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Coordinated Calculus

Nathan Wakefield, Christine Kelley, Marla Williams, Michelle Haver, Lawrence Seminario-Romero, Robert Huben, Aurora Marks, Stephanie Prahl, Based upon Active Calculus by Matthew Boelkins

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Section 3.2 Global Optimization

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Supplemental Videos.

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The main topics of this section are also presented in the following videos:
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We have seen that we can use the first derivative of a function to determine where the function is increasing or decreasing, and the second derivative to know where the function is concave up or concave down. This information helps us determine the overall shape and behavior of the graph as well as whether the function has relative extrema.
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Remember the difference between a relative maximum and a global maximum: there is a relative maximum of f at x=p if f(p)≥f(x) for all x near p, while there is a global maximum at p if f(p)≥f(x) for all x in the domain of f.
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For instance, in Figure 3.18 at the right, we see a function f that has a global maximum at x=c and a relative maximum at x=a, since f(c) is greater than or equal to f(x) for every value of x, while f(a) is only greater than or equal to the value of f(x) for x near a. Since the function appears to decrease without bound, f has no global minimum, although there is a relative minimum at x=b.
Figure 3.18. A function f with a global maximum, but no global minimum.
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When determining global extrema, the long-run behavior of the function, that is, the behavior of the function as x goes to ∞ or −∞, must be considered. In many instances, the long run behavior can be determined using basic limit properties. For now, if basic limit properties are not sufficient, determine the long-run behavior by looking at a graph of the function using any online graphing calculator. There are more tools for determining limiting behavior of functions that will be considered in a later section; then visualizing a graph won’t be necessary.
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Our emphasis in this section is on finding the global extreme values of a function (if they exist), either over its entire domain or on some restricted portion.

Example 3.19.

