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Calculus

http://matrix.skku.ac.kr/Cal-Book1/Ch1/

http://matrix.skku.ac.kr/Cal-Book1/Ch2/

Chapter 4. Applications of Differentiation

4.1 Extreme values of a function   http://youtu.be/mXVU8OqIHJY

문제풀이 by 김태영  http://youtu.be/_V4MryNEzWY

4.2 The Shape of a Graph    http://youtu.be/cZrAF_77On4

4.3 The Limit of Indeterminate Forms and L’Hospital’s Rule

문제풀이 by 신종희  http://youtu.be/gR2luDDPsMY

4.4 Optimization Problems   http://youtu.be/k0NtkmZFnh8

문제풀이 by 이승철  http://youtu.be/AELEV2ElaeQ

4.5 Newton’s Method  http://youtu.be/VxCfl2JzMYU

문제풀이 by 이승철  http://youtu.be/fdBHQ46g9RE

4.1 Extreme values of a function

Basic concept of extreme values

Let be a function defined on a domain . A function has an absolute maximum (or global maximum) at if for all in . The number is called the absolute maximum value of on . Similarly, has an absolute minimum (or global minimum) at if for all in ; the number is called the absolute minimum value of on .

Let us now consider only values of near . A function has a local maximum (or relative maximum) at if when is near c, that is, there exists some open interval containing such that for all in . The number is called the local maximum value of on . Similarly, has a local minimum (or relative minimum) at if when is near , and the number is called the local minimum value of on .

By the definition, the global maximum and global minimum values are also local maximum and local minimum values of , respectively. However, the converse is not true in general. In general, the local maximum and local minimum values are called the extreme values of a function .

Figure 1 shows the graph of a function defined on with local maxima (plural in Latin) at and local minima at . Note that the function has the absolute maximum value on and the absolute minimum value on

Figure 1

EXAMPLE 1

Let . Since for all and for any integer , has maximum value of 1 which is achieved at infinitely many points (, ...) infinitely many times. Similarly, since for any integer , has minimum value of infinitely many times, achieved at infinitely many points (, ...).

EXAMPLE 2

Figure 2

Let . Then has no extreme values as we see in Figure 2.

As shown in the examples above, some functions have extreme values whereas others do not. In the following theorem we obtain some conditions under which a function has extreme values.

THEOREM 1 The Extreme Value Theorem

If is continuous on a closed interval , then attains an absolute maximum value and an absolute minimum value at some numbers and in .

If either hypothesis (continuity or closed interval) of the Extreme Value Theorem is not satisfied, then the function need not possess an absolute maximum value nor an absolute minimum value. For instance, the function in Figure 3 has minimum value but no maximum value. The function in Figure 4 has no maximum and minimum values.

Graph of                      Graph of

Figure 3                                           Figure 4

Let us now consider extreme values for a differentiable function. As illustrated in Figure 1, the tangent lines of at the points , , , are horizontal. It suggests that if has a local maximum or minimum at and exists then . In general, we have the following theorem.

THEOREM 2 Fermats Theorem

If is differentiable at and has an extreme value at , then　.

This theorem can be easily proven by applying the definition of extreme values and of a derivative. (See Exercise 28.)

The converse of Fermat’s theorem is not true in general, that is, although , does not necessarily have extreme values at . For example, has neither local maximum nor local minimum value at 0 although .

We should note that may have an extreme value at where does not exist. For instance, has its local minimum value at 0, although does not exist in Figure 5. Also see the function at which has a removable discontinuity in Figure 1.

Figure 5

A critical number (point) of a function is a number in the domain of such that either or does not exist.

To find extreme values of , we should find critical numbers of .

EXAMPLE 3

Find the critical numbers of .

Solution.  Differentiating gives

Therefore, if and does not exist when . Thus, the critical numbers are 2 and 0.

To find the absolute maximum and minimum values of a differentiable function on a closed interval , we suggest the following three steps:

Step 1. Find the values of at the critical numbers of in .

Step 2. Find the values of at the endpoints of the interval.

Step 3. The largest of the values from Steps 1 and 2 is the absolute  maximum value; the smallest of these values is the absolute minimum value. This method is called “The Closed Interval Method.”

EXAMPLE 4

Let , . Find the absolute maximum and absolute minimum values of on the given interval.

Solution.  (Step 1) Differentiating gives

Therefore, if and does not exist when . Thus, the critical numbers on the given interval are 1 and 0. At these critical numbers, , .

(Step 2) At the endpoints of the interval, ,

.

(Step3) Comparing these four function values, the absolute  maximum value is and the absolute minimum value is .

Figure 6

The Mean Value Theorem

We will now study how the signs of the first and second derivatives and influence the extreme values of . The Mean Value Theorem will be useful not only presently but also for explaining why some of the basic properties of calculus are true. We begin with Rolle’s theorem. (For proof refer to Exercise 29.)

THEOREM 3 Rolles Theorem

Let be a function satisfying the three conditions:

(ⅰ) is continuous on the closed interval ;

(ⅱ) is differentiable on the open interval ;

(ⅲ) .

Then there exists a number in such that .

We can see that this theorem is reasonable by interpreting it geometrically.

Geometrically, Rolle’s Theorem tells us that at some point between and the tangent line of the graph of is horizontal whenever

satisfies Rolle’s hypotheses. See Figure 7(a).

Figure 7

Rolle’s theorem can be used to prove another well known theorem in calculus the Mean Value Theorem.

THEOREM 4 Rolles Theorem

Let be a function satisfying the following two conditions:

(ⅰ) is continuous on the closed interval ;

(ⅱ) is differentiable on the open interval .

Then there exists number in such that

1.

or, equivalently,

2.                .

Proof  Let us define a new function as

3. .

