Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 38 docx

A continuous function on a closed interval is guaranteed to attain an absolute maximum value and an absolute minimum value.. x 1 –1 Figure 10.16 Summary If f has any local or global extr

x

(i) f (x) = |x| on (– ∞, ∞)

f

x

(ii) f (x) = x on [–1, 2]

Figure 10.12 Local Extrema

Suppose x = c is an interior critical point of a continuous function f How can we tell if,

at x = c, f has a local maximum, local minimum, or neither?

One approach is to look at the sign of f to determine whether f changes from

increasing to decreasing across x = c This type of analysis is referred to as the first derivative test. If f is continuous, x = c is an interior critical point of f , and f is differentiable on an open interval around c (even if f is not differentiable specifically at c), then:

if fchanges sign from negative to positive at x = c, then f has a local minimum at c;

if fchanges sign from positive to negative at x = c, then f has a local maximum at c;

if fdoes not change sign across x = c, then f does not have a local extremum at c

graph of f

sign of f ′

graph of f sign of f ′

c

Figure 10.13

Suppose x = c is an interior critical point of f but f is not continuous You might wonder

if there is a first derivative test we can apply The answer is no Look carefully at the graphs presented in Figure 10.14

k

x

1 –1

f

x

Figure 10.14 Global Extrema

Suppose we’ve rounded up the usual suspects for extrema; we’ve identified the critical points

of f Is there an easy way to identify the global maximum and minimum values? First we

have to figure out whether or not the function has a global maximum If we know it does,

we can calculate the value of the function at each of the critical points The largest value is the global maximum value (The corresponding x gives you the absolute maximum point.) Sometimes you can exclude a few candidates For instance, a local minimum will never be

a global maximum

Critical point x f (x)

Compare the

values of f

at its critical points

List all critical

points of f.

Some functions don’t have absolute maximum and minimum values Think about the functions f (x) = 1/x and h(x) = x2, each on its natural domain The former has neither

a maximum nor a minimum and the latter has only a minimum Or consider g(x) = x and h(x) = x2on the interval (0, 1) On this open interval neither of these functions has a global maximum or a global minimum

g

x

(b) g(x) = x on (0, 1)

f

x

h

x

1

Figure 10.15

Think about what went wrong in the cases above and see what criteria might guarantee

a global extrema To begin with, let’s look at a function whose domain is a closed

inter-val, [a, b] This will fix the problems encountered in the case of functions f , g, and h above

Now consider the functions j (x) =1xfor x = 0 and 0 for x = 0 or k(x) = (x2)for x = 0 and 1 for x = 0, both restricted to the closed interval [−1, 1] (See Figure 10.16.) To rule out these cases, we insist that the function be continuous A continuous function on a closed interval is guaranteed to attain an absolute maximum value and an absolute minimum value This is the Extreme Value Theorem discussed in Chapter 7 (It should seem reasonable, but its proof is difficult and beyond the scope of this text.)

x

1 –1

Figure 10.16

Summary

If f has any local or global extrema, these will occur at critical points of f x0is a critical point of f if x0is in the domain of f , and:

f(x0) = 0, or

f(x0)is undefined, or

x0is an endpoint of the domain of f

Absolute extrema: If f is continuous on a closed interval [a, b], then f attains an absolute maximum and absolute minimum on [a, b] (Extreme Value Theorem) These can be found by evaluating f at each of its critical points Where the value of f is greatest, f has an absolute maximum; where the value of f is least, f has an absolute minimum When looking for absolute extrema in general, you must treat the situations

on a case-by-case basis Stand back and take a bird’s-eye view of the function to see if you expect global extrema If f has any absolute maximum or minimum values, they will occur at the critical points

Local extrema:Suppose x0is a critical point of f If f is continuous at x0and fexists

in an open interval around x0, although not necessarily at x0itself, then we can apply the

first derivative test If the sign of fchanges at x0, then f has a local extremum at x0 Plot sign information on a number line to distinguish between maxima and minima If the sign of fdoes not change on either side of x0, then f has neither a local maximum nor a local minimum at x0

