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Calculus Volume 1

4.3 Maxima and Minima

Calculus Volume 14.3 Maxima and Minima
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 4.3.1. Define absolute extrema.
  • 4.3.2. Define local extrema.
  • 4.3.3. Explain how to find the critical points of a function over a closed interval.
  • 4.3.4. Describe how to use critical points to locate absolute extrema over a closed interval.

Given a particular function, we are often interested in determining the largest and smallest values of the function. This information is important in creating accurate graphs. Finding the maximum and minimum values of a function also has practical significance because we can use this method to solve optimization problems, such as maximizing profit, minimizing the amount of material used in manufacturing an aluminum can, or finding the maximum height a rocket can reach. In this section, we look at how to use derivatives to find the largest and smallest values for a function.

Absolute Extrema

Consider the function f(x)=x2+1f(x)=x2+1 over the interval (,).(,). As x±,x±, f(x).f(x). Therefore, the function does not have a largest value. However, since x2+11x2+11 for all real numbers xx and x2+1=1x2+1=1 when x=0,x=0, the function has a smallest value, 1, when x=0.x=0. We say that 1 is the absolute minimum of f(x)=x2+1f(x)=x2+1 and it occurs at x=0.x=0. We say that f(x)=x2+1f(x)=x2+1 does not have an absolute maximum (see the following figure).

The function f(x) = x2 + 1 is graphed, and its minimum of 1 is seen to be at x = 0.
Figure 4.12 The given function has an absolute minimum of 1 at x=0.x=0. The function does not have an absolute maximum.

Definition

Let ff be a function defined over an interval II and let cI.cI. We say ff has an absolute maximum on II at cc if f(c)f(x)f(c)f(x) for all xI.xI. We say ff has an absolute minimum on II at cc if f(c)f(x)f(c)f(x) for all xI.xI. If ff has an absolute maximum on II at cc or an absolute minimum on II at c,c, we say ff has an absolute extremum on II at c.c.

Before proceeding, let’s note two important issues regarding this definition. First, the term absolute here does not refer to absolute value. An absolute extremum may be positive, negative, or zero. Second, if a function ff has an absolute extremum over an interval II at c,c, the absolute extremum is f(c).f(c). The real number cc is a point in the domain at which the absolute extremum occurs. For example, consider the function f(x)=1/(x2+1)f(x)=1/(x2+1) over the interval (,).(,). Since

f(0)=11x2+1=f(x)f(0)=11x2+1=f(x)

for all real numbers x,x, we say ff has an absolute maximum over (,)(,) at x=0.x=0. The absolute maximum is f(0)=1.f(0)=1. It occurs at x=0,x=0, as shown in Figure 4.13(b).

A function may have both an absolute maximum and an absolute minimum, just one extremum, or neither. Figure 4.13 shows several functions and some of the different possibilities regarding absolute extrema. However, the following theorem, called the Extreme Value Theorem, guarantees that a continuous function ff over a closed, bounded interval [a,b][a,b] has both an absolute maximum and an absolute minimum.

This figure has six parts a, b, c, d, e, and f. In figure a, the line f(x) = x3 is shown, and it is noted that it has no absolute minimum and no absolute maximum. In figure b, the line f(x) = 1/(x2 + 1) is shown, which is near 0 for most of its length and rises to a bump at (0, 1); it has no absolute minimum, but does have an absolute maximum of 1 at x = 0. In figure c, the line f(x) = cos x is shown, which has absolute minimums of −1 at ±π, ±3π, … and absolute maximums of 1 at 0, ±2π, ±4π, …. In figure d, the piecewise function f(x) = 2 – x2 for 0 ≤ x < 2 and x – 3 for 2 ≤ x ≤ 4 is shown, with absolute maximum of 2 at x = 0 and no absolute minimum. In figure e, the function f(x) = (x – 2)2 is shown on [1, 4], which has absolute maximum of 4 at x = 4 and absolute minimum of 0 at x = 2. In figure f, the function f(x) = x/(2 − x) is shown on [0, 2), with absolute minimum of 0 at x = 0 and no absolute maximum.
Figure 4.13 Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of (,).(,). Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a bounded interval.
Theorem 4.1

Extreme Value Theorem

If ff is a continuous function over the closed, bounded interval [a,b],[a,b], then there is a point in [a,b][a,b] at which ff has an absolute maximum over [a,b][a,b] and there is a point in [a,b][a,b] at which ff has an absolute minimum over [a,b].[a,b].