By the assumptions (ⅰ) and (ⅱ) for , we see that is continuous on and is differentiable on . In addition, it can be easily seen that . Thus satisfies Rolle’s hypotheses. By Rolle’s Theorem, there exists a number in such that that is,

and so               .

We note that the hypotheses of the Mean Value Theorem do not require to be differentiable at either or . It is enough to be continuous at and . Geometrically, the Mean Value Theorem states that there is at least one point on the graph of where the slope of the tangent line is parallel to the secant line joining the points and . (See Figure 7(b).)

In general, the Mean Value Theorem says that there is a number at which the instantaneous rate of change is equal to the average rate of change over an interval. This enables us to obtain information about a function from information about its derivative. For instance, if an object moves in a straight line with position function , then the velocity at is and the average velocity between and is . Thus by the Mean Value Theorem we see that the instantaneous velocity is equal to that average velocity at some time between and .

Extreme Value Tests at a Critical Number

Recall that by Fermat’s Theorem if has a local maximum or minimum at then must be a critical number of . But not every critical number gives rise to an extreme value. Thus we need a test that will tell us whether or not has a local maximum or minimum at a critical number. First, it is important to use the derivative to determine intervals on which a function is increasing or decreasing.

Definition of Increasing and Decreasing Test

(ⅰ) If on an interval then is increasing on that interval.

(ⅱ) If on an interval then is decreasing on that interval.

This can be easily proven by the Mean Value Theorem (see Exercise 30).

The following two tests are useful in determining whether a critical number gives rise to a local maximum or minimum.

The First Derivative Test

Let be a critical number of a function that is differentiable on some open interval containing , except possibly at .

(ⅰ) If changes from positive to negative at , then has a local maximum at .

(ⅱ) If changes from negative to positive at , then has a local minimum at .

(ⅲ) If does not change sign at , then has neither local maximum nor local minimum at .

Proof  (i) Let us choose sufficiently near to such that . By  the Mean Value Theorem, there exist with

and  such that

and .

Since for and for , is increasing on and decreasing on . Hence has a local maximum at . This proves the first case of the theorem. The second and third cases can be proven in a similar way. (See Exercise 31.)

EXAMPLE 5

Figure 8

Find the local minimum and maximum values of .

SolutionDifferentiating gives

Therefore, if and does not exist when and . Thus, the critical numbers are and changes from negative to positive at , so is a local minimum value by the First Derivative Test. Similarly, changes from positive to negative at 0, so is a local maximum value, and, changes from negative to positive at , so is also a local minimum. (See Figure 8.)

The Second Derivative Test

Let be a critical number such that and exists in an open interval containing .

(ⅰ) If , then has a local minimum at .

(ⅱ) If , then has a local maximum at .

Proof  (i) Since , by the Increasing and Decreasing Test, is  increasing on some open interval containing . Hence for and for . Thus by the First Derivative Test (ⅱ), has a local minimum at . This proves the first case of the theorem. The second case can be proved in a similar way. (See Exercise 32).

CAS EXAMPLE 6

Find all critical numbers for the following function.

.

Solution

var('x')

f(x)=9*x^4 + 26*x^3 - 27*x^2 - 60*x +4

df(x)=diff(f) # 6*(x - 1)*(2*x + 5)*(3*x + 2) : factored

solve(df==0,x)

Answer: [x == (-5/2), x == (-2/3), x == 1]

EXAMPLE 7

Find the local minimum and maximum values of .

SolutionDifferentiating gives . Thus , are critical numbers such that . Since , we have and . Thus by the Second Derivative Test, is a local maximum value and is a local minimum value.

Figure 9

4.1 EXERCISES (Extreme Values of a Function)

1-4. Determine if the following statement is true or false. Explain

1. If is a continuous function, then has a maximum at only one point

in .

Solution.  False. Consider that has more than one maximum.

2. One can apply the Mean Value Theorem to on .

Solution.  False, since is not differentiable on the open interval .

3. If a continuous function has a extreme value on , then has a absolute

maximum or minimum value on .

Solution.  False.   Draw the graph of on .

4. For a continuous function , has only one zero provided is strictly

decreasing.

Solution.  True. We may prove it By the Way Of Contradiction (BWOC).

5-8. Find all critical numbers of the given function.

CAS 5.

Solution.

var('x')

f=9*x^4-62*x^3+159*x^2-180*x

df=diff(f)

solve(df==0,x)

Answer : [x == 2, x == (3/2), x == (5/3)]

6.

Solution.

Therefore critical numbers of are

CAS 7.

Solution.  var('x')

f=abs(x)+abs(x^2 –1)

df=diff

solve(df==0,x)

Answer : [x == 0, x^2 == -1/2*(abs(x^2 - 1) - 2*abs(x))/abs(x)]

CAS 8.

Solution.

var('x')

f=x^2*exp(x)

print f

plot(f, xmin=-3, xmax=3, ymax=5)

To find critical numbers, we differentiate .

df=diff(f)

print df

Answer : x^2*e^x + 2*x*e^x #

Therefore, and (since )

solve((df/e^x)==0, x)

Answer : [x == -2, x == 0]

print f(-2)

plot(f, xmin=-3, xmax=-1, ymax=0.6)

print f(0)

plot(f, xmin=-1, xmax=1, ymax=0.6)

Critical numbers of are or .

9-13. Find all local maxima and minima if any exist.