Let f(x)=2+31+(x+1)2.
  1. Determine all of the critical numbers of f.
  2. Construct a first derivative sign chart for f and thus determine all intervals on which f is increasing or decreasing.
  3. Does f have a global maximum? If so, why, and what is its value and where is the maximum attained? If not, explain why.
  4. Determine limx→∞f(x) and limx→−∞f(x).
  5. Explain why f(x)>2 for every value of x.
  6. Does f have a global minimum? If so, why, and what is its value and where is the minimum attained? If not, explain why.
Hint.
  1. What is true about f′(c) when c is a critical number of f?
  2. Remember that on a given interval of the first derivative sign chart for f, the derivative f′ should have the same sign everywhere, so you only need to test at a single point on the interval.
  3. What does the chart from (b) tell you? Thinking about the algebraic structure of f(x) can also support your reasoning.
  4. What happens to 1+(x+1)2 for large values of x? What happens to its reciprocal?
  5. 1+(x+1)2>0 for every value of x.
  6. Think about the long-run behavior of f that you found in (d).
Answer.
  1. −1 is the only critical number of f.
  2. f is increasing on (−∞,−1) and decreasing on (−1,∞).
  3. f has a global maximum of 5 at the point (−1,5).
  4. limx→±∞f(x)=2.
  5. 31+(x+1)2>0 for every value of x.
  6. f has a horizontal asymptote at y=2 but no global minimum.
Solution.
  1. We first notice that the domain of f is all real numbers, since the denominator of f(x) is always at least 1:
    (x+1)2≥(01+(x+1)2≥(1.
    To differentiate f, we will first rewrite f(x) as
    f(x)=2+3(1+(x+1)2)−1
    and then use the chain rule to find
    f′(x)=(3((−1)(1+(x+1)2)−2(2(x+1)(1)))=(−6(x+1)(1+(x+1)2)2.
    The denominator of f′(x) is always at least 1 for the same reason that the denominator of f(x) is always at least 1, so there are no points x=c for which f′(c) does not exist. Thus the only critical number(s) of f will occur when f′(c)=0. This happens when
    −6(x+1)=(0x+1=(0x=(−1.
    Thus the only critical number of f is c=−1, with f′(−1)=0.
  2. As stated in (a), the denominator of f′(x) is always at least 1, so in particular, it is always positive. On our first derivative sign chart, the only critical number we have is c=−1, which splits the domain of f and f′ into the intervals (−∞,−1) and (−1,∞). Thinking of the factors of f′(x) as −6, (x+1), and 1(1+(x+1)2)2, the signs of these at x=−2 are “−−+”, with the final product positive. Likewise, the signs of these factors at x=0 are “−++”, making the final product negative. Therefore f′(x)>0 on the interval x<−1, and f′(x)<0 on the interval x>−1. Hence f is increasing on (−∞,−1) and decreasing on (−1,∞).
  3. According to the first derivative test, the point (−1,f(−1)) is a relative maximum of f. Since this is the only critical point of the function and f is continuous everywhere, there is nowhere else that f can change direction. It follows that this point is actually a global maximum.
    From an algebraic standpoint, we have already observed that 1+(x+1)2≥1 for every value of x. Consequently,
    0<31+(x+1)2≤3
    for every value of x, and this quotient achieves its maximum of 3 when x=−1. Hence
    2<2+31+(x+1)2≤5
    for every value of x. Since f(−1)=5, we conclude that the global maximum of f is 5 at x=−1.
  4. The symmetry of f about x=−1 allows us to compute both limits simultaneously. We first note that limx→±∞(1+(x+1)2)=∞ and limx→±∞11+(x+1)2=0. Thus
    limx→±∞f(x)=limx→±∞(2+31+(x+1)2)=2+0=2.
    This tells us the graph of y=f(x) has a horizontal asymptote of y=2.
  5. In (c), we stated the inequality
    2<2+31+(x+1)2≤5.
    Since the quotient in the formula for f(x) is always strictly greater than zero, adding 2 to the quotient to obtain f(x) will always yield a value strictly greater than 2.
  6. Since f(x)>2 for every value of x, a global minimum of f (if it existed) would necessarily be larger than 2. But we showed in (d) that limx→±∞f(x)=2, meaning that f gets arbitrarily close to 2 as x gets large. Moreover, f′ is always positive for x<−1 and f′ is always negative for x>−1, so a lowest point of f is never attained.
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Subsection 3.2.1 Global Optimization

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In both Figure 3.18 and Example 3.19 above, we were interested in finding the global minimum and global maximum for f on its entire domain. At other times, we might focus on some restriction of the domain.
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For example, rather than considering f(x)=2+31+(x+1)2 for every value of x, perhaps instead we are only interested in those x for which 0≤x≤4, and we would like to know which values of x in the interval [0,4] produce the largest possible and smallest possible values of f. We are accustomed to critical numbers playing a key role in determining the location of extreme values of a function; now, by restricting the domain to an interval it makes sense that the endpoints of the interval will also be important to consider, as we see in the following example. When limiting ourselves to a particular interval, we will often refer to the absolute maximum or minimum value, rather than the global maximum or minimum.

Example 3.20.