EXAMPLE 10.4 Let f (x) = x3− 12x + 3 Find all local extrema of f

SOLUTION Look for critical points

f(x) = 3x2− 12

fis defined everywhere, and the function is defined for all real numbers, therefore the only critical points are the zeros of f

f(x) = 3x2− 12 = 0 3(x2− 4) = 0

x = ±2

To determine whether these points are extrema, we set up a number line and determine the sign of fin each of the three intervals into which the critical points partition the line

graph of f

So f has a local minimum at x = 2 and a local maximum at x = −2

EXAMPLE 10.5 Suppose f is continuous on (−∞, ∞) f= 0 at x = −1 and at x = 2 fdoes not exist at

x = −3 and at x = 0 Classify the critical points of f as best you can given the information below regarding the sign of f

sign of f ′

f ′ undef f ′ undef

SOLUTION Correlating the sign of the derivative with information about the slope of the function gives

us the following number line

graph of f sign of f ′

This gives us the following information about the extrema of f :

at x = −3, f has a local maximum;

at x = −1, f has a local minimum;

at x = 0, f has a local maximum;

at x = 2, f has neither a local maximum nor minimum

There is not enough information to determine whether or not f has an absolute minimum value It has an absolute maximum at either x = −3 or x = 0, or at both There is not enough information to make a stronger statement

P R O B L E M S F O R S E C T I O N 1 0 1

In Problems 1 through 16, for each function:

(a) Find all critical points on the specified interval.

(b) Classify each critical point: Is it a local maximum, a local minimum, an absolute maximum, or an absolute minimum?

(c) If the function attains an absolute maximum and/or minimum on the specified interval, what is the maximum and/or minimum value?

1 f (x) = x3− 3x + 2 on (−∞, ∞)

2 f (x) = x3− 3x + 2 on [−5, 5]

3 f (x) = x3− 3x + 2 on [0, 3]

4 f (x) = x3− 3x + 2 on (0, 3)

5 f (x) = −2x3+ 3x2+ 12x + 5 on (−∞, ∞)

6 f (x) = −2x3+ 3x2+ 12x + 5 on [−3, 4]

7 f (x) = x5− 20x + 5 on (−∞, ∞)

8 f (x) = x5− 20x + 5 on [−2, 0]

9 f (x) = x5− 20x + 5 on [0, 2]

10 f (x) = 3x4− 8x3+ 3 on (−∞, ∞)

11 f (x) = 3x4− 8x3+ 3 on [−1, 1]

12 f (x) = 3x4− 8x3+ 3 on (0, 3)

13 f (x) = x33 + 2x +3x on its natural domain Why is x = 0 not a critical point?

14 f (x) = x33 + 2x +3x on [−3, 0)

15 f (x) = x33 + 2x +3x on (0, 3]

16 f (x) = x21+4 on (−∞, ∞)

17 Let f (x) = ex

x

(a) Find all critical points of f

(b) Identify all local extrema

(c) Does f have an absolute maximum value? If so, where is it attained? What is its value?

(d) Does f have an absolute minimum value? If so, where is it attained? What is its value?

(e) Answer parts (c) and (d) if x is restricted to (0, ∞)

18 Let f (x) = x2e−x

(a) Find all critical points of f

(b) Classify the critical points

(c) Does f take on an absolute maximum value? If so, where? What is it?

(d) Does f take on an absolute minimum value? If so, where? What is it?

In Problems 19 through 26, find and classify all critical points Determine whether or not f attains an absolute maximum and absolute minimum value If it does, determine the absolute maximum and/or minimum value.

19 f (x) = (x2− 4)ex

20 f (x) = x10x2 +1

21 f (x) = x10x2 +12

22 f (x) = exx

23 f (x) = xx−12

+3

24 f (x) = x4−x2 +9

25 f (x) = x2x3

+1

26 f (x) = e2xx

10.2 CONCAVITY AND THE SECOND DERIVATIVE

In this section we will take another look at concavity to see what it can tell us when we are analyzing critical points Recall that where fis increasing the graph of f is concave up, and where fis decreasing the graph of f is concave down Let’s see what bearing this has

on optimization

Suppose x0is a stationary point: f(x0) = 0 Then f has a horizontal tangent line at

x = x0

If fis increasing at x0, then locally the graph must look like Figure 10.17(a);

f has a local minimum at x0

If fis decreasing at x0, then locally the graph must look like Figure 10.17(b);

f has a local maximum at x0

(x0, f(x0))

