHAPTER 3 APPLICATIONS OF THE DERIVATIVE. 3.1 Maxima & Minima Maxima: point whose function value...

CHAPTER 3

APPLICATIONS OF THE DERIVATIVE

3.1

• Maxima & Minima

• Maxima: point whose function value is greater than or equal to function value of any other point in the interval

• Minima: point whose function value is less than or equal to function value of any other point in the interval

• Extrema: Either a maxima or a minima

Where do extrema occur?• Peaks or valleys (either on a smooth curve, or

at a cusp or corner)f’(c)=0 or f’(c) is undefined

• Discontinuties• Endpoints of an interval• These are known as the critical points of the

function• Once you know you have a critical point, you

can test a point on either side to determine if it’s a max or min (or maybe neither…just a leveling off point)

3.2

• Monotonicity and Concavity

Let f be defined on an interval I (open, closed, or neither). Then f is

a) INCREASING on I if,

b) DECREASING on I if,

c) MONTONIC on I if it is ether increasing or decreasing

)()(

)()(

2121

2121

xfxfxx

xfxfxx

Monotonicity Theorem

• Let f be continuous on an interval I and differentiable at every interior point of I.

a) If f’(x)>0 for all x interior to I, then f is increasing on I

b) If f’(x)<0 for all x interior to I, then f is decreasing on I.

Concave UP vs. Concave DOWN

Let f be differentiable on an open interval I.

a) If f’ is increasing on I, f is concave up(the graph appears to be curved up, as a container that would hold water)

b) If f’ is decreasing on I, f is concave down

(the graph appears to be curved down, as if a container were dumping water out)

Point of Inflection

• Where concavity changes: goes from concave up to concave down (or vice versa)

• f’ is neither increasing or decreasing, the change in f’ = 0, thus f’’=0

Find inflection points & determine concavity for f(x)

• Inflection pts: x=-2,0,1• Concave up: (-2,0), (1,infinity)• Concave dn: (-inf.,-2), (0,1)

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3.3

• Local Extrema and Extrema on Open Intervals

First Derivative Test

Let f be continuous on an open interval (a,b) that contains a critical point c.

a) If f’(x)>0 for all x in (1,c) and f’(x)<0 for all x in (c,b), then f(c) is a local max. value.

b) If f’(x)<0 for all x in (1,c) and f’(x)>0 for all x in (c,b), then f(c) is a local min. value.

c) If f’(x) has the same sign on both sides of c, then f(c) is not a local extreme value.

Second Derivative Test

• Let f’ and f’’ exist at every point in an open interval (a,b) containing c, and suppose that f’(c)=0.

a) If f’’(c)<0, then f© is a local max. value of f.

b) If f’’(c)>0, then f© is a local min. value of f.

3.4

• Practical Problems

• Optimization problems – finding the “best” or “least” of “most cost effective”, etc. often involves finding the extrema of the function

• Use either 1st or 2nd derivative test

Example

• A fence is to be constructed using three lengths of fence (the 4th side of the enclosure will be the side of the barn).

• I have 120 yd. of fencing and the barn is 150’ long. In order to enclose the largest possible area, what dimensions of fence should be used?

• (continued on next slide)

Example continued• Area is to be optimized: A = l x w• Perimeter = 120 yd = 360’=2l + w• w = 360’ – 2l

• So, 2 lengths of 90’ and a width of 180’. HOWEVER, the barn is only 150’ wide, so in order to enclose the greatest area, we won’t use a critical point of the function, rather we will evaluate the area using the endpoints of the interval, with w=150’. The length = 55’ and the area enclosed = 8250 sq. ft.

180)90(2360

90,43600

4360'

2360)2360( 2

w

ll

lArea

llllArea

3.5

• Graphing Functions Using Calculus

Critical points & Inflections points

• If f’(x) = 0, function levels off at that point (check on either side or use 2nd deriv. test to see if max. or min.)

• If f’(x) is undefined: cusp, corner, discontinuity, or vertical asymptote (look at behavior and limits of function on either side)

• If f’’(x) = 0: inflection point, curvature changes

3.6

• Mean Value Theorem for Derivatives• If f is continuous on a closed interval

[a,b] and differentiable on the open interval (a,b), then there is at least one number c in (a,b) where

))((')()()(')()(

abcfafbfcfab

afbf

Example: Find a point within the interval (2,5) where the instantaneous velocity is the same as

the average velocity between t=2 and t=5.

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1425

)5)2(2(5)5(2

25

)2()5(

52)(22

2

tttstv

ftssv

tts

ave

If functions have the same derivatives, they differ by a constant.

• If F’(x) = G’(x) for all x in (a,b), then there is a constant C such that F(x) = G(x) + C for all x in (a,b).

3.7

• Solving Equations Numerically– Bisection Method– Newton’s Method– Fixed-Point Algorithm

Bisection Method

• Let f(x) be a continuous function, and let a and b be numbers satisfying a<b and f(a) x f(b) < 0. Let E denote the desired bound for the error (difference between the actual root and the average of a and b).

• Repeat steps until the solution is within the desired bound for error.

• Continue next slide.

Bisection Method

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nnn

nn

nnn

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mbandaasetmfafIf

abhCalculate

STOPmfifmfCalculate

bamCalculate

11

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2/)(:)1

Newton’s Method• Let f(x) be a differentiable function and let

x(1) be an initial approximation to the root r of f(x) = 0. Let E denote a bound for the error. Repeat the following step for n = 1,2,… until the difference between successive error terms is within the error.

)('

)(1

n

nnn xf

xfxx

Fixed-Point Algorithm• Let g(x) be a continuous function and let x(1) be

an initial approximation to the root ro of x = g(x). Let E denote a bound for the error (difference between r and the approximation). Repeat the following step for n – 1,2,… until the difference between succesive approximations are within the error.

)(1 nn xgx

3.8 Antiderivatives

• Definition: We call F an antiderivative of f on an interval if F’(x) = f(x) for all x in the interval.

Power Rule

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xdxx

rr

1

1

Integrate = AntidifferentiateIndefinite integral = Antiderivative

• Constants can be moved out of the integral

• Integral of a sum is the sum of the integrals

• Integral of a difference is the difference of the integrals

3.9 Introduction to Differential Equations

• An equation in which the unknown is a function and that involves derivates (or differentials) of this unknown function is called a differential equation.

• We will work with only first-order separable differential equations.

Example

• Solve the differential equation and find the solution for which y = 3 when x = 1.

CxyCxy

CxCxy

Cxyy

CxCyy

dxxdyy

dxxdyy

y

x

dx

dy

22

23

22

322

22

1

2

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21)2()1(

22

2

)2()1(

)2()1(

1

2

Example continued

• If x = 1 and y = 3, solve for C.

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21

2

2

2

xy

CC

Cxy