Let g(x)=13x3−2x+2.
  1. Find all critical numbers of g that lie in the interval −2≤x≤3.
  2. Use a graphing utility to construct the graph of g on the interval −2≤x≤3.
  3. From the graph, determine the x-values at which the absolute minimum and absolute maximum of g occur on the interval [−2,3].
  4. How do your answers change if we instead consider the interval −2≤x≤2?
  5. What if we instead consider the interval −2≤x≤1?
Hint.
  1. Check that each critical number you find satisfies −2≤x≤3.
  2. desmos.com
     12 
    desmos.com/calculator
    is a great choice.
  3. On the graph, look for the lowest and highest possible values of the function.
  4. Ask yourself the same questions as (a)-(c), simply using the new interval.
  5. Ask yourself the same questions as (a)-(c), simply using the new interval.
Answer.
  1. x=±2≈±1.414.
  2. Below is the graph of y=g(x) replicated three times, with critical points and endpoints of each interval marked. On the left, we have the interval [−2,3] for (c); the middle shows the interval [−2,2] for (d); the right displays the interval [−2,1] for (e).
  3. On [−2,3], g has a global maximum at x=3 and a global minimum at x=2.
  4. On [−2,2], g has a global maximum at x=−2 and a global minimum at x=2.
  5. On [−2,3], g has a global maximum at x=−2 and a global minimum at x=1.
Solution.
  1. Since g′(x)=x2−2, the critical numbers of g are x=±2≈±1.414, both of which lie in the interval −2≤x≤3.
  2. Below is the graph of y=g(x) replicated three times, with critical points and endpoints of each interval marked. On the left, we have the interval [−2,3] for (c); the middle shows the interval [−2,2] for (d); the right displays the interval [−2,1] for (e).
  3. On [−2,3], g has a global maximum at x=3 and a global minimum at x=2.
  4. On [−2,2], g has a global maximum at x=−2 and a global minimum at x=2.
  5. On [−2,3], g has a global maximum at x=−2 and a global minimum at x=1.
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Example 3.20 showed how the absolute maximum and absolute minimum of a function on a closed, bounded interval [a,b] depend not only on the critical numbers of the function, but also on the values of a and b. In fact, the interval we choose has nearly the same influence on extreme values as the function under consideration. For instance, Figure 3.21 below depicts again the graph of y=g(x) that we looked at in Example 3.20.
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Figure 3.21. A function g considered on three different intervals.
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From left to right, the interval under consideration is changed from [−2,3] to [−2,2] to [−2,1].
  • There are two critical numbers on the interval [−2,3]; the absolute minimum is at one critical number and the absolute maximum is at the right endpoint.
  • On the interval [−2,2], both critical numbers are again in the interval. One critical value is still the absolute minimum; however, the absolute maximum is now at the other critical number rather than at the right endpoint.
  • Only one critical number lies on the interval [−2,1]. That critical number gives the absolute maximum; the absolute minimum is at the right endpoint.
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These observations demonstrate several important facts that hold more generally.
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The Extreme Value Theorem.

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If f is a continuous function on a closed and bounded interval [a,b], then f attains both an absolute minimum and absolute maximum on [a,b]. That is, there is some value xm in [a,b] such that f(xm)≤f(x) for all x in [a,b]. Similarly, there is a value xM in [a,b] such that f(xM)≥f(x) for all x in [a,b]. Letting m=f(xm) and M=f(xM), it follows that m≤f(x)≤M for all x in [a,b].
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The Extreme Value Theorem tells us that on any closed and bounded interval [a,b], a continuous function has to achieve both an absolute minimum and an absolute maximum. The theorem does not tell us where these extreme values occur, only that they must exist. As we saw in Example 3.20, the only possible locations for relative extrema are at the endpoints of the interval or at a critical number. It is important to consider only the critical numbers that lie within the interval.
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We now have the following approach to finding the absolute maximum and minimum of a continuous function f on the interval [a,b]:
  1. Find all critical numbers of f that lie in the interval.
  2. Evaluate the function f at each critical number in the interval and at each endpoint of the interval.
  3. From among these function values, the smallest is the absolute minimum of f on the interval, while the largest is the absolute maximum.

Example 3.22.