(x0, f(x0))

Figure 10.17

We know that

f>0 ⇒ fis increasing ⇒ f is concave up;

f<0 ⇒ fis decreasing ⇒ f is concave down

or or

or or

Therefore, if f(x0) = 0 and f(x0)exists, we can look at the sign of f(x0)and draw the following conclusions:













If f(x0) >0, then f is concave up around x0and therefore has a local minimum

at x0

If f(x0) <0, then f is concave down around x0and therefore has a local maximum

at x0

If f(x0) = 0, we cannot draw any conclusions The function could have a maximum

at x0, a minimum at x0, or neither

This set of criteria is referred to as the second derivative test.

EXERCISE 10.1 Look at the case where f(x0) = 0 and f(x0) = 0 In the three examples below, determine

whether the function has a maximum at x0= 0, a minimum at x0= 0, or neither a maximum nor a minimum at x0= 0 Sketch the graphs of these functions on the set of axes provided below as an illustration of why f(x0) = 0 and f(x0) = 0 does not allow us to classify x0

f

x

(ii) f(x) = x4

f

x

(iii) f(x) = –x4

f

x (i) f(x) = x3

Figure 10.18

D e f i n i t i o n

A point of inflection is a point in the domain of f at which f changes concavity,

i.e., a point at which fchanges sign

CAUTIONThe fact that f(x0) = 0 does not necessarily mean that x = x0is an inflection point Look back at the graphs you’ve drawn in Exercise 10.1 x = 0 is a point of inflection for f (x) = x3but not for f (x) = x4

f

x

–1 –2 –3

x = –2, x = –1, x = 1 and

x = 3.5 are all points

of inflection.

Figure 10.19

Putting It All Together

Suppose x = 3 is an interior critical point of a continuous function f and we are interested

in classifying this critical point, that is, in determining whether x = 3 is a local maximum,

a local minimum, or neither What can we do?

Method (i) First Derivative Test.

We can look at the sign of f, the first derivative, on either side of 3 If fchanges sign around 3, then 3 is a local maximum or local minimum point

Method (ii) Second Derivative Test.

If f(3) = 0, then we can look at the sign of f(3)

If f(3) > 0, then f has a local minimum at x = 3

If f(3) < 0, then f has a local maximum at x = 3

If f(3) = 0, we have insufficient information to draw a conclusion In this case, turn

to method (i)

If x0is an endpoint, then the only label it can get is absolute maximum or absolute minimum; otherwise it remains without a label By looking at the sign of fin the vicinity

of x0we can determine whether it is a potential absolute maximum or potential absolute

minimum We need a bird’s-eye view to determine whether there is an absolute extremum.

If we expect one, we must compare the value of f at x0with its value at other candidates The first derivative test is widely applicable It requires determining the sign of f

on intervals Because the second derivative test only requires evaluating fat a point, it

is simple to apply if fis easy to calculate For instance, consider Example 10.3 where

V (x) = 4x3− 48x2+ 144x, V(x) = 12x2− 96x + 144 and V(x) = 24x − 96 The only interior critical point of V on [0, 6] is x = 2, and V(2) = 24 × 2 − 96 < 0 We conclude that V has a local max at x = 2 The second derivative test, however, can only be applied at

a point at which the first derivative is zero, i.e., at a stationary point The graphs in Figure 10.20 illustrate why this is the case

x0

x0

x0

f" > 0 except at the critical point

f" < 0 except at the critical point

f" changes sign

at the critical point

concavity gives us no useful information on classifying endpoints

Figure 10.20 The second derivative test can’t be applied unless f= 0 at the critical point

f positive / negative increasing /decreasing concave up / concave down

P R O B L E M S F O R S E C T I O N 1 0 2

In Problems 1 through 12:

(a) Find all critical points.

(b) Find f Use the second derivative, wherever possible, to determine which critical points are local maxima and which are local minima If the second derivative test fails or is inapplicable, explain why and use an alternative method for classifying the critical point.