CAS 9.

http://matrix.skku.ac.kr/cal-lab/cal-4-1-9.html

Solution.  No local minimum and maximum when

var('x');

f(x)=sqrt(5-x)

P = plot(f,x, xmax=5, xmin=0, linestyle="--", color='red')

Q = text("\$f(x)=sqrt(5-x) \$", (1,1), fontsize=20)

show(P+Q)

CAS 10.

http://matrix.skku.ac.kr/cal-lab/cal-4-1-10.html

Solution.  Local minimum , but no local maximum.

var('x')

f=abs(x)

plot(f,-1,1)

CAS 11.

http://matrix.skku.ac.kr/cal-lab/cal-4-1-11.html

Solution.

var('x');

f=abs(sin(x))

plot(f,0,pi)

df=diff(f)

df

sin(x)*cos(x)/abs(sin(x))

plot(df,0,pi)

plot(df,0,pi)

Answer : [x == 0, x == 1/2*pi, x == pi]

12.

Solution.

if , , then has a local

minimum at . if , , then has a local

minimum at .  if , , then has a local

maximum at .

13. ,

Solution.        then has a local

maximum at

14-17. Find the intervals where the function is increasing or decreasing.

14.

Solution.  Increasing on , decreasing on

. See graph of

var('x')

f=x/(x^2+1)

plot(f,-2,2)

df=diff(f);

show(df)

plot(df,-2,2)

CAS 15.

http://matrix.skku.ac.kr/cal-lab/cal-4-1-15.html

Solution.  Increasing on , decreasing on

var('x')

f=2*exp(-(x-1)^2)

F=plot(f,-5,5)

df=diff(f);

show(F+plot(df,-5,5, linestyle="--", color='red'))

print solve(df==0,x)

16.

Solution.

var('x')

f=(x^2-1)*exp(x^2)

g=plot(f, -2, 2, ymax=3, ymin=-1)

df=diff(f)

show(g)

show(df)

Answer :  Increasing on , decreasing on

17. Suppose and for all . Using the Mean

Value Theorem show for all .

Solution.  By the Mean Value Theorem,

and

Since , we have  .

18-20. Prove the inequality.

18.

Solution.

So, for .

var('x')

p1=plot(x*exp(x))

p2=plot(exp(x) -1)

p3=plot(x)

show(p1+p2+p3)

19. ,

Solution.  Let , , . Then ,

, .

From the inequality and the Mean Value Theorem,

, .

Therefore for all .

20.

Solution.    and

var('x, y')

p1=plot(-1/(1-x)^2)

p2=plot(1/(1+x))

show(p1+p2, ymax=2, ymin=-2)

So,   for all .

CAS 21. Let . Find the number of zeros and give a proper interval containing all zeros.

Solution.  7 zeros in

var('x,y')

p1=plot.((x-1+ 10*sin(x)),x,-8,13,color='blue');

show(p1, aspect_ratio=1, ymax=20)

22. Show that for .

Solution.

Then, .

23-24. Prove the inequality using the Mean Value Theorem.

23.

Solution.  Let , then by the Mean Value Theorem there exists

in such that   .

Since  for , we have  .

24.

Solution.  Let , then there exists in such that

.

Since, , we have .

25. Prove that the inequality for all and using the Mean Value Theorem.

Solution.  Let , , then there exists in such that

Thus,.

26. Let be a continuous function. Prove that the only functions satisfying for all are of the form , by the Mean Value Theorem.

Solution.  Applying Mean Value Theorem to on , there exists in

such that   for all .

This implies   .

Now set , , then for all .

27.Suppose is continuous on and differentiable on . Prove

that there exists such that .

Solution.  Let , then .

By Rolle’s theorem,   there exist such that , so

.

28. Prove Fermat’s Theorem: If is differentiable at and has an extreme value at , then .

Solution. Suppose, for the sake of definiteness, that has a local maximum at . Then, if is sufficiently close to . This implies that if is sufficiently close to 0, with being positive or negative, then and therefore . We can divide both sides of this inequality by a positive number.

Thus, if and is sufficiently small, we have .

Taking the right-hand limit of both sides of this inequality,

we get    Since exists, we have

and so we have shown that .

If , then the direction of the inequality (1) is reversed. We have

So, taking the left-hand limit, we have

We have shown that and also that . Since both of these inequalities must be true, the only possibility is that .

29. Prove Rolle’s Theorem.

http://en.wikipedia.org/wiki/Rolle's_theorem

Solution.  Since , we can use the above result due to Fermat’s Theorem to conclude this. Suppose that the maximum is obtained at an interior point of . We shall examine the above right- and left-hand limits separately. For a real such that is in , the value is smaller or equal to because attains its maximum at .

Therefore, for every , , hence  ,

where the limit exists by assumption (it may be negative infinity).

Similarly, for every , the inequality is reversed because the denominator is now negative and we get  , hence  ,

where the limit might be positive infinity.

Finally, because the above right- and left-hand limits agree, the derivative of at must be zero.

30. Prove the Increasing and Decreasing Test.

Solution.  (a) Let and be any two numbers in the interval with .

According to the definition of an increasing function we have to show that .

Applying the Mean Value theorem to on the interval , there exists such that  . By assumption, and , we get . In particular, . Since and were arbitrary, we conclude that whenever . (b) is left

31. Prove the cases (ii), (iii) of the First Derivative Test.

http://en.wikipedia.org/wiki/First_derivative_test

Solution.  (ii) Let us choose sufficiently near to such that . By the Mean Value Theorem, there exist with and such that and .

This implies for and for . In particular is decreasing on and increasing on . Hence has a local minimum at .

(iii) Let us choose sufficiently near such that . By the Mean Value Theorem, there exist with and such that and .

Since for and for . Hence has neither local maximum nor local minimum at .

32. Prove the case (b) of the Second Derivative Test.

Solution.  (b) Suppose we have . Then

Thus, for sufficiently small we get      which means that if so that is decreasing to the left of , and that if so that is increasing to the right of . Now, by the first derivative test we know that has a local maximum at .

