chapter 4 applications of differentiation 1

Max and Min ValuesLocal maximum and minimum valuesThe number fc is a■ local maximum value of f on D if fc ≥ fx when x is near c.■ local minimum value of f on D if fc ≤ fx when x is near

4 APPLICATIONS OF DIFFERENTIATION

4.1 Maximum and Minimum Values

Applied Project: The Calculus of Rainbows

4.2 The Mean Value Theorem

4.3 How Derivatives Affect the Shape of a Graph4.4 Indeterminate Forms and l’Hospital’s Rule

Writing Project: The Origins of l’Hospital’s Rule

4.9 Computer Algebra (Maple\Matlab)

4.1 Max and Min Values

Absolute Maximum and Minimum Values

Let c be a number in the domain D of a function f Then f(c) is the

• absolute (global) maximum value of f on D if f(c) ≥ f(x) for every

4.1 Max and Min Values

Local maximum and minimum values

The number f(c) is a

■ local maximum value of f on D if f(c) ≥ f(x) when x is near c.■ local minimum value of f on D if f(c) ≤ f(x) when x is near c.

4.1 Max and Min Values

Example Determine

whether the given function has extreme values at

indicated points

4.1 Max and Min ValuesExistence of extreme values on closed interval

If f is a continuous function on closed interval [a, b] or on a union of finitely many closed interval, then f must have an absolute maximum

value and an absolute minimum value on this interval.

4.1 Max and Min Values

Fermat’s Theorem

If f has a local maximum or minimum at c, and if f ’(c) exists, then

f ’(c) = 0.

4.1 Max and Min Values

4.1 Max and Min ValuesNecessary conditions of extreme values

If f has extreme value at c, then c must be one of the following

1 Critical points of f (f ’(c) = 0 or f ’(c) does not exist),

2 Endpoints of the domain of f

4.1 Max and Min ValuesEx1. Find the absolute maximum and minimum values of the function

f(x) = x3 – 3x2 + 1 on [– ½, 4].

4.1 Max and Min ValuesEx2. The Hubble Space Telescope was deployed on April 24, 1990,by the space shuttle Discovery A model for the velocity of the shuttle

during this mission, from liftoff at t = 0 until the solid rocket boosterswere jettisoned at t = 126 s, is given by

v(t) = 0.001302t3 – 0.09029t2 + 23.61t – 3.083

(in feet per second) Using this model, estimate the absolute maximumand minimum values of the acceleration of the shuttle between liftoffand the jettisoning of the boosters.

4.1 Max and Min ValuesFunctions not defined on closed, finite intervals

If f is continuous on I = (a, b) (or (– ∞, ∞)), and if

(1) If there exists u in I | f(u) > max (L, R) => f has a max value on I.(2) If there exists v in I | f(v) < min (L, R) => f has a min value on I.

Lx

4.1 Max and Min Values

Example 1 Show that f(x) = x + 4/x has an abs.min value on (0, ∞),

and find that min value.

Example 2 Let Find and classify the critical points of f.f(x)xex2

4.2 The Mean Value Theorem

Rolle’s Theorem Let f be a function that satisfies the following

three hypotheses:

1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f(a) = f(b)

Then there is a number c in (a, b) such that f’ (c) = 0.

4.2 The Mean Value Theorem

4.2 The Mean Value TheoremPhysical meaning: If the moving particle is in the same place at

two different instants t = a and t = b, then there is some instant t = cbetween a and b at which the velocity of the object is zero (In

particular, you can see that this is true when a ball is thrown directlyupward).

4.2 The Mean Value TheoremEx1. Prove that the equation x3 + x – 1 = 0 has exactly one real root.

Step 1 Prove that a root exists by using the Intermediate ValueTheorem.

Step 2 Argue by contradiction that f(x) = x3 + x – 1 = 0 has more than

one real solution, then using the Rolle’s theorem to give thecontradiction.

4.2 The Mean Value Theorem

The Mean Value Theorem

Let f be a function that satisfies the following two hypotheses:1 f is continuous on the closed interval [a, b].

2 f is differentiable on the open interval (a, b).Then there is a number c in (a, b) such that

' or f(b)f(a)f'(c)(ba)

4.2 The Mean Value Theorem

There is at least one point on the graph where the tangent line is parallel to the secant line AB.