1 f (x) = x3− 6x + 1

2 f (x) = −x3+ 3π2x

3 f (x) = x3+92x2− 12x +32

4 f (x) = x5− 5x

5 f (x) = 2x4+ 64x

6 f (x) = x6+ x4

7 f (x) = x4+ 4x3+ 2

8 f (x) = 4x−1+ 2x2

9 f (x) = ex− x

10 f (x) = xex− ex

11 f (x) = x55 − x4+43x3+ 2

12 f (x) = 3x4− 8x3+ 6x + 1

13 Suppose that f is a continuous function and that f (3) = 2, f(3) = 0, and f(3) = 3

At x = 3, does f have a local maximum, a local minimum, neither a local maximum nor a local minimum, or is it impossible to determine? Explain your answer

14 Suppose that f is a continuous function and that f (4) = 1, f(4) = 0, and f(4) = 0

At x = 4, f could have a local maximum, a local minimum, or neither Sketch three graphs, all satisfying the conditions given, one in which f has a local minimum at

x = 4, one in which f has a local maximum at x = 4, and one in which f has neither

a local maximum nor a local minimum at x = 4

15 Without using the graphing capabilities of your graphing calculator, sketch the follow-ing graphs Label the x-coordinates of all peaks and valleys Label exactly, not usfollow-ing

a numerical approximation (If the x-coordinate is√

2, it should be labeled√

2, not 1.41421.)

Below the sketch of f , sketch f(x), labeling the x-intercepts of the graph of f (You can use your graphing calculator to check your answers.)

(a) f (x) = x(x − 9)(x − 3) (Start by looking at the x-intercepts Then look at the sign

of f(x)in order to determine where the graph of f is increasing and where it is decreasing.)

(b) f (x) = −2x(x − 9)(x − 3) (Conserve your energy! Think!) (c) f (x) = −2x(x − 9)(x − 3) + 18 (Conserve your energy! Think!)

16 (a) The function g with domain (−∞, ∞) is continuous everywhere We are told that

g(√ 5) = 0 Some of the scenarios below would allow us to conclude that g has a local minimum at x =√5 Identify all such scenarios.

i g(√ 5) = 0, g(2) = 1, g(3) = 1

ii g(√ 5) < 0 and g(x) >0 for x >√

5

iii g(√

5) > 0

iv g(√

5) < 0

v g(x) >0 for x <√

5 and g(x) <0 for x >√

5

vi g(x) <0 for x <√

5 and g(x) >0 for x >√

5 vii g(√

5) > 0 and g(√

5) = 0 (b) The function h with domain [−8, −3] has the following characteristics

his continuous at every point in its domain

h(x) <0 for −8 < x < −4 and h(x) >0 for (−4, −3)

h(−4) is undefined

What can you conclude about the local and absolute extrema of h? Please say as much as you can given the information above

17 The graph of f(not f , but f) is a parabola with x-intercepts of −π and 2π and a

y-intercept of −2

(a) Draw a graph of f (b) Write an equation for f This equation should have no unknown constants (c) On the graph you drew in part (a), go back and label the x- and y-coordinates of the vertex

(d) Find f(x)

(e) This part of the question asks about f , not f

i Where does f have a local maximum? Explain your reasoning clearly and briefly

ii Where does f have a local minimum? Explain your reasoning clearly and briefly

iii Does f have an absolute maximum or minimum value? Explain

iv The function f has a single point of inflection What is the x-coordinate of this point of inflection? Suppose you are told that the y-coordinate of the point

of inflection is −1 Find the equation of the tangent line to the graph of f at its point of inflection

18 Consider the function f (x) = x5− 2x4− 7 restricted to the domain [−1, 1] Your reasoning for the questions below must be fully explained and be independent of a graphing calculator

(a) Find the absolute maximum value of f (x) on the interval [−1, 1] or explain why this is not possible

(b) Find the absolute minimum value of f (x) on the interval [−1, 1] or explain why this is not possible

(c) Find the absolute minimum value of f (x) on the open interval (−1, 1) or explain why this is not possible