4.2 Shapes of Curves

In section 4.1, we discussed how the signs of the first and second derivatives , can be used to determine the extreme values of a function . Moreover, we have studied the following basic concepts of calculus that will be useful in plotting graphs of functions:

◦ Domain and range                  ◦ Symmetry

◦ x-intercepts and y-intercepts       ◦ Continuity

◦ Critical numbers                    ◦ Vertical and horizontal asymptotes

◦ Infinite limits                         ◦ Absolute maximum and minimum

◦ Limits as                  ◦ Increasing and decreasing

◦ Local maximum and minimum

To determine the shape of a graph, here we will revisit those concepts above and we will study more basic concepts of calculus such as

◦ Concavity                           ◦ Inflection points

◦ Oblique asymptotes

Concavity

First, we will see how locating the intervals in which increases or decreases can be used to determine where the graph of is concave upward or concave downward.

Let be differentiable on an open interval . The graph of is called concave upward on if is increasing on the interval, and is called concave downward on if is decreasing on the interval.

Geometrically, it means that if the graph of lies above all of its tangents on an interval , then it is concave upward on (see Figure 1(a)), and if the graph of lies below all of its tangents on , it is concave downward on (see Figure 1(b)).

Figure 1 (a) Concave Upward       (b) Concave downward

Concavity Test

Let be a function whose second derivative exists on an open interval .

(ⅰ) If for all in , then the graph of is concave upward on .

(ⅱ) If for all in , then the graph of is concave downward on .

A proof of this test follows directly from the Increasing and Decreasing Test and the definition of concavity. (See Exercise 24.)

EXAMPLE 1

Figure 2 The graph of

Determine the open intervals on which the graph of is concave upward and concave downward.

Solution.  Differentiating twice gives

Clearly, for and for . Hence . There is no point at which , but by the concavity test the graph of is concave upward on and is concave downward on . (See Figure 2.)

Inflection Point

As shown in Figure 2, we see that the curve changes from concave upward to concave downward at . A point where the graph of a function has a tangent line and where the concavity changes is called an inflection point. There are four types of inflection points illustrated in Figure 3.

(a)                       (b)                      (c)                      (d)

Figure 3

Oblique Asymptotes

In some cases, a function has a slant asymptote. Assume that can be written as

.

If or , then is called an equation of the oblique asymptote (or slant asymptote) of the graph of . We note that if is a rational function such that

then the graph of has an oblique asymptote

EXAMPLE 2

Figure 4

(a) If then and .

Hence is the oblique asymptote in Figure 4.

var('x, y')

p1=plot(x + 2 + 1/x)

p2=plot(x + 2)

show(p1+p2, ymax=10, ymin=-10)

Figure 5

(b) If then and .

Hence is the oblique asymptote in Figure 5.

var('x, y')

p1=plot(x + e^(-x))

p2=plot(x)

show(p1+p2, ymax=4, ymin=-1, xmax=1.5)

How to Sketch a Graph

Below is the list of steps we will use to sketch the graph of a function.

1. Find the domain and range of the function.

2. Find the -intercepts, symmetry of the graph.

3. Find vertical, horizontal, oblique asymptotes of the graph.

4.Compute and to find critical numbers, inflection points, extreme values, intervals for increasing and decreasing, and intervals for concave upward and concave downward.

5. Sketch the shape of a graph, consistent with all of the information obtained.

EXAMPLE 3

Sketch the graph of the function .

Solution.  We will follow the above guidelines:

1. The domain is .

2.The graph has two intercepts , . Since , the graph is symmetric about the origin.

3. It has no oblique asymptotes, but it has one vertical asymptote and one horizontal asymptote .

4. Computing the derivatives gives

and .

The function has three critical numbers , , and two possible inflection points . The other information for the graph is shown in the table.

 Information undefined undefined undefined vertical asymptote increasing,  concave downward 0 local maximum decreasing,  concave downward 0 inflection point decreasing,  concave upward

5. With symmetry about the origin, the graph is shown in Figure 6.

■

CAS EXAMPLE 4

Let ,

(a) Find the vertical and horizontal asymptotes.

(b) Sketch the graph using and .

Solution. We now show the graph of as a blue curve.

C=plot(f, -5, 5, rgbcolor='blue')

show(C)

Then we find and

:

Figure 7

x=var('x')

f=sech(x)

df=diff(f,x)

ddf=diff(df,x)

df, ddf

solve(f==0,x)

df= -tanh(x)*sech(x), ddf=tanh(x)^2*sech(x)+(tanh(x)^2 –1)*sech(x))

[ ]    # No solution

We see that there is no such that . Sage input/output

limit(f, x = infinity)

0

limit(f, x = -infinity)

0

So is a horizontal asymptote, both to the right and to the left.

We now plot and as red and green curves (respectively).

A=plot(df,-5,5,rgbcolor='red')

show(A)

Figure 8

B=plot(ddf,-5,5,rgbcolor='green')

show(B)

Figure 9

Notice how each curve’s values are the slopes of tangent lines to the subsequent curve.

Also interpret the sign of as increasing or decreasing for and the sign of as

concavity of . We have:

show(A+B+C)

Figure 10

(a)

(b)

 < < 0 < < 1 0 0 0

4.2 EXERCISES (Shapes of Curves)

1-5. Find the local maximum and minimum values of . In addition, find the intervals on which is increasing and decreasing, and the intervals of concavity and the inflection points, sketch a graph of .

1.

Solution.

(a) local maximum: local minimum:

(b) increasing on decreasing on

(c) inflection point at   concave down on

concave up on

var('x')

f=2*x^3 +3*x^2 –36*x +7

plot(f, (x-6, 6))

CAS 2.

http://matrix.skku.ac.kr/cal-lab/cal-4-2-2.html

Solution.  (a) local minimum:

(b) increasing on

decreasing on

(c) inflection point at   concave down on

concave up on

var('x')

f=x^4-34*x^2+225

plot(f,-6,6)

df=diff(f,x)

df

solve(df==0,x)

Answer : [x == -sqrt(17), x == sqrt(17), x == 0]

solve(dff==0,x)

Answer : [x == -1/3*sqrt(3)*sqrt(17), x == 1/3*sqrt(3)* sqrt(17)]

3.

