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Global optimization

Introduction

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.

Recall the difference between a relative maximum and a global maximum: there is a relative maximum of \(f\) at \(x = p\) if \(f(p) \ge f(x)\) for all \(x\) near \(p\), while there is a global maximum at \(p\) if \(f(p) \ge f(x)\) for all \(x\) in the domain of \(f\). For instance, in Figure, \(f\) has a global maximum at \(x = c\) and a relative maximum at \(x = a\), since \(f(c)\) is greater than \(f(x)\) for every value of \(x\), while \(f(a)\) is only greater than the value of \(f(x)\) for \(x\) near \(a\). Since the function appears to decrease without bound, \(f\) has no global minimum, though clearly \(f\) has a relative minimum at \(x = b\).

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.

Exploration

Global Optimization

In Figure and Preview Activity, 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.

For example, rather than considering \(f(x) = 2 + \frac{3}{1+(x+1)^2}\) for every value of \(x\), perhaps instead we are only interested in those \(x\) for which \(0 \le x \le 4\), 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 activity. 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.

In Activity, we saw 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 \(b\). These observations demonstrate several important facts that hold more generally. First, we state an important result called the Extreme Value Theorem.

If \(f\) is a continuous function on a closed interval \([a,b]\), then \(f\) attains both an absolute minimum and absolute maximum on \([a,b]\).

The Extreme Value Theorem tells us that on any closed 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, but rather only that they must exist. As we saw in Activity, the only possible locations for relative extremes are at the endpoints of the interval or at a critical number.

Condensed — the full section is in Boelkins, Active Calculus.

Moving toward applications

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 and Activity 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 to this point in Chapter.

Condensed — the full section is in Boelkins, Active Calculus.

Summary

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 extremes over the function's entire domain, we again use a first or second derivative sign chart. If we are working to find absolute extremes 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. Thus, we simply evaluate the function at each endpoint and each critical number in the interval, and compare the results to decide which is largest (the absolute maximum) and which is smallest (the absolute minimum).

