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Applied Calculus

Kevin Gonzales, Eric Hopkins, Catherine Zimmitti, Cheryl Kane, Modified to fit Applied Calculus from Coordinated Calculus by Nathan Wakefield et. al., Based upon Active Calculus by Matthew Boelkins

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Section 4.3 Absolute Extrema

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This section corresponds to 3.6 Absolute Extrema in the workbook.
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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. In this section we will explore when a function has an absolute maximum or minimum.
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Absolute Maximum and Minimum.

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Given a function f, we say that f(c) is an absolute maximum of f provided that f(c)≥f(x) for all x in the domain of f. Similarly, we call f(c) an absolute minimum of f whenever f(c)≤f(x) for all x in the domain of f.
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The difference between a relative maximum and an absolute maximum: there is a relative maximum of f at x=c if f(c)≥f(x) for all x near c, while there is an absolute maximum at c if f(c)≥f(x) for all x in the domain of f.
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For instance, in Figure 4.3.1 at the right, we see a function f that has an absolute 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 absolute minimum, although there is a relative minimum at x=b.
Figure 4.3.1. A function f with an absolute maximum, but no absolute minimum.
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Our emphasis in this section is on finding the absolute extreme values of a function (if they exist), either over its entire domain or on some restricted portion.

Example 4.3.2.

Consider the function h given by the graph in Figure 4.1.5 below. Use the graph to answer each of the following questions.
Figure 4.3.3. The graph of y=h(x) on the interval [−3,3].
  1. Identify all of the values of c for which h(c) is a relative maximum of h.
  2. Identify all of the values of c for which h(c) is a relative minimum of h.
  3. Does h have an absolute maximum on the interval [−3,3]? If so, what is the value of this absolute maximum?
  4. Does h have an absolute minimum on the interval [−3,3]? If so, what is its value?
  5. Identify all values of c for which h′(c)=0.
  6. Identify all values of c for which h′(c) does not exist.
  7. True or false: every relative maximum and minimum of h occurs at a point where h′(c) is either zero or does not exist.
  8. True or false: at every point where h′(c) is zero or does not exist, h has a relative maximum or minimum.
Hint.
  1. Start by looking for x-values where the graph switches from increasing to decreasing. Are there any other points that might work?
  2. Start by looking for x-values where the graph switches from decreasing to increasing. Are there any other points that might work?
  3. Is there a highest point on the graph?
  4. Is there a lowest point on the graph?
  5. Remember that horizontal lines have a slope of 0.
  6. Think back to earlier chapters. What characteristics of h or its graph might make it non-differentiable at a certain point?
  7. Compare your answers to (a) and (b) with those to (e) and (f).
  8. Compare your answers to (a) and (b) with those to (e) and (f).
Answer.
  1. c=−2 and c=1.
  2. c=0.
  3. h(1)=2.
  4. h(−3)=−3.
  5. c=−2.5 and c=1.
  6. c=−2, c=0, and c=1.5.
  7. True.
  8. False.
Solution.
  1. The graph shows that h(x)≤−1=h(−2) for every x near −2 and h(x)≤2=h(1) for every x near 1. We notice that these are points where h switches from increasing to decreasing. Since there are no other points on the graph where the y-coordinate is larger than everything nearby, it follows that the only relative maxima of h on this domain occur at c=−2 and c=1.
  2. The graph shows that h(x)≥−2=h(0) for every x near 0, a point where h switches from decreasing to increasing. Since this is the only point shown where this occurs, we say that the only relative minimum of h on this domain occurs at c=0.
  3. The highest point shown on the graph is the point (1,2). Since h is continuous on [−3,3], we don’t have to worry about any points on the interval being higher and not showing up on the displayed grid. Hence we say the absolute maximum of h on the interval [−3,3] is 2 and occurs at x=1.
  4. The lowest point shown on the graph is the point (−3,−3). Again, since h is continuous on [−3,3], we don’t have to worry about any points on the interval being lower and not showing up in the window that is displayed. Thus we say the absolute minimum of h on the interval [−3,3] is −3 and occurs at x=−3.
    It is worth noting that this was not a point that we found as a local minimum. The difference is that here we are restricting our focus to the interval that is shown, effectively cutting off the graph at its endpoints. Earlier, when we were looking for relative minima, we made the assumption that the graph of y=h(x) continues outside of the depicted interval by following the trends displayed and continuing downward on both sides.
  5. When looking for points (c,h(c)) at which h′(c)=0, we are looking for points of horizontal tangency. The graph flattens out at (−2.5,−2) and (1,2), but nowhere else
     1 
    Note that at (1.5,1) it looks as though the graph is flat to the right of this point. However, the derivative is undefined here because of the cusp, so this is not a point of horizontal tangency
    . Thus h′(c)=0 exactly when c=−2.5 and c=1.
  6. Recall that h′(c) only exists when all of the following are true:
    • h(c) is defined,
    • limx→ch(x) exists,
    • limx→ch(x)=h(c), and
    • lima→0h(c+a)−h(c)a exists.
    Assuming h is continuous at c (i.e. the first three points hold), the main consequence of the final point is that a corner or a cusp at (c,h(c)) makes h non-differentiable at c.
    With all this in mind, we can say that h′(c) does not exist at c=−2, c=0, or c=1.5 because h has a cusp at these points.
  7. This is true. Every relative extremum of h occurs at a point where h′(c) is either zero or does not exist.
  8. This is false. Some points at which h′(c) is zero or undefined are not relative extrema. In particular, h′(−2.5) is zero, but the graph is non-decreasing near (−2.5,−2). Similarly, h′(1.5) is undefined, but the graph is non-increasing near (1.5,1). Neither of these points is a local maximum or minimum.
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Subsection 4.3.1 Finding Absolute Extrema