Solution.

(a) local maximum: local minimum:

(b) increasing on decreasing on

(c) inflection point at concave down on

concave up on

4.

Solution.  (a) local maximum :  No  local minimum :

(b) increasing on decreasing on

(c) inflection point : No  concave up on

CAS 5.

http://matrix.skku.ac.kr/cal-lab/cal-4-2-5.html

Solution.  (a) Maximum : 1, Minimum : 0

(b) interval of increase : interval of decrease :

(c) inflection point at   concave down on

var('x')

f=tanh(x^2)

plot(f,-3*pi/2,3*pi/2)

df=diff(f)

df

plot(df,-3*pi/2,3*pi/2)

6-11. Find the inflection points of In addition, find intervals in which the graph of is concave upward or concave downward.

6.

Solution.  (a) inflection point at   (b) concave down on

concave up on

7.

Solution.

(a) inflection point at

(b) concave up on    concave down on

8.,

Solution.  (a) inflection point at

(b) concave down on   concave up on

CAS 9.

Solution.

var('x')

f(x)=sqrt(x)*(x-1);

plot(f(x), -5, 5)

var('x')

f(x)=sqrt(x)*(x-1)

plot(f(x), -5, 5)

df=diff(f(x));

dff=diff(df, x)

show(df)

plot(df, -5, 5)

print solve(df==0, x)

print solve(dff==0, x)

Answer : [x == (1/3)], [x == (-1/3)]

(a) critical point at ,  inflection point at

(b) concave down on concave up on

10.

Solution.

(a) inflection point at

(b) concave up on   concave down on

var('x')

f=plot(e^(x)*sin(x), x, ymax=8, ymin=-1, xmax=3)

g=plot(e^(x)*(sin(x)+cos(x)), x,linestyle="--", color='red', ymax=8, ymin=-1, xmax=3)

h=plot(2*e^(x)*cos(x), x,linestyle="--", color='green', ymax=8, ymin=-1, xmax=3) show(f+g+h)

11.

Solution. (a) inflection points : and

(b) concave down on concave up on

12.

Solution. vertical asymptote

horizontal asymptote

13.

Solution. vertical asymptote:   horizontal asymptote:

var('x')

plot((x/(x^2 - 3))^2, -5, 5, thickness=2, color='goldenrod', figsize=4, ymax=20)

14.

Solution.  horizontal asymptote

15.

http://matrix.skku.ac.kr/cal-lab/cal-4-2-15.htm

Solution.  vertical asymptote:   horizontal asymptote: No

var('x')

p1=plot(ln(sec(x)),-pi/2,pi/2,ymax=6,aspect_ratio=1,tick_formatter=pi)

l1=line([(-pi/2,0),(-pi/2,6)],linestyle="--",color='red')

l2=line([(pi/2,0),(pi/2,6)],linestyle="--",color='red')

p1+l1+l2

16-18. Sketch the graph of using the following information.

(a) Find the local maximum and minimum values of .

(b) Find the intervals of increase or decrease.

(c) Find the inflection points of and intervals of concavity.

(d) Find the vertical and horizontal asymptotes.

16.

Solution.  (a) local maximum: No  local minimum: No

(b) decreasing on

(c) inflection point at concave up on

concave down on

(d) vertical asymptote:   horizontal asymptote:

17.

Solution.    ,

(a) local minimum at

(b) increasing on decreasing on

(c) inflection point: No concave up on

(d) vertical asymptote:   horizontal asymptote: No

18. ,

Solution. and and ;

and

(a) local minimum at   local maximum :

(b) increasing on   decreasing on

(c) inflection point at   concave up on

concave down on

(d) vertical asymptote: No  horizontal asymptote:

19.Find all the values of a such that has different roots.

Solution.  Let . Then will have one local maximum and one local minimum.

There will be roots if and only if the maximum is positive and the minimum is negative.

and

Note that and . Hence has local maximum at and local minimum

at . Therefore, if there will be roots.

20. Find the minimum constant for which for all real .

Solution.

,     local minimum at

Therefore, .

21. Find so that has two inflection points at and .

Solution.

and so ,

and so

22.Let .

(a) Find and .

(b) Find the vertical and horizontal asymptotes of .

(c) Sketch the graph of using (a) and (b).

Solution. (a)

(b) ,   vertical asymptote horizontal asymptote

(c)

23.Let and be increasing functions. Prove that is increasing function.

Solution.   For , ,

Therefore, is increasing.

24.Prove the Concavity Test.

Solution.  (a) By Increasing Test  If on an interval then is increasing on .

So is concave upward on .  Part

(b) is proved similarly.

CAS 25. Use CAS to find and when  .

http://matrix.skku.ac.kr/cal-lab/cal-4-2-25.html

Solution.

var('x')

f=6*x^(1/3)+3*x^(4/3)

df=diff(f,x)

dff=diff(df,x)

Answer : [df == 4*x^(1/3) + 2/x^(2/3), dff == 4/3/x^(2/3) - 4/3/x^(5/3)]

CAS 26. Let . Find the local maximum, minimum values and inflection points of

. Sketch the graph of .