Find the exact absolute maximum and minimum of each function on the stated interval.
  1. h(x)=xe−x on [0,3]
  2. p(t)=sin⁡(t)+cos⁡(t) on [−π2,π2]
  3. q(x)=x2x−2 on [3,7]
  4. f(x)=4−e−(x−2)2 on (−∞,∞)
  5. h(x)=xe−ax on [0,2a]; assume a>0
  6. f(x)=b−e−(x−a)2 on (−∞,∞); assume a,b>0
Hint.
  1. After computing h′(x), factor to write the derivative as a product.
  2. The sine and cosine functions share the same value at π4±kπ for any integer k.
  3. Upon finding q′(x), factor its numerator.
  4. Remember that e−(x−2)2 is never zero. Also, notice that (−∞,∞) is not a bounded interval.
  5. After differentiating, remove a factor of e−ax.
  6. Compare to part (d).
Answer.
  1. Absolute maximum: e−1; absolute minimum: 0.
  2. Absolute maximum: 2; absolute minimum: −1.
  3. Absolute maximum: 9.8; absolute minimum: 8.
  4. Absolute minimum: 3; no absolute maximum.
  5. Absolute maximum: (ae)−1; absolute minimum: 0.
  6. Absolute minimum: b−1; no absolute maximum.
Solution.
  1. For h(x)=xe−x, we know that h′(x)=e−x+xe−x(−1)=e−x(1−x). Therefore the only critical number of h is x=1. Next, we compute h(1), h(0), and h(3). Observe that
    • h(0)=0
    • h(1)=e−1≈0.36788
    • h(3)=3e−3≈0.14936
    Thus, on [0,3], the absolute maximum of h is e−1 and the absolute minimum is 0.
  2. Given p(t)=sin⁡(t)+cos⁡(t), it follows that p′(t)=cos⁡(t)−sin⁡(t), so p′(t)=0 implies that cos⁡(t)=sin⁡(t). The sine and cosine functions have the same value at π4±kπ for any integer k. The only time this occurs in [−π2,π2] is at x=π4, thus this is the only critical number of p in the given interval. Now
    • p(−π2)=sin⁡(−π2)+cos⁡(−π2)=−1+0=−1
    • p(π4)=sin⁡(π4)+cos⁡(π4)=22+22=2≈1.41421
    • p(π2)=sin⁡(π2)+cos⁡(π2)=1+0=1
    Therefore, on [−π2,π2], the absolute maximum of p is 2 and the absolute minimum is −1.
  3. With q(x)=x2x−2, we have
    q′(x)=2x(x−2)−x2(x−2)2=x2−4x(x−2)2=x(x−4)(x−2)2.
    Hence the critical numbers of q are x=0 and x=4. Only the latter critical number lies in the interval [3,7], and thus we evaluate q and find
    • q(3)=91=9
    • q(4)=162=8
    • q(7)=495=9.8
    We now see that on [3,7] the absolute maximum of q is 9.8 and the absolute minimum is 8.
  4. We first observe that we are working on the domain of all real numbers rather than on a closed bounded interval. Hence the extreme value theorem does not apply and we need to think about the overall behavior of the function. First, since f(x)=4−e−(x−2)2, we see by the chain rule that f′(x)=−e−(x−2)2(−2(x−2))=2(x−2)e−(x−2)2. Since e−(x−2)2 is always positive (and in particular is never zero), it follows that the only critical number of f is x=2. Furthermore, with f′(x)=2(x−2)e−(x−2)2, we see that f′(x)<0 for x<2, and f′(x)>0 for x>2. Thus f is decreasing for x<2 and increasing for x>2. The first derivative test then tells us that f has an absolute minimum at x=2 and f does not have an absolute maximum.
  5. This will be similar to part (a). We start by differentiating h, finding h′(x)=e−ax(1−ax). Thus the only critical number of h is 1a, which is on the interval [0,2a]. Evaluating h at the interval endpoints and the critical number, we find
    • h(0)=0
    • h(1a)=1ae−1≈0.368a>0
    • h(2a)=2ae−2≈0.271a>0
    Thus on [0,2a], the absolute maximum of h is 1ae−1=(ae)−1 and the absolute minimum is 0.
  6. This is similar to part (d), including the fact that we are looking at all real numbers and not a closed bounded domain. We first find that f′(x)=2(x−a)e−(x−a)2 and the only critical number of f is x=a. Since f′(x)<0 for x<a and f′(x)>0 for x>a, the first derivative test tells us that f has a minimum at x=a but has no maximum. Finally, the value of the absolute minimum is f(a)=b−1.
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Subsection 3.2.2 Moving Toward Applications

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We conclude this section with an example of an applied optimization problem. It highlights the role that a closed, bounded domain can play in finding absolute extrema.