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We have seen how to find the relative extrema of a function f(x) using the critical points. We will now examine how to find the absolute extrema. We will start by examining cases of a function f(x) restricted to a closed domain [a,b]. 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.

Example 4.3.4.

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
     2 
    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 an absolute maximum at x=3 and an absolute minimum at x=2.
  4. On [−2,2], g has an absolute maximum at x=−2 and an absolute minimum at x=2.
  5. On [−2,3], g has an absolute maximum at x=−2 and an absolute 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 an absolute maximum at x=3 and an absolute minimum at x=2.
  4. On [−2,2], g has an absolute maximum at x=−2 and an absolute minimum at x=2.
  5. On [−2,3], g has an absolute maximum at x=−2 and an absolute minimum at x=1.
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Example 4.3.4 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 4.3.5 below depicts again the graph of y=g(x) that we looked at in Example 4.3.4.
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Figure 4.3.5. 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 has both an absolute minimum and absolute maximum on [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 4.3.4, 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 4.3.6.

Find the absolute maximum and minimum of each function on the stated interval.
  1. f(x)=x4−98x2+10 on [−6,15]
  2. h(x)=xe−x on [0,3]
  3. q(x)=x2x−2 on [3,7]
  4. h(x)=xe−ax on [0,2a]; assume a>0
Answer.
  1. Absolute maximum: 28585; Absolute minimum: −2391.
  2. Absolute maximum: e−1; Absolute minimum: 0.
  3. Absolute maximum: 9.8; absolute minimum: 8.
  4. Absolute maximum: (ae)−1; absolute minimum: 0.
Solution.
  1. Given f(x)=x4−98x2+10, it follows that f′(x)=4x3−196x, solving f′(x)=0=4x(x2−49) which gives three critical values: x=0, x=7 and x=−7.
    Since we are looking at the interval −6≤x≤15 we will only test the critical values within this interval, which are x=0,7.
    Now we plug in these two critical values, along with then endpoints x=−6 and x=15 to find the absolute maximum and absolute minimum:
    f(−6)=(−6)4−98(−6)2+10=−2222
    f(0)=(0)4−98(0)+10=10
    f(7)=(7)4−98(7)2+10=−2391
    f(15)=(15)4−98(15)2+10=28585
    Thus the absolute maximum of f(x) is 28585 which occurs at x=15, and the absolute minimum of f(x) is −2391 which occurs at x=7.
  2. 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.
  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 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.
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Up to this point, we have seen how to find the absolute maximum and minimum of a function on a closed interval. But what if we are not restricted to a closed domain [a,b]?
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In general, a function on an open domain, say (0,∞) or (−∞,∞) may or may not have an absolute maximum or minimum. Will have to find the critical points and then examine the long term behavior of the function.