http://matrix.skku.ac.kr/cal-lab/cal-4-2-26.html

Solution.  local maximum:   local minimum: No  inflection point:

var('x')

f=(x+2)/((x+5)^(2))

df=diff(f, x);

dff=diff(f, x, 2);

print "The first derivative is:“

show(df(x))

print "The second derivative is:“

show(dff(x))

s=solve(df(x)==0, x, solution_dict=true)

plot(f(x), -10, 2, thickness=1.2, color='blue', figsize=5, ymin=-10, dpi=120)

print "f has only one critical point at x =", s[0][x]

print "f has local maximum at x=", s[0][x]

print "The maximum value of f  is ", f(s[0][x])

s1=solve(dff(x)==0, x, solution_dict=true); s1

print "The inflection point of f is at ", s

f has only one critical point at x = 1

f has local maximum at x= 1

The maximum value of f at x=1 is  1/12

The inflection point of f is at x = 4

CAS 27. Let . Find the local maximum, minimum values and inflection

points of .

http://matrix.skku.ac.kr/cal-lab/cal-4-2-27.html

Solution.  local maximum: local minimum: No

inflection point:

var('x')

f(x)=(ln(x))/(x^(1/4))

df=diff(f,x)

dff=diff(df,x)

plot(f,x,0,900)

4.3 The Limit of Indeterminate Forms and

L’Hosptal’s Rule

Some functions have certain limits which appear to be one of the following indeterminate forms.

In some cases, we can use L’Hospital’s Rule to find these limits.

LHospitals Rule

Let and be differentiable functions on an open interval containing and near (except possibly at ).

If has an indeterminate form of type or , then

provided the limit on the right side exists (or is or ).

The result follows immediately from

1).

We note that indeterminate forms of type , , , can be converted into a form to or by suitable algebraic transformations.

EXAMPLE 1

Type 1.

EXAMPLE 2

Type 2.

.

■

EXAMPLE 3

Type 3.

.

EXAMPLE 4

Type 4.

.   ■

EXAMPLE 5

Type 5.

Find .        ■

Solution. Let . Then

.

Since , . Hence .

EXAMPLE 6

Type 6.

Find .

Solution.  Let . Then

.

Hence .

EXAMPLE 7

Type 7.

Find .

Solution.  Let . Then

.

Hence

CAS EXAMPLE 8

Evaluate the limit of these indeterminate forms.

(a)                       (b)

Solution. (a)

var('x')

f=sin(sin(x))/x

plot(f, (x,-1,1))

limit(f, x=0)

(b)

var('x')

f=(1-1/x)^(sqrt(x))

show(plot(f, 0,250))

print limit(f, x=infinity)

4.3 EXERCISES (The Limit of Indeterminate Forms and L’Hospital’s Rule)

1-11. Evaluate the limits of the given indeterminate forms.

1.

Solution.  Apply L’Hospital Rule twice.

2.

Solution.

3.

Solution.  By L’Hospital Rule,   .

4.

Solution.

5.

Solution.  Let .

Note and .

By L’Hospital’s Rule,

()

CAS 6.

Solution.  var('x')

f=ln(x)/(x-sqrt(x))

plot(f,x,0.5,1.5,ymin=0)

limit(f,x=1)

7.

Solution.

8.

Solution.  By L’Hospital’s Rule .

9.

Solution.

CAS 10.

Solution.

var('x')

f(x)=sin(x)/sqrt(x)

limit(f(x),x=0)

plot(f(x),x,0,5)

11.

Solution.

12. Let . Use L’Hospital’s Theorem to show that .

Solution.   ,

since is constant and the denominator grows without bound.

13. For what values of and is the following equation correct?

Solution.  ,

By L’Hospital’s Rule for a form of type , .

and we may take to be any value.

14.

Solution.  By L’Hospital’s, Rule for a form of type

var('x')

f(x)=(x^3-2*x^2+2*x-1)/(x^3-x^2)

show(plot(f,(x,-0.5,1.5), ymax=2)+plot(1,(x,-0.5,1.5),linestyle='--'))

limit(f(x), x=1)

A      NSWER. 1

15.

Solution.    Note .

Therefore, by L’Hospital’s Rule,

16-17. Find the following limit.

16.

Solution. (by L’Hospital’s Rule) or

17.

Solution.

(by L’Hospital’s Theorem) or

18-27. Find the limit by using L’Hospital’s rule. If you cannot apply L’Hospital’s rule, explain why and then find the limit by another method.

18.

Solution.  , limit does not exist.

19.

Solution.

(by L’Hospital’s Rule)

var('x');

f(x)=ln(x)/x;

show(plot(f(x),x,0,100, ymax=2, ymin=-2))

limit(f, x=Infinity)

20.

Solution.  Apply L’Hospital’s rule,

CAS 21.  .

http://matrix.skku.ac.kr/cal-lab/cal-4-3-21.html

Solution.

var('x')

f(x)=x^3/exp(x);

limit(f(x),x=0)

plot(f(x),x,-5,5, ymax=2, ymin=-2)

22.

Solution.

23.

Solution.

24.

Solution.  does not exist.

CAS 25.

Solution.

26.

Solution.  var('n, a');

a=2

f=log((1 + a/n)^n)

plot(f, (n,0,40)).show(ymax=3)

limit(f,n=oo)

27.

Solution.

28. The function is defined by

and  .

(a) Find .    (b) Does exist?

Solution.  (a)            (b)    does not exist.

29. Let be a continuous function with and . Find .

Solution.  and . Hence, by L’Hospital’s Theorem

.

30.Let be the angle of a sector of a circle. Find where and

are the area of the segment shaded region between the chord and arc, the area of the triangle respectively.

Solution.  ,

4.4 Mathematical Optimization Problems

In this section, we apply calculus to various optimization problems which occur in the real life. There are many applications to optimizing: maximizing areas, volumes, and minimizing distances, times. An example of such a problem is maximum profit and minimum cost in business.

Guidelines for Solving Optimization Problems

Step 1.Understand the Problem:First understand the given problem, read carefully. What is unknown? What are the given quantities and the given conditions?

Step 2.Draw a Picture with Variables: Drawing a picture about the given and required quantities such as for areas, for volume, for height and for time is very useful.