Example 3.23.

A 20 cm piece of wire is cut into two pieces. One piece is used to form a square and the other to form an equilateral triangle. How should the wire be cut to maximize the total area enclosed by the square and triangle? To minimize the area?
Hint.
Drawing and labeling a picture to model the situation may be useful.
Answer.
The area is maximized when the wire is uncut and all used for the square; the area is minimized when approximately 11.3 cm of wire are used for the triangle and approximately 8.7 cm are used for the square.
Solution.
We begin by sketching a picture that illustrates the situation. The variable in the problem is where we decide to cut the wire, so we label the cut point at a distance x cm from one end of the wire and note that the remaining portion of the wire then has length 20−x cm.
Figure 3.24 below shows how the wire is used to form the two regular polygons. From the x cm segment of wire, we obtain an equilateral triangle with three sides of length x3; the square shaped from the remaining 20−x cm of wire will have four sides of length 20−x4.
Figure 3.24. A 20 cm piece of wire cut into two pieces, one of which forms an equilateral triangle while the other yields a square.
At this point, we note that there are obvious restrictions on x: in particular, 0≤x≤20. In the extreme cases, all of the wire is being used to make just one figure. For instance, if x=0 then all 20 cm of wire are used to make a square that is 5×5.
Now, our overall goal is to find the minimum and maximum areas that can be enclosed. Because the height h of an equilateral triangle is 3 times half the length of the base b, the area of the triangle is
AΔ=12bh=12⋅x3⋅x36=3x236.
The area of the square is A◻=(20−x4)2. Therefore, the total area function is
A(x)=3x236+(20−x4)2.
Remember that we are considering this function only on the restricted domain [0,20].
Differentiating A(x), we have
A′(x)=3x18+2(20−x4)(−14)=318x+18x−52.
When we set A′(x)=0, we find that x=18043+9≈11.3007 is the only critical number of A in the interval [0,20].
Evaluating A at the critical number and endpoints, we see that
  • A(0)=25 cm2
  • A(18043+9)=3(18043+9)24+(20−18043+94)2≈10.8741 cm2
  • A(20)=336(400)=10093≈19.2450 cm2
Thus, the absolute minimum occurs when x≈11.3007 cm and results in the minimum area of approximately 10.8741 square centimeters. The absolute maximum occurs when we invest all of the wire in the square (and none in the triangle), resulting in 25 square centimeters of area. These results are confirmed by a plot of y=A(x) on the interval [0,20], as shown below in Figure 3.25.
Figure 3.25. A plot of the area function from Example 3.23.

Example 3.26.