4.2 The Mean Value TheoremPhysical meaning: The Mean-value theorem tell us that, at some

time t = c in (a, b), the instantaneous velocity is equal to the average

velocity over this time period.

4.2 The Mean Value TheoremEx1. Suppose that f(0) = –3 and f ’(x) ≤ 5 for all values of x How largecan f(2) possibly be?

1 If f ’(x) = 0 for all x in (a, b), then f (x) = C for all x in (a, b);

2 If f ’(x) = g’(x) for all x in (a, b), then f(x) – g(x) = C for all x in (a, b).

4.2 The Mean Value TheoremEx1. Proof that

Solution Let

Show that F ’(x) = 0 Then F(x) = F(1) =/2.

tan1  1 

F()tan1 cot1

4.3 How Derivatives Affect the Shape of a Graph

Increasing/Decreasing Test

(a) If f ’(x) > 0 on an interval, then

f is increasing on that interval.

4.3 How Derivatives Affect the Shape of a GraphThe first derivative tests

4.3 How Derivatives Affect the Shape of a Graph

4.3 How Derivatives Affect the Shape of a Graph

Ex2. Find the local maximum and minimum values of the function

g(x) = x + 2sinx, 0 ≤ x ≤ 2.

4.3 How Derivatives Affect the Shape of a Graph

If the graph of f lies above all of its tangents on an interval I,

then it is called concave upward on I.

If the graph of f lies below all of its tangents on I, it is called

4.3 How Derivatives Affect the Shape of a GraphConcavity Test

(a) If f ’’(x) > 0 for all x in I, then the graph of f is concave upward on I.(b) If f ’’(x) < 0 for all x in I, then the graph of f is concave downward

on I.

f is concave up on I if it is differentiable there and the

derivative f ’ is an increasing function on I.

f is concave down on I if it is differentiable there and the

derivative f ’ is an decreasing function on I.

4.3 How Derivatives Affect the Shape of a Graph

TEST FOR CONCAVITY

f ’’(x)f ’’(x)

Miss SmileMr Frown

4.3 How Derivatives Affect the Shape of a Graph

Ex3. Figure shows a population graph for Cyprian honeybees raisedin an apiary How does the rate of population increase change overtime?

P (thousands)

3691215 18 tin weeks

When is this ratehighest? Over whatintervals is P concaveupward or concavedownward?

4.3 How Derivatives Affect the Shape of a Graph

Definition A point P on a curve y = f(x) is called an inflection point

if is continuous there and the curve changes its concavity at P.

In view of the Concavity Test, there is a point of inflection at anypoint where the second derivative changes sign.

4.3 How Derivatives Affect the Shape of a Graph

Ex4. Sketch a possible graph of a function that satisfies the followingthree conditions:

(1) f ’(x) > 0 on (– ∞, 1), f ’(x) < 0 on (1, ∞),

(2) f ’’(x) > 0 on (– ∞, – 2) and (2, ∞), f ’’(x) < 0 on (– 2, 2) and(3) f (x) ‒> – 2 when x ‒> – ∞, f (x) ‒> 0 when x ‒> ∞.

4.3 How Derivatives Affect the Shape of a GraphThe Second Derivative Test

Suppose f ’’ is continuous near c.

(a) If f ’(c) = 0 and f ’’(c) > 0, then has a local minimum at c.(b) If f ’(c) = 0 and f ’’(c) < 0, then has a local maximum at c.

f (c) f (x) P

cf (c)

f (x) xP

4.3 How Derivatives Affect the Shape of a Graph

Ex5. Discuss the curve y = x4 – 4x3 with respect to concavity, points ofinflection, and local maxima and minima Use this information to sketchthe curve.

Solution f(x) = x4 – 4x3, f ’(x) = 4x3 – 12x2, f ”(x) = 12x2 – 24x.

f ’(x) = 0  4x2 (x – 3) = 0 => x = 0, x = 3,

f ”(x) = 0  12x (x – 2) = 0 => x = 0, x = 2

4.3 How Derivatives Affect the Shape of a Graph

04 6

4.3 How Derivatives Affect the Shape of a Graph

Ex6 Sketch the graph of the

function y = x2/3(6 – x)1/3.

min

4.3 How Derivatives Affect the Shape of a Graph

4.3 How Derivatives Affect the Shape of a Graph



4.4.Indeterminate Forms and l’Hospital’s Rule

Def 1.