The proof of the extreme value theorem is beyond the scope of this text. Typically, it is proved in a course on real analysis. There are a couple of key points to note about the statement of this theorem. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. If the interval II is open or the function has even one point of discontinuity, the function may not have an absolute maximum or absolute minimum over I.I. For example, consider the functions shown in Figure 4.13(d), (e), and (f). All three of these functions are defined over bounded intervals. However, the function in graph (e) is the only one that has both an absolute maximum and an absolute minimum over its domain. The extreme value theorem cannot be applied to the functions in graphs (d) and (f) because neither of these functions is continuous over a closed, bounded interval. Although the function in graph (d) is defined over the closed interval [0,4],[0,4], the function is discontinuous at x=2.x=2. The function has an absolute maximum over [0,4][0,4] but does not have an absolute minimum. The function in graph (f) is continuous over the half-open interval [0,2),[0,2), but is not defined at x=2,x=2, and therefore is not continuous over a closed, bounded interval. The function has an absolute minimum over [0,2),[0,2), but does not have an absolute maximum over [0,2).[0,2). These two graphs illustrate why a function over a bounded interval may fail to have an absolute maximum and/or absolute minimum.

Before looking at how to find absolute extrema, let’s examine the related concept of local extrema. This idea is useful in determining where absolute extrema occur.

Local Extrema and Critical Points

Consider the function ff shown in Figure 4.14. The graph can be described as two mountains with a valley in the middle. The absolute maximum value of the function occurs at the higher peak, at x=2.x=2. However, x=0x=0 is also a point of interest. Although f(0)f(0) is not the largest value of f,f, the value f(0)f(0) is larger than f(x)f(x) for all xx near 0. We say ff has a local maximum at x=0.x=0. Similarly, the function ff does not have an absolute minimum, but it does have a local minimum at x=1x=1 because f(1)f(1) is less than f(x)f(x) for xx near 1.

The function f(x) is shown, which curves upward from quadrant III, slows down in quadrant II, achieves a local maximum on the y-axis, decreases to achieve a local minimum in quadrant I at x = 1, increases to a local maximum at x = 2 that is greater than the other local maximum, and then decreases rapidly through quadrant IV.
Figure 4.14 This function ff has two local maxima and one local minimum. The local maximum at x=2x=2 is also the absolute maximum.

Definition

A function ff has a local maximum at cc if there exists an open interval II containing cc such that II is contained in the domain of ff and f(c)f(x)f(c)f(x) for all xI.xI. A function ff has a local minimum at cc if there exists an open interval II containing cc such that II is contained in the domain of ff and f(c)f(x)f(c)f(x) for all xI.xI. A function ff has a local extremum at cc if ff has a local maximum at cc or ff has a local minimum at c.c.

Note that if ff has an absolute extremum at cc and ff is defined over an interval containing c,c, then f(c)f(c) is also considered a local extremum. If an absolute extremum for a function ff occurs at an endpoint, we do not consider that to be a local extremum, but instead refer to that as an endpoint extremum.

Given the graph of a function f,f, it is sometimes easy to see where a local maximum or local minimum occurs. However, it is not always easy to see, since the interesting features on the graph of a function may not be visible because they occur at a very small scale. Also, we may not have a graph of the function. In these cases, how can we use a formula for a function to determine where these extrema occur?

To answer this question, let’s look at Figure 4.14 again. The local extrema occur at x=0,x=0, x=1,x=1, and x=2.x=2. Notice that at x=0x=0 and x=1,x=1, the derivative f(x)=0.f(x)=0. At x=2,x=2, the derivative f(x)f(x) does not exist, since the function ff has a corner there. In fact, if ff has a local extremum at a point x=c,x=c, the derivative f(c)f(c) must satisfy one of the following conditions: either f(c)=0f(c)=0 or f(c)f(c) is undefined. Such a value cc is known as a critical point and it is important in finding extreme values for functions.