A piece of cardboard that is 10×15 (each measured in inches) is being made into a box without a top. To do so, squares are cut from each corner of the box and the remaining sides are folded up. Assuming that the box needs to be at least 1 inch deep and no more than 3 inches deep, answer the following questions.
  1. Draw a labeled diagram that shows the given information. What variable should we introduce to represent the choice we make in creating the box? Label the diagram appropriately with the variable and write a sentence to state what the variable represents.
  2. Determine a formula for the function V that tells us the volume of the box and depends on the variable defined in (a).
  3. What is the domain of the function V? That is, in the context of this problem, what values of x make sense for the input of V? Are there additional restrictions provided in the problem?
  4. Determine all critical numbers of the function V.
  5. Evaluate V at each of the endpoints of the domain and at any critical numbers that lie in the domain.
  6. What is the maximum possible volume of the box? What is the minimum volume? Justify your answers using calculus.
Hint.
  1. Consider letting the length of one side of the removed squares be represented by x.
  2. Remember that the volume of a box is length × width × height.
  3. Read the given information carefully and think about the picture.
  4. Note that since V is a cubic function, V′ is quadratic.
  5. Which critical numbers satisfy 1≤x≤3?
  6. Evaluate the function at appropriate points.
Answer.
  2. V(x)=x(10−2x)(15−2x)=4x3−50x2+150x.
  3. 1≤x≤3.
  4. x=25±576≈6.371,1.962.
    • V(1)=104 in3.
    • V(25−576)≈132.038 in3.
    • V(3)=108 in3.
  6. The maximum volume is about 132 cubic inches; the minimum volume is 104 cubic inches.
Solution.
  1. From each corner of the cardboard, we will cut a square with side length x inches. The result is the following picture:
  2. After cutting out the four x×x squares from the corners of the cardboard and folding up the edges, the resulting box is x inches tall, 10−2x inches wide, and 15−2x inches long. Thus the volume of the box (in cubic inches) is given by
    V(x)=x(10−2x)(15−2x)=4x3−50x2+150x.
  3. At first glance, we see that V is a polynomial and so is defined for every real number x. However, because x represents a length in this problem, the smallest x can be is 0. Moreover, since the shortest side of the cardboard is initially only 10 inches and cutting the squares reduces each side by 2x inches, the largest x can be is 5. We note that both of these extremes would result in a “box” with zero volume, as the first would have no height and the second would have no width. Finally, the problem says the height needs to be between 1 and 3 inches, so we will further restrict the domain of V to 1≤x≤3.
  4. Since V′(x)=12x2−100x+150, it follows that the critical numbers (where V′(x)=0) are
    x=25±576≈6.371,1.962.
  5. Only the latter critical number is in the relevant domain of V, so we consider
    • V(1)=104 in3,
    • V(25−576)≈132.038 in3,
    • V(3)=108 in3.
  6. The maximum possible volume of the box is about 132.038 cubic inches, occurring when squares with side length around 1.962 inches are cut from each corner of the cardboard. The minimum possible volume of the box is 104 in3 and occurs with a box that is 1 inch tall.
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Example 3.23 and Example 3.26 illustrate standard steps that we undertake in almost every applied optimization problem: we draw a picture to demonstrate the situation, introduce one or more variables to represent quantities that are changing, find a function that models the quantity to be optimized, and then decide on an appropriate domain for that function. Once that is done, we are in the familiar situation of finding the absolute minimum and maximum of a function over a particular domain, so we apply the calculus ideas that we have been studying up to this point in Chapter 3.
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Subsection 3.2.3 Summary

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  • To find relative extreme values of a function, we use a first derivative sign chart and classify all of the function’s critical numbers. If instead we are interested in absolute extreme values, we first decide whether we are considering the entire domain of the function or a particular interval.
  • In the case of finding global extrema over the function’s entire domain, we again use a first or second derivative sign chart. If we are working to find absolute extrema on a restricted interval, then we first identify all critical numbers of the function that lie in the interval.
  • For a continuous function on a closed, bounded interval, the only possible points at which absolute extreme values occur are the critical numbers and the endpoints. Hence we evaluate the function at each endpoint and at each critical number in the interval and then compare the results to decide which is largest (the absolute maximum) and which is smallest (the absolute minimum).
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Exercises 3.2.4 Exercises

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1. Finding Global Extrema.

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Consider the function f(x)=2x3+12x2−72x+10,−6≤x≤3.
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Find the absolute minimum value of this function.
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Answer:
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Find the absolute maximum value of this function.
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Answer:
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2. Finding Global Extrema.

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Consider the function f(x)=x4−32x2+2,−3≤x≤9.
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Find the absolute minimum value of this function.
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Answer:
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find the absolute maximum value of this function.
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Answer:
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3. Analyzing Function Behavior.

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Let f(x)=2+31+(x+1)2.
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Determine all critical values of f. If there is more than one, enter the values as a comma-separated list.
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Critical value(s) =
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Construct a first derivative sign chart for f and thus determine all intervals on which f is increasing or decreasing. If there is more than one, enter the intervals as a comma-separated list. Use interval notation: for example, (-17,20) is the interval −17<x<20, and (-inf, 40) is the interval x<40.
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Interval(s) where f is increasing:
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Interval(s) where f is decreasing:
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Does f have a global maximum? If so, enter its value. If not, enter DNE.
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Global maximum =
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Determine the following limits.
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limx→∞f(x)=
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limx→−∞f(x)=
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Explain why f(x)>2 for every value of x.
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Does f have a global minimum? If so, enter its value. If not, enter DNE.
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Global minimum =
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4. Analyzing Extrema From a Graph.