Step 3. Express the Quantity as a Function: Express the quantity that is to be maximized or minimized as a function with one variable. If the function is of more than one variable then use mathematical ideas or given conditions to obtain a function of one variable. Do not forget to include the domain of this function.

Step 4. Determine the absolute maximum or minimum value of the function.

Step 5. Finally, interpret your solution obtained from step 4 in the context of  the original problem.

EXAMPLE 1

A waterway is to be made so that its cross section is a trapezoid with equally sloping sides. The width and two sides of the waterway are of the same length and equal to 5m. How should the sloping angle be chosen so that the waterway carries the maximum amount of water?

Figure 1

http://matrix.skku.ac.kr/cal-lab/cal-4-4-example-1.html

Solution.

Figure 2

Figure 3

1.Unknown is the cross-sectional area of the waterway, and the shape of cross-section is given as the trapezoid with sides sloping by the angle , with the length of each side equal to 5m.

2.The cross-sectional area is same as the area of the rectangle.

3. In order to maximize the cross-sectional area, we express the area as,

.

4. Our goal is to find a value of that maximizes .

Since and , we obtain , that is, .( is a critical number that does not make physical sense in the context of the problem.)

Figure 4

EXAMPLE 2

A closed cylindrical can is to be made to hold 5 L (Liter) of oil.
What dimensions will use the least amount of metal to make the can?

Solution.  We will follow the strategy above.

Figure 5

1.Unknowns are the dimension of a cylindrical can; knowns are the shape of the can is given as a cylinder together with a volume of 5L.

2. Let be the radius and the height (both in centimeters) of the can. Draw the picture labeled and as shown in Figure 4.

3. In order to minimize the amount of the metal, we express the total surface area of the cylinder consisting of top, bottom and sides as a function with and . So the surface area is

.

To eliminate a variable we use the fact that the volume is 5L which is equal to 5000cm3. Thus which gives . Therefore,

,

4.Our goal is to find a value of that minimizes the value of .

Since  the critical number of intersect as we search for an extreme value is . Applying the First Derivative Test, we see that has an absolute minimum at . The corresponding value of is .

5.Therefore, the cylindrical can of 5L that uses the least metal has height equal to twice the radius with cm and cm.

EXAMPLE 3

Car B is 30km directly east of car A and begins moving west at 90km/h. At the same moment car A begins moving north at 60km/h. What will be the minimum distance between the two cars, and at what time does the minimum distance occur?

Solution.  1. Unknown is the distance between the two cars A and B. The speeds of the cars are given as 60km/h and 90km/h, respectively.

2. Let , be the distance car A and B travel in hours respectively, and let be the distance between two cars A and B after hours as shown in Figure 6.

Figure 6

3. We express the distance as a function with and by the Pythagorean theorem. Thus we have,

.

To express as a function of , we recall that if travel is at a constant rate then  (distance traveled) = (speed) (time). So substituting and into gives .

4. Our goal is to find that minimizes . Since, , we have . If then km. and km.

5.Therefore the shortest distance between the two cars is km.

*Application to Economics

We consider an optimization problem arising in economic theory. This problem is about maximizing profit and minimizing average cost.

Let be the cost of producing units of a certain product. Then the marginal cost is the rate of change of with respect to , that is, the marginal cost function is . The average cost function is given by

which represents the cost per unit when units are produced. To minimize the average cost, differentiating gives

.

Thus when , that is,

.

Therefore by the First Derivative Test we see that if the marginal cost is equal to the average cost then the average cost is a minimum. This principle is plausible because if the marginal cost is smaller than the average cost, then we should produce more, thereby lowering average cost. Similarly, if the marginal cost is larger than the average cost, then we should produce less in order to raise the average cost.

EXAMPLE 4

A company estimates that the cost in dollars of producing items is

.

(a) Find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items.

(b) At what production level will the average cost be lowest, and what is this minimum average cost?

Solution. (a) The average cost function is

.

The marginal cost function is .

The following table gives the cost, average cost, and marginal cost:

 1000 2000 3000 5,600.00 10,600.00 17,600.00 5.60 5.30 5.87 4.00 6.00 8.00

(b) To minimize the average cost we must have marginal cost = average                 cost .

Hence .

This equation simplifies to , that is, and .

Since , is concave upward on its entire domain. Thus the minimum average cost at

is  .

4.4 EXERCISES (Mathematical Optimization Problems)

1. Find the point on the curve that is closest to the point .

Solution.  The distance from to an arbitrary point on the curve is         and the square of the distance is

,          .

Graphing on gives us a zero at , and so . The point on that is closest to is .

2. A shop sells 200 MP3 players per week while each costs . According to the market research, sales will increase by 20 MP3 players per week for each discount. How much should they discount to maximize profits?

Solution.  If discount is made for each MP3 player sale,

.

The profit is maximized when .  They should discount by \$125.

var('x')

f=100*(-2*x^2 + 50*x + 700)

df=diff(f, x);

show (df)

solve (df==0, x)

3. The height of a safe is meters and its bottom is in the shape of a square whose side is meters. It costs Korean won per to make the top and the bottom, and Korean won per to make the side. Find the maximum volume of the safe which can be made by using won.

Solution.

and so

4.  Two particles have locations at time on the -plane given by and . Find the minimum distance between and .

Solution.

Therefore, .

Therefore, the minimum distance between the two particle is 0.

5. A closed cylindrical can is to hold  of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can.

http://matrix.skku.ac.kr/cal-lab/cal-4-4-5.html

Solution.

var('x,y')

f=2*pi*x^2 + 1000/x

P= plot(f,x, xmin=-10, xmax=10, ymin=-60, ymax=60)

df=diff(f, x)

Q=plot(df, x, xmin=-10, xmax=10, ymin=-60,

ymax=60,linestyle="--", color='red')

print solve(df==0, x)   # (250/pi)^(1/3)

print numerical_approx( (250/pi)^(1/3), 100) # 4.3

show(P+Q)

6. Determine where the point between and should be located to maximize the angle .

Solution.  Let , , .