Definition

Let cc be an interior point in the domain of f.f. We say that cc is a critical point of ff if f(c)=0f(c)=0 or f(c)f(c) is undefined.

As mentioned earlier, if ff has a local extremum at a point x=c,x=c, then cc must be a critical point of f.f. This fact is known as Fermat’s theorem.

Theorem 4.2

Fermat’s Theorem

If ff has a local extremum at cc and ff is differentiable at c,c, then f(c)=0.f(c)=0.

Proof

Suppose ff has a local extremum at cc and ff is differentiable at c.c. We need to show that f(c)=0.f(c)=0. To do this, we will show that f(c)0f(c)0 and f(c)0,f(c)0, and therefore f(c)=0.f(c)=0. Since ff has a local extremum at c,c, ff has a local maximum or local minimum at c.c. Suppose ff has a local maximum at c.c. The case in which ff has a local minimum at cc can be handled similarly. There then exists an open interval II such that f(c)f(x)f(c)f(x) for all xI.xI. Since ff is differentiable at c,c, from the definition of the derivative, we know that

f(c)=limxcf(x)f(c)xc.f(c)=limxcf(x)f(c)xc.

Since this limit exists, both one-sided limits also exist and equal f(c).f(c). Therefore,

f(c)=limxc+f(x)f(c)xc,f(c)=limxc+f(x)f(c)xc,
4.4

and

f(c)=limxcf(x)f(c)xc.f(c)=limxcf(x)f(c)xc.
4.5

Since f(c)f(c) is a local maximum, we see that f(x)f(c)0f(x)f(c)0 for xx near c.c. Therefore, for xx near c,c, but x>c,x>c, we have f(x)f(c)xc0.f(x)f(c)xc0. From Equation 4.4 we conclude that f(c)0.f(c)0. Similarly, it can be shown that f(c)0.f(c)0. Therefore, f(c)=0.f(c)=0.

From Fermat’s theorem, we conclude that if ff has a local extremum at c,c, then either f(c)=0f(c)=0 or f(c)f(c) is undefined. In other words, local extrema can only occur at critical points.

Note this theorem does not claim that a function ff must have a local extremum at a critical point. Rather, it states that critical points are candidates for local extrema. For example, consider the function f(x)=x3.f(x)=x3. We have f(x)=3x2=0f(x)=3x2=0 when x=0.x=0. Therefore, x=0x=0 is a critical point. However, f(x)=x3f(x)=x3 is increasing over (,),(,), and thus ff does not have a local extremum at x=0.x=0. In Figure 4.15, we see several different possibilities for critical points. In some of these cases, the functions have local extrema at critical points, whereas in other cases the functions do not. Note that these graphs do not show all possibilities for the behavior of a function at a critical point.

This figure has five parts a, b, c, d, and e. In figure a, a parabola is shown facing down in quadrant I; there is a horizontal tangent line at the local maximum marked f’(c) = 0. In figure b, there is a function drawn with an asymptote at c, meaning that the function increases toward infinity on both sides of c; it is noted that f’(c) is undefined. In figure c, a version of the absolute value graph is shown that has been shifted so that its minimum is in quadrant I with x = c. It is noted that f’(c) is undefined. In figure d, a version of the function f(x) = x3 is shown that has been shifted so that its inflection point is in quadrant I with x = c. Its inflection point at (c, f(c)) has a horizontal line through it, and it is noted that f’(c) = 0. In figure e, a version of the function f(x) = x1/3 is shown that has been shifted so that its inflection point is in quadrant I with x = c. Its inflection point at (c, f(c)) has a vertical line through it, and it is noted that f’(c) is undefined.
Figure 4.15 (a–e) A function ff has a critical point at cc if f(c)=0f(c)=0 or f(c)f(c) is undefined. A function may or may not have a local extremum at a critical point.