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Identify the marked points as being an absolute maximum or minimum, a relative maximum or minimum, or none of the above. (Select all that apply.)
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A.
  • Relative minimum
  • Relative maximum
  • Absolute minimum
  • Absolute maximum
  • None of the above
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B.
  • Relative minimum
  • Absolute minimum
  • Absolute maximum
  • Relative maximum
  • None of the above
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C.
  • Relative maximum
  • Absolute maximum
  • Relative minimum
  • Absolute minimum
  • None of the above
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D.
  • Relative minimum
  • Relative maximum
  • Absolute minimum
  • Absolute maximum
  • None of the above
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E.
  • Absolute maximum
  • Relative maximum
  • Relative minimum
  • Absolute minimum
  • None of the above
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F.
  • Relative maximum
  • Absolute maximum
  • Absolute minimum
  • Relative minimum
  • None of the above
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G.
  • Absolute maximum
  • Relative minimum
  • Absolute minimum
  • Relative maximum
  • None of the above
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5. Conditions for When Global Extrema May Occur.

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Based on the given information about each function, decide whether the function has a global maximum, a global minimum, neither, both, or that it is not possible to say without more information. Assume that each function is twice differentiable and defined for all real numbers, unless noted otherwise. In each case, write one sentence to explain your conclusion.
  1. f is a function such that f″(x)<0 for every x.
  2. g is a function with two critical numbers a and b (where a<b), and g′(x)<0 for x<a, g′(x)<0 for a<x<b, and g′(x)>0 for x>b.
  3. h is a function with two critical numbers a and b (where a<b), and h′(x)<0 for x<a, h′(x)>0 for a<x<b, and h′(x)<0 for x>b. In addition, limx→∞h(x)=0 and limx→−∞h(x)=0.
  4. p is a function differentiable everywhere except at x=a and p″(x)>0 for x<a and p″(x)<0 for x>a.
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6. Finding Extrema on Closed and Bounded Intervals.

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For each family of functions that depends on one or more parameters, determine the function’s absolute maximum and absolute minimum on the given interval.
  1. p(x)=x3−a2x on [0,a]; assume a>0.
  2. r(x)=axe−bx on [12b,2b]; assume a>0,b>1.
  3. w(x)=a(1−e−bx) on [b,3b]; assume a,b>0.
  4. s(x)=sin⁡(kx) on [π3k,5π6k]; assume k>0.
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7. Conditions for Where Global Extrema May Occur.

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For each of the functions described below (each continuous on [a,b]), state the location of the function’s absolute maximum and absolute minimum on the interval [a,b], or say there is not enough information provided to make a conclusion. Assume that any critical numbers mentioned in the problem statement represent all of the critical numbers the function has in [a,b]. In each case, write one sentence to explain your answer.
  1. f′(x)≤0 for all x in [a,b].
  2. g has a critical number at c such that a<c<b and g′(x)>0 for x<c and g′(x)<0 for x>c.
  3. h(a)=h(b) and h″(x)<0 for all x in [a,b].
  4. p(a)>0, p(b)<0, and for the critical number c such that a<c<b, p′(x)<0 for x<c and p′(x)>0 for x>c.
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8. Using the Extreme Value Theorem.

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Let s(t)=3sin⁡(2(t−π6))+5. Find the exact absolute maximum and minimum of s on the provided intervals by testing the endpoints and by finding and evaluating s at all relevant critical numbers of s.
  1. [π6,7π6]
  2. [0,π2]
  3. [0,2π]
  4. [π3,5π6]
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