Consider a limit of the form

If both f(x) -> 0 and g(x) -> 0 as x -> a (a - finite or infinite), then

this type of limit is called an indeterminate form of type 0/0.

If both f(x) -> ∞ (– ∞ ) and g(x) -> ∞ (– ∞ ) as x -> a (a - finite or

infinite) then this type of limit is called an indeterminate form oftype ∞/∞.

xfax

4.4.Indeterminate Forms and l’Hospital’s Rule

L’Hospital’s Rule

Suppose f and g are differentiable and g ’(x) ≠ 0 on an open interval Ithat contains a (except possibly at a) Suppose that

is an indeterminate form of type 0/0 or ∞/∞ Then

if the limit of the right side exists (or is ∞ or – ∞).

axa

4.4.Indeterminate Forms and l’Hospital’s Rule

f and g approach 0 as x -> a If

we were to zoom in toward the

point (a, 0), the graphs would

start to look almost linear.

L’Hospital’s Rule might be true visually

f and g are linear

axa

4.4.Indeterminate Forms and l’Hospital’s Rule

Examples. Calculate the following limits

1 

 xx

1 sin

 

  

4.4.Indeterminate Forms and l’Hospital’s Rule

vào Ví dụ 2

4.4.Indeterminate Forms and l’Hospital’s Rule

INDETERMINATE PRODUCTS

The limit of the form

where f(x) -> 0 and g(x) -> ∞ (or – ∞) as x -> a (a - finite or infinite)

is called an indeterminate form of type 0 ∞.

Solution: Write the product as a quotient or

Use the L’Hospital’s rule.



4.4.Indeterminate Forms and l’Hospital’s Rule

Ex6. Evaluate



4.4.Indeterminate Forms and l’Hospital’s Rule

lim lnlim1

x

4.4.Indeterminate Forms and l’Hospital’s Rule

INDETERMINATE DIFFERENCES

The limit of the form

where f(x) -> ∞ and g(x) -> ∞ as x -> a (a - finite or infinite) is called

an indeterminate form of type ∞ – ∞.

Solution: Convert the difference into a quotient (using commondenominator, rationalization, factoring out a common factor, etc.) thenuse the L’Hospital’s rule.



4.4.Indeterminate Forms and l’Hospital’s Rule

Ex7. Evaluate

( 

4.4.Indeterminate Forms and l’Hospital’s Rule



4.4.Indeterminate Forms and l’Hospital’s Rule

Examples. Evaluate the following limits

 



4.4.Indeterminate Forms and l’Hospital’s Rule

 

Solution

4.4.Indeterminate Forms and l’Hospital’s Rule

 

 

 

 

 



4.5.Summary of Curve Sketching

Some particular aspects of curve sketching:(1) domain, range, and symmetry in Chapter 1;(2) limits, continuity, and asymptotes in Chapter 2;(3) derivatives and tangents in Chapters 2 and 3;

(4) extreme values, intervals of increase and decrease, concavity,points of inflection, and l’Hospital’s Rule in this chapter.

We put all above information together to sketch graphs that reveal the important features of functions.

4.5.Summary of Curve Sketching

Guidelines for Sketching a Curve y = f(x) by Hands

(1) Domain

(2) Intercepts: the y-intercepts = f (0), the x-intercepts: 0 = f(x).

(3) Symmetry: odd and even functions, periodic function.(4) Asymptotes: horizontal (x -> ±∞),

vertical (y -> ±∞ when x -> a),

slant (f(x) – (ax + b) -> 0 when x -> ±∞)

4.5.Summary of Curve Sketching

(5) Intervals of Increase and Decrease: f’(x)

(6) Local maximum and minimum: first derivative test(7) Concavity and points of inflection: concavity test(8) Sketch the curve

4.5.Summary of Curve Sketching

Example. Sketch the graph of

(1) Domain: (–∞, ∞).

(2) The x- and y-intercepts: 0, 0.

(3) Symmetry: odd function, graph is symmetric about the origin.

(4) Asymptotes: oblique asymptote: y = x.

xf

inf

4.6 Graphing with Caculus and Caculators

See from page 318 to page 325

4.7 Optimization ProblemsProcedure for Solving Extreme-Value Problems

S1. Find out what is given and what must be found.