Later in this chapter we look at analytical methods for determining whether a function actually has a local extremum at a critical point. For now, let’s turn our attention to finding critical points. We will use graphical observations to determine whether a critical point is associated with a local extremum.

Example 4.12

Locating Critical Points

For each of the following functions, find all critical points. Use a graphing utility to determine whether the function has a local extremum at each of the critical points.

  1. f(x)=13x352x2+4xf(x)=13x352x2+4x
  2. f(x)=(x21)3f(x)=(x21)3
  3. f(x)=4x1+x2f(x)=4x1+x2

Solution

  1. The derivative f(x)=x25x+4f(x)=x25x+4 is defined for all real numbers x.x. Therefore, we only need to find the values for xx where f(x)=0.f(x)=0. Since f(x)=x25x+4=(x4)(x1),f(x)=x25x+4=(x4)(x1), the critical points are x=1x=1 and x=4.x=4. From the graph of ff in Figure 4.16, we see that ff has a local maximum at x=1x=1 and a local minimum at x=4.x=4.
    The function f(x) = (1/3) x3 – (5/2) x2 + 4x is graphed. The function has local maximum at x = 1 and local minimum at x = 4.
    Figure 4.16 This function has a local maximum and a local minimum.
  2. Using the chain rule, we see the derivative is
    f(x)=3(x21)2(2x)=6x(x21)2.f(x)=3(x21)2(2x)=6x(x21)2.

    Therefore, ff has critical points when x=0x=0 and when x21=0.x21=0. We conclude that the critical points are x=0,±1.x=0,±1. From the graph of ff in Figure 4.17, we see that ff has a local (and absolute) minimum at x=0,x=0, but does not have a local extremum at x=1x=1 or x=−1.x=−1.
    The function f(x) = (x2 − 1)3 is graphed. The function has local minimum at x = 0, and inflection points at x = ±1.
    Figure 4.17 This function has three critical points: x=0,x=0, x=1,x=1, and x=−1.x=−1. The function has a local (and absolute) minimum at x=0,x=0, but does not have extrema at the other two critical points.
  3. By the chain rule, we see that the derivative is
    f(x)=(1+x24)4x(2x)(1+x2)2=44x2(1+x2)2.f(x)=(1+x24)4x(2x)(1+x2)2=44x2(1+x2)2.

    The derivative is defined everywhere. Therefore, we only need to find values for xx where f(x)=0.f(x)=0. Solving f(x)=0,f(x)=0, we see that 44x2=0,44x2=0, which implies x=±1.x=±1. Therefore, the critical points are x=±1.x=±1. From the graph of ff in Figure 4.18, we see that ff has an absolute maximum at x=1x=1 and an absolute minimum at x=−1.x=−1. Hence, ff has a local maximum at x=1x=1 and a local minimum at x=−1.x=−1. (Note that if ff has an absolute extremum over an interval II at a point cc that is not an endpoint of I,I, then ff has a local extremum at c.)c.)
    The function f(x) = 4x/(1 + x2) is graphed. The function has local/absolute maximum at x = 1 and local/absolute minimum at x = −1.
    Figure 4.18 This function has an absolute maximum and an absolute minimum.

Checkpoint 4.12

Find all critical points for f(x)=x312x22x+1.f(x)=x312x22x+1.

Locating Absolute Extrema

The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. As shown in Figure 4.13, one or both of these absolute extrema could occur at an endpoint. If an absolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absolute extremum is a local extremum. Therefore, by Fermat’s Theorem, the point cc at which the local extremum occurs must be a critical point. We summarize this result in the following theorem.

Theorem 4.3

Location of Absolute Extrema

Let ff be a continuous function over a closed, bounded interval I.I. The absolute maximum of ff over II and the absolute minimum of ff over II must occur at endpoints of II or at critical points of ff in I.I.

With this idea in mind, let’s examine a procedure for locating absolute extrema.

Problem-Solving Strategy: Locating Absolute Extrema over a Closed Interval

Consider a continuous function ff defined over the closed interval [a,b].[a,b].