Example 4.3.7.

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 an absolute 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 an absolute 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 an absolute 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 absolute 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 an absolute 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, so the maximum value that f can attain is 5. Since f(−1)=5, we conclude that the absolute 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, an absolute 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. So, there is no value larger than 2 that is an absolute minimum for f, and thus f has no absolute minimum.
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The Extreme Value Theorem on an Open Domain.

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Suppose f(x) is a continuous function on a open interval, for example (0,∞) or (∞,∞). If f(x) has exactly one critical value x=c on the interval, then:
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If the function has a relative maximum at x=c, then f(x) has an absolute maximum at x=c and no absolute minimum.
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If the function has a relative minimum at x=c, then f(x) has an absolute minimum at x=c and no absolute maximum.

Example 4.3.8.

Find the absolute maximum and minimum of each function on the stated interval.
  1. f(x)=4x2−16x+4 on (−∞,∞)
  2. h(x)=4x+1x on (−∞,0)
  3. f(x)=4−e−(x−2)2 on (−∞,∞)
  4. f(x)=b−e−(x−a)2 on (−∞,∞); assume a,b>0
Answer.
  1. Absolute minimum of: −12 at x=2, no absolute maximum.
  2. Absolute maximum: −4; no absolute minimum.
  3. Absolute minimum: 3; no absolute maximum.
  4. Absolute minimum: b−1; no absolute maximum.
Solution.
  1. Given f(x)=4x2−16x+4, it follows that f′(x)=8x−16, solving f′(x)=0 gives a single critical value: x=2
    Then f″(x)=8 which is always positive, that is f(x) is always concave up and thus by the second derivative test f(x) has a relative minimum at x=2. Since this is the only critical value, it follows that f(x) has an absolute minimum of f(2)=4(2)2−16(2)+4=−12 at x=2, and consequently has no absolute maximum.
  2. For h(x)=4x+1x, we know that h′(x)=4−x−2=4−1x2. To solve h′(x)=4−1x2=0 add the 1x2 to both sides to get 4=1x2, then multiply both sides by x2 and divide by 4 to get x2=1x2. Solve by taking the square root and we get two critical values x=±12. Since we are looking at the interval (−∞,0) we only have a single critical value x=−12=−0.5.
    Then h″(x)=2x−3 and h″(−.5)<0 x=−0.5 is a relative maximum. Since there is a single critical value the function h(x)=4x+1x has an absolute maximum of h(−0.5)=4(−0.5)+1−0.5=−4 at x=−0.5 and no absolute minimum.
  3. 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.
  4. 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 4.3.2 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 must consider the interval on which we are finding the extreme values (is it a closed interval, an open interval, or the entire domain?).
  • 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).
  • In the case of finding absolute extrema on an open interval, we first identify all of the critical numbers of the function on that interval, and then use a first or second derivative sign chart to investigate the behavior of the function on that interval.
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Exercises 4.3.3 Exercises

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1. Finding absolute Extrema.

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Consider the function f(x)=2x3+24x2−120x+6,−10≤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 absolute Extrema.

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Consider the function f(x)=x4−18x2+1,−2≤x≤7.
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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 maximum
  • Absolute minimum
  • Absolute maximum
  • Relative minimum
  • None of the above
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B.
  • Relative maximum
  • Absolute maximum
  • Relative minimum
  • Absolute minimum
  • None of the above
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C.
  • Absolute maximum
  • Absolute minimum
  • Relative maximum
  • Relative minimum
  • None of the above
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D.
  • Relative maximum
  • Absolute minimum
  • Relative minimum
  • Absolute maximum
  • None of the above
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E.
  • Relative maximum
  • Relative minimum
  • Absolute maximum
  • Absolute minimum
  • None of the above
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F.
  • Relative minimum
  • Absolute minimum
  • Absolute maximum
  • Relative maximum
  • None of the above
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G.
  • Absolute minimum
  • Relative minimum
  • Absolute maximum
  • Relative maximum
  • None of the above
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5. Conditions for When absolute Extrema May Occur.

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Based on the given information about each function, decide whether the function has an absolute maximum, an absolute 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.
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7. Conditions for Where absolute 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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