S2. Make a diagram if appropriate.

S3. Use symbols to denote variables.

S4. Express the quantity Q to be minimized (maximized) as a function

of variables

4.7 Optimization Problems

S5. If Q depends on n variables, where n > 1, find (n – 1) constrains.

S6. Use these constrains to express Q as a function of only one

S7. Find the required extreme value of Q.

4.7 Optimization ProblemsEx1. A rectangular animal enclosure is to be constructed having oneside along an existing long wall and the other three sides fenced If100m of fence are available, what is the largest possible area for theenclosure?

xA = xy

4.7 Optimization ProblemsEx2. A cylindrical can is to be made to hold 1 L of oil Find thedimensions that will minimize the cost of the metal to manufacture thecan.

4.7 Optimization ProblemsEx3. Find the point on the parabola y2 = 2x that is closest to the

point (1, 4).

4.7 Optimization ProblemsEx4. A lighthouse L is located on a small island 5km north of a pointA on a straight east-west shoreline A cable is to be laid from L to pointB on the shoreline 10km east of A The cable will be laid through thewater in a straight line from L to a point C on the shoreline between Aand B, and from there to B along the shoreline The part of the cablelying in the water costs $5,000/km and the part along the shorelinecosts $3,000/km.

(a) Where should C be chosen to minimize the total cost of the cable?(b) Where should C be chosen if B is only 3km from A?

4.7 Optimization Problems

5 km

x km(10 – x) km

4.7 Optimization Problems

Applications to Business and Economics

(1) cost function: C(x) – the cost of producing x units of a certain

(2) marginal cost: C’(x).

(3) demand (price) function: p(x) – the price per unit that the

company can charge if it sells x units.

4.7 Optimization ProblemsEx5. A store has been selling 2,000 chairs per month at $50 each Amarket survey indicates that, for each $5 rebate offered to buyers, thenumber of chairs sold will increase by 500 a month.

(a) Find the demand function and revenue function.

(b) How much of a rebate should the store offer to maximize itsrevenue?

4.7 Optimization ProblemsSolution.

(a) Suppose x is the number of chairs sold per month.Increase in number of chairs: x – 2000

4.8 Newton’s Method

PROBLEMS

4.8 Newton’s Method

We use Newton’s method for finding root of the equation f(x) = 0.

f(x1)



4.8 Newton’s Method

(4) Similar formulas produce x4 from x3, then x5 from x4, and so on.

The formula producing xnfrom xn – 1 is

If the numbers xnbecome closer and closer to r as n becomes large,then we say that the sequence converges to r and write

 

lim

4.8 Newton’s Method

1. When using Newton’s method to solve an equation of the form g(x)= h(x), we must rewrite it in the form f(x) = g(x) – h(x) = 0, and applythe Newton’s method to f.

4.8 Newton’s MethodEx1. Given equation: x3 – 2x – 5 = 0 Initial guess x1= 2.

(a) Find the third approximation x3 to the real root of this equation.(b) Find the real root correct to four decimal places

• f(x) = x3 – 2x – 5, f’ (x) = 3x2 – 2, using the first derivative test, it iseasy to see that the equation has only one real solution.

• As f(2) f(3) < 0 then this solution lies somewhere between 2 and 3.

Hence our initial guess is seemed to be appropriate.

4.8 Newton’s MethodEx2. Use Newton’s method to find (2)1/6 correct to eight decimalplaces.

4.9 AntiderivatiesDef 1. An anti-derivative of a function f on an interval I is anotherfunction F satisfying

4.9 Antiderivaties

FEW SIMPLE INDEFINITE INTEGRALS

4.9 Antiderivaties

FEW SIMPLE INDEFINITE INTEGRALS

4.9 Antiderivaties

Differential Equations and Initial-Value Problems

Def 3.A differential equation is an equation involving derivatives

of an unknown function.

Ex1 a) y’ + 2y = 1,b) xy” – sin(x) = x

c) y’’’ – 3xy” + 2y = 0…

4.9 Antiderivaties

 Solution of a DE: Any function whose derivatives satisfy the DE. Order of DE: The order of the highest-order derivative in DE. General solution: A function that contains all solutions of DE

Aparticular solution: One of the possible solutions of DE.

An initial-value prob: DE + sufficiently enough prescribed values.