  1. Evaluate ff at the endpoints x=ax=a and x=b.x=b.
  2. Find all critical points of ff that lie over the interval (a,b)(a,b) and evaluate ff at those critical points.
  3. Compare all values found in (1) and (2). From Location of Absolute Extrema, the absolute extrema must occur at endpoints or critical points. Therefore, the largest of these values is the absolute maximum of f.f. The smallest of these values is the absolute minimum of f.f.

Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions.

Example 4.13

Locating Absolute Extrema

For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur.

  1. f(x)=x2+3x2f(x)=x2+3x2 over [1,3].[1,3].
  2. f(x)=x23x2/3f(x)=x23x2/3 over [0,2].[0,2].

Solution

  1. Step 1. Evaluate ff at the endpoints x=1x=1 and x=3.x=3.
    f(1)=0andf(3)=−2f(1)=0andf(3)=−2

    Step 2. Since f(x)=−2x+3,f(x)=−2x+3, ff is defined for all real numbers x.x. Therefore, there are no critical points where the derivative is undefined. It remains to check where f(x)=0.f(x)=0. Since f(x)=−2x+3=0f(x)=−2x+3=0 at x=32x=32 and 3232 is in the interval [1,3],[1,3], f(32)f(32) is a candidate for an absolute extremum of ff over [1,3].[1,3]. We evaluate f(32)f(32) and find
    f(32)=14.f(32)=14.

    Step 3. We set up the following table to compare the values found in steps 1 and 2.
    xx f(x)f(x) Conclusion
    00 00
    3232 1414 Absolute maximum
    33 −2−2 Absolute minimum

    From the table, we find that the absolute maximum of ff over the interval [1, 3] is 14,14, and it occurs at x=32.x=32. The absolute minimum of ff over the interval [1, 3] is −2,−2, and it occurs at x=3x=3 as shown in the following graph.
    The function f(x) = – x2 + 3x – 2 is graphed from (1, 0) to (3, −2), with its maximum marked at (3/2, 1/4).
    Figure 4.19 This function has both an absolute maximum and an absolute minimum.
  2. Step 1. Evaluate ff at the endpoints x=0x=0 and x=2.x=2.
    f(0)=0andf(2)=43430.762f(0)=0andf(2)=43430.762

    Step 2. The derivative of ff is given by
    f(x)=2x2x1/3=2x4/32x1/3f(x)=2x2x1/3=2x4/32x1/3

    for x0.x0. The derivative is zero when 2x4/32=0,2x4/32=0, which implies x=±1.x=±1. The derivative is undefined at x=0.x=0. Therefore, the critical points of ff are x=0,1,−1.x=0,1,−1. The point x=0x=0 is an endpoint, so we already evaluated f(0)f(0) in step 1. The point x=−1x=−1 is not in the interval of interest, so we need only evaluate f(1).f(1). We find that
    f(1)=−2.f(1)=−2.

    Step 3. We compare the values found in steps 1 and 2, in the following table.
    xx f(x)f(x) Conclusion
    00 00 Absolute maximum
    11 −2−2 Absolute minimum
    22 −0.762−0.762

    We conclude that the absolute maximum of ff over the interval [0, 2] is zero, and it occurs at x=0.x=0. The absolute minimum is −2, and it occurs at x=1x=1 as shown in the following graph.
    The function f(x) = x2 – 3x2/3 is graphed from (0, 0) to (2, −0.762), with its minimum marked at (1, −2).
    Figure 4.20 This function has an absolute maximum at an endpoint of the interval.
Checkpoint 4.13

Find the absolute maximum and absolute minimum of f(x)=x24x+3f(x)=x24x+3 over the interval [1,4].[1,4].

At this point, we know how to locate absolute extrema for continuous functions over closed intervals. We have also defined local extrema and determined that if a function ff has a local extremum at a point c,c, then cc must be a critical point of f.f. However, cc being a critical point is not a sufficient condition for ff to have a local extremum at c.c. Later in this chapter, we show how to determine whether a function actually has a local extremum at a critical point. First, however, we need to introduce the Mean Value Theorem, which will help as we analyze the behavior of the graph of a function.