Then ,

.

Solving , we get .

Since we have .

7. Find the largest area of a triangle which is inscribed in a circle of radius .

Solution.  Let and be as shown in the figure. The area of the triangle is

Then,

Now .

The maximum area occurs when with the height and the base .

The largest area of a triangle is .

8. Let be the volume of the right circular cone and be the volume of the right circular cylinder that can be inscribed in the cone. Find the ratio when the cylinder has the greatest volume.

Solution.  Let and be the radius of the base  and height of the right circular cone let. Let and as show in the figure. .

By similar triangles, , so .

The volume of the cylinder is .

Now . So .

The maximum clearly occurs when and then the volume is

.

9.  Consider an ellipse . Find the area of the rectangle of greatest area that can be inscribed in the ellipse.

Solution.  Without loss of generality, choose on the ellipse in first quadrant.

Let = area of the rectangle and = area of the ellipse.

Since ,

Thus, .

10. A closed cylindrical can is to be made to hold 100 of oil. What dimensions will use the least metal?

Solution.  Let be the radius and the height (both in centimeters) of the can. The total surface area of the cylinder consisting of top, bottom and sides as a function with and is . We eliminate the variable by using the given condition . Therefore   , and

.

The only critical number is . Applying the First Derivative Test, we see that has an absolute minimum at . The corresponding value of is             .

11. A cone-shaped paper cup is to be made to hold of water. Find the height and radius of the cup that minimizes the amount of paper needed to make the cup.

Solution.  The volume and surface area of a cone with radius and height are given by and    . We will minimize subject to .

so

, so and hence has an absolute minimum at these values of and .

12. A pipe is being carried horizontally around a corner from a hallway wide into a hallway 2 wide. What is the longest length the pipe may have?

Solution. Let be the length of the line going from wall to wall touching the inner corner . As or , we have and there will be an angle that makes a minimum. A pipe of this length will just fit around the corner. From the diagram,

where and .

when  .

Then and , so the longest pipe has length

13. Find the length of the shortest ladder that reaches over an 8ft high fence to a wall that is 3ft behind the fence.

Solution.  ,

when

when , when ,

so has an absolute minimum when , and the shortest ladder has length  , .

(*Label theta and in the figure.)

4.5 Newton’s Method     http://youtu.be/VxCfl2JzMYU

The simple formulas for solving linear equations and quadratic equations are well-known. Also, there are somewhat more complicated formulas for equations of degree 3 and 4. But there is no such formula for equations of degree or greater. Likewise, there is no formula that will enable us to find the exact solutions of a transcendental equation such as . That means if there is no exact formula for solving , so we turn to numerical techniques. One of these techniques is called Newton’s method. This method is based on the idea of a linear approximation with tangent lines to the graph of .

Procedure for Newton’s Method

Figure 1

Let us consider the graph of and we are trying to solve . We start with the initial approximation , which may be obtained by guessing, or examining the graph of . Then we use the tangent line to the curve at the point to approximate the curve and look at the -intercept of , labeled . The equation of the tangent line is

.

Thus, we can obtain -intercept , by replacing ;

.

If , we can solve this equation for :

Under certain conditions, is usually a better approximation to the solution than . Then we repeat this procedure with replaced by , using the tangent line at . This gives a third approximation:

Continuing this process, we obtain a sequence of approximations , , , as shown in Figure 1. In general, if then we have

1.

The number becomes closer and closer to the solution if the sequence converges as . We note that if then the sequence may not converge. In this case, we have to choose a different initial .

EXAMPLE 1

Compute the third approximation of the root of the equation , starting with .

http://math1.skku.ac.kr/home/pub/1028

Solution.  Apply Newton’s method with

and

Choose, .

Note that                 .

With we have

.

Then with we obtain

It turns out this third approximation is an approximated root. We may

not know this unless we compute more iterates.

Newton’s method

EXAMPLE 2

Find correct to four decimal places using Newton’s method.

Solution.  Finding is equivalent to finding the positive root of the equation  , so take .

Then . Newton’s formula becomes  .

Using as the initial approximation, we obtain

Since and agree to four decimal places, we conclude that to four decimal places.

Newtons Method by Sage:

In Sage define

newton(x)=x-(x^5-40)/(5x^4)

x1=2

x2=newton(x1);N(x2)

2.1000

x3=newton(x2);N(x3)

2.09135

x4=newton(x3);N(x4)

2.0913

Root Finding Using Bisection:

Newton’s method

4.5 EXERCISES (Newton’s Method)

1-3. Compute , the third approximation to the root of the given equation using Newton's method with the specified initial approximation .

1.

Solution.

2.

Solution.

3.

Solution.

4.  Calculate the second approximation to the root of the equation by using Newton’s method with initial approximation .

http://matrix.skku.ac.kr/cal-lab/cal-Newton-method.html

Solution.

5-6. Approximate the given number using Newton’s method.

5. .

http://matrix.skku.ac.kr/2012-LAwithSage/interact/4/3.html

Solution.  Let  .  Let and find by Newton’s method with .

6.

Solution.

7.

Solution.

8.

Solution.

9.

Solution.

(ⅰ)

(ⅱ)

10.

Solution.

(ⅰ)

(ⅱ)

11. Apply Newton’s method to the equation  to derive the following reciprocal algorithm: .

Hence compute .

Solution.

(ⅰ)

12. Use Newton’s method to find the absolute minimum value of the function .

Solution.

Newton-Raphson Root Finding:

Calculus

Sang-Gu Lee and Dr. Jae Hwa Lee and 박수영

*This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03035865).