Section 4.3 Exercises

90.

In precalculus, you learned a formula for the position of the maximum or minimum of a quadratic equation y=ax2+bx+c,y=ax2+bx+c, which was m=b(2a).m=b(2a). Prove this formula using calculus.

91.

If you are finding an absolute minimum over an interval [a,b],[a,b], why do you need to check the endpoints? Draw a graph that supports your hypothesis.

92.

If you are examining a function over an interval (a,b),(a,b), for aa and bb finite, is it possible not to have an absolute maximum or absolute minimum?

93.

When you are checking for critical points, explain why you also need to determine points where f'(x)f'(x) is undefined. Draw a graph to support your explanation.

94.

Can you have a finite absolute maximum for y=ax2+bx+cy=ax2+bx+c over (,)?(,)? Explain why or why not using graphical arguments.

95.

Can you have a finite absolute maximum for y=ax3+bx2+cx+dy=ax3+bx2+cx+d over (,)(,) assuming a is non-zero? Explain why or why not using graphical arguments.

96.

Let mm be the number of local minima and MM be the number of local maxima. Can you create a function where M>m+2?M>m+2? Draw a graph to support your explanation.

97.

Is it possible to have more than one absolute maximum? Use a graphical argument to prove your hypothesis.

98.

Is it possible to have no absolute minimum or maximum for a function? If so, construct such a function. If not, explain why this is not possible.

99.

[T] Graph the function y=eax.y=eax. For which values of a,a, on any infinite domain, will you have an absolute minimum and absolute maximum?

For the following exercises, determine where the local and absolute maxima and minima occur on the graph given. Assume the graph represents the entirety of each function.

100.
The function graphed starts at (−4, 60), decreases rapidly to (−3, −40), increases to (−1, 10) before decreasing slowly to (2, 0), at which point it increases rapidly to (3, 30).
101.
The function graphed starts at (−2.2, 10), decreases rapidly to (−2, −11), increases to (−1, 5) before decreasing slowly to (1, 3), at which point it increases to (2, 7), and then decreases to (3, −20).
102.
The function graphed starts at (−3, −1), increases rapidly to (−2, 0.7), decreases to (−1, −0.25) before decreasing slowly to (1, 0.25), at which point it decreases to (2, 0.7), and then increases to (3, 1).
103.
The function graphed starts at (−2.5, 1), decreases rapidly to (−2, −1.25), increases to (−1, 0.25) before decreasing slowly to (0, 0.2), at which point it increases slowly to (1, 0.25), then decreases rapidly to (2, −1.25), and finally increases to (2.5, 1).

For the following problems, draw graphs of f(x),f(x), which is continuous, over the interval [−4,4][−4,4] with the following properties:

104.

Absolute maximum at x=2x=2 and absolute minima at x=±3x=±3

105.

Absolute minimum at x=1x=1 and absolute maximum at x=2x=2

106.

Absolute maximum at x=4,x=4, absolute minimum at x=−1,x=−1, local maximum at x=−2,x=−2, and a critical point that is not a maximum or minimum at x=2x=2

107.

Absolute maxima at x=2x=2 and x=−3,x=−3, local minimum at x=1,x=1, and absolute minimum at x=4x=4

For the following exercises, find the critical points in the domains of the following functions.

108.

y=4x33xy=4x33x

109.

y=4xx2y=4xx2

110.

y=1x1y=1x1

111.

y=ln(x2)y=ln(x2)

112.

y=tan(x)y=tan(x)

113.

y=4x2y=4x2

114.

y=x3/23x5/2y=x3/23x5/2

115.

y=x21x2+2x3y=x21x2+2x3

116.

y=sin2(x)y=sin2(x)

117.

y=x+1xy=x+1x

For the following exercises, find the local and/or absolute maxima for the functions over the specified domain.

118.

f(x)=x2+3f(x)=x2+3 over [−1,4][−1,4]

119.

y=x2+2xy=x2+2x over [1,4][1,4]

120.

y=(xx2)2y=(xx2)2 over [−1,1][−1,1]

121.

y=1(xx2)y=1(xx2) over (0,1)(0,1)

122.

y=9xy=9x over [1,9][1,9]

123.

y=x+sin(x)y=x+sin(x) over [0,2π][0,2π]

124.

y=x1+xy=x1+x over [0,100][0,100]

125.

y=|x+1|+|x1|y=|x+1|+|x1| over [−3,2][−3,2]

126.

y=xx3y=xx3 over [0,4][0,4]

127.

y=sinx+cosxy=sinx+cosx over [0,2π][0,2π]

128.

y=4sinθ3cosθy=4sinθ3cosθ over [0,2π][0,2π]

For the following exercises, find the local and absolute minima and maxima for the functions over (,).(,).

129.

y=x2+4x+5y=x2+4x+5

130.

y=x312xy=x312x

131.

y=3x4+8x318x2y=3x4+8x318x2

132.

y=x3(1x)6y=x3(1x)6

133.

y=x2+x+6x1y=x2+x+6x1

134.

y=x21x1y=x21x1

For the following functions, use a calculator to graph the function and to estimate the absolute and local maxima and minima. Then, solve for them explicitly.

135.

[T] y=3x1x2y=3x1x2

136.

[T] y=x+sin(x)y=x+sin(x)

137.

[T] y=12x5+45x4+20x390x2120x+3y=12x5+45x4+20x390x2120x+3

138.

[T] y=x3+6x2x30x2y=x3+6x2x30x2

139.

[T] y=4x24+x2y=4x24+x2

140.

A company that produces cell phones has a cost function of C=x21200x+36,400,C=x21200x+36,400, where CC is cost in dollars and xx is number of cell phones produced (in thousands). How many units of cell phone (in thousands) minimizes this cost function?

141.

A ball is thrown into the air and its position is given by h(t)=−4.9t2+60t+5m.h(t)=−4.9t2+60t+5m. Find the height at which the ball stops ascending. How long after it is thrown does this happen?

For the following exercises, consider the production of gold during the California gold rush (1848–1888). The production of gold can be modeled by G(t)=(25t)(t2+16),G(t)=(25t)(t2+16), where tt is the number of years since the rush began (0t40)(0t40) and GG is ounces of gold produced (in millions). A summary of the data is shown in the following figure.

The bar graph shows gold (in millions of troy ounces) per year, starting in 1848 and ending in 1888. In 1848, the bar graph shows 0.05; in 1849, 0.5; in 1850, 2; in 1851, 3.6; in 1852, 3.9; in 1853, 3.3; in 1854, 3.4; in 1855, 2.6; in 1856, 2.75; in 1857, 2.1; in 1858, 2.2; in 1859, 2.15; in 1860, 2.1; in 1861, 2; in 1862, 1.8; in 1863, 1.1; in 1864, 1.15; in 1865, 0.9; in 1866, 0.85; in 1867, 0.9; in 1868, 0.85; in 1869, 0.9; in 1870, 0.85; in 1871, 0.85; in 1872, 0.75; in 1873, 0.7; in 1874, 0.8; in 1875, 0.75; in 1876, 0.7; in 1877, 0.73; in 1878, 0.9; in 1879, 0.95; in 1880, 1; in 1881, 0.95; in 1882, 0.85; in 1883, 1.1; in 1884, 0.6; in 1885, 0.55; in 1886, 0.65; in 1887, 0.6; and in 1888, 0.55.
142.

Find when the maximum (local and global) gold production occurred, and the amount of gold produced during that maximum.

143.

Find when the minimum (local and global) gold production occurred. What was the amount of gold produced during this minimum?

Find the critical points, maxima, and minima for the following piecewise functions.

144.

y={x24x0x1x241<x2y={x24x0x1x241<x2

145.

y={x2+1x1x24x+5x>1y={x2+1x1x24x+5x>1

For the following exercises, find the critical points of the following generic functions. Are they maxima, minima, or neither? State the necessary conditions.

146.

y=ax2+bx+c,y=ax2+bx+c, given that a>0a>0

147.

y=(x1)a,y=(x1)a, given that a>1a>1

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