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For each of the following prompts, sketch a graph of a function that has the stated combination of properties, or explain why such a combination is impossible.
1. One relative maximum and one relative minimum
2. One relative maximum but no global maximum
3. One global minimum but no relative minimum
4. Two relative maxima but no relative minimum
5. Two relative maxima but no global minimum
Based on the given information about each function, decide whether the function has 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 \(f''(x) \lt 0\) for every \(x\).
2. \(g\) is a function with two critical numbers \(a\) and \(b\) (where \(a \lt b\)), and \(g'(x) \lt 0\) for \(x \lt a\), \(g'(x) \lt 0\) for \(a \lt x \lt b\), and \(g'(x) \gt 0\) for \(x \gt b\).
3. \(h\) is a function with two critical numbers \(a\) and \(b\) (where \(a \lt b\)), and \(h'(x) \lt 0\) for \(x \lt a\), \(h'(x) \gt 0\) for \(a \lt x \lt b\), and \(h'(x) \lt 0\) for \(x \gt b\). In addition, \(\lim_{x \to \infty} h(x) = 0\) and \(\lim_{x \to -\infty} h(x) = 0\).
4. \(p\) is a function differentiable everywhere except at \(x = a\) and \(p''(x) \gt 0\) for \(x \lt a\) and \(p''(x) \lt 0\) for \(x \gt a\).
Kuratidza mhinduro
1. It's not possible to say without more information: for instance, the functions \(f(x) = e^{-x}\), \(f(x) = e^{-x}\), and \(f(x) = x^2\) are each always concave up, but only one of them has a global minimum. We can say for sure that by being always concave up, \(f\) cannot have a global maximum, and it also cannot have both a global maximum and a global minimum. It might have a global minimum, or it might have neither.
2. By being twice differentiable, we know that \(g'(x) = 0\) any time \(x\) is a critical number of \(g\). Thus, given the signs of \(g'(x)\) on the stated intervals, \(g\) is decreasing before \(x = a\) and decreasing on \(a \lt x \lt b\), with a horizontal tangent line at \(x = a\). In addition, \(g\) is increasing for \(x \gt b\). Thus, we know that \(g\) has a global minimum at \(x = b\).
3. Since \(h\) is always decreasing for \(x \lt a\), always increasing for \(a \lt x \lt b\), and always decreasing for \(x \gt b\), we know that \(h\) has a local minimum at \(x = a\) and a local maximum at \(x = b\). Moreover, these local extremes are both global extremes since \(h(x) to 0\) as \(x \to \pm \infty\). In particular, since \(h(x)\) decreases to \(0\) for \(x \gt b\) and increases on \(a \lt x \lt b\), it follows that \(h(b)\) is the largest value of \(h(x)\) on \((a, \infty)\), and that \(h(b) \gt 0\). Similarly, since \(h(x)\) decreases on \((-\infty, a)\) and approaches \(0\) as \(x \to -\infty\), and \(h(x)\) increases on \(a \lt x \lt b\), it must be the case that \(h(a)\) is the least value of \(h(x)\) on \((-\infty, b)\) and that \(h(b) \lt 0\). Hence, \(h(x)\) is always negative on \(x \lt a\) and \(h(x)\) is always positive on \(x \gt b\). From all of these observations, it follows that \(h\) has a global minimum at \(x = a\) and a global maximum at \(x = b\).
4. First, observe that \(p\) is defined for all real numbers by assumption, so \(p(a)\) has some finite value. However, \(p\) is not differentiable at \(x = a\) and \(p\) is concave up to the left of \(x = a\) and concave down to the right of \(x = a\). This indicates that \(p\) has a relative maximum at \(x = a\). Not enough information is provided, however, to determine if this relative maximum is a global maximum, nor if \(p\) happens to have a global minimum at some other point.
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. \(p(x) = x^3 - a^2x\), \([0,a]\) (\(a \gt 0\))
2. \(r(x) = axe^{-bx}\), \([\frac{1}{2b}, \frac{2}{b}]\) (\(a \gt 0, b \gt 1\))
3. \(w(x) = a(1-e^{-bx})\), \([b, 3b]\) (\(a, b \gt 0\))
4. \(s(x) = \sin(kx)\), \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\) (\(k \gt 0\))
Kuratidza mhinduro
1. Given \(p(x) = x^3 - a^2x\) on \([0,a]\), we first find the critical number(s) of \(p\) that lie in \([0,a]\). Since \(p'(x) = 3x^2 - a^2 = 3\left(x^2 - \frac{a^2}{3} \right)\), we know \(p\) has critical numbers at \(x = \pm \frac{a}{\sqrt{3}}\). The negative critical number lies outside the interval, while the positive one lies within the interval since \(0 \lt \frac{1}{\sqrt{3}} \lt 1\). Evaluating \(p\) at the endpoints and the critical number within the interval, we find that \(p(0) = 0\), \(p(a) = a^3 - a^2 \cdot a = 0\), and \(p\left( \frac{a}{\sqrt{3}} \right) = \left( \frac{a}{\sqrt{3}} \right)^3 - a^2 \cdot \left( \frac{a}{\sqrt{3}} \right) = \frac{a^3}{3\sqrt{3}} - \frac{a^3}{\sqrt{3}} = -\frac{2a^3}{3\sqrt{3}}\). Hence, the absolute maximum of \(p\) on \([0,a]\) is \(p(0) = p(a) = 0\) and the absolute minimum is \(p\left( \frac{a}{\sqrt{3}} \right) = -\frac{2a^3}{3\sqrt{3}}\).
2. Given \(r(x) = axe^{-bx}\) on \(\left[\frac{1}{2b}, \frac{2}{b}\right]\), we first find the critical numbers of \(r\) that lie in the given interval. Note that by the product rule, \(r'(x) = ax e^{-bx} (-b) + e^{-bx} a\). Factoring, \(r'(x) = ae^{-bx}(-bx + 1)\), and thus the only critical number of \(r\) occurs where \(x = \frac{1}{b}\). Note that \(\frac{1}{2b} \lt \frac{1}{b} \lt b\) since \(b \gt 1\), and thus the critical number lies within the interval.
  Evaluating \(r\) at the endpoints and the critical number, we find
  \(r\left( \frac{1}{2b} \right) = a \cdot \frac{1}{2b} \cdot e^{-1/2} \approx 0.607 \frac{a}{2b} = 0.3035 \frac{a}{b}\)
  \(r\left( \frac{2}{b} \right) = a \cdot \frac{2}{b} \cdot e^{-2} \approx 0.135 \frac{2a}{b} = 0.270 \frac{a}{b}\)
  \(r\left( \frac{1}{b} \right) = a \cdot \frac{1}{b} \cdot e^{-1} \approx 0.368 \frac{a}{b}\)
  The absolute max of \(r\) is therefore \(r\left( \frac{1}{b} \right) \approx 0.368 \frac{a}{b}\) and the absolute min is \(r\left( \frac{2}{b} \right) \approx 0.270 \frac{a}{b}\).
3. For \(w(x) = a(1-e^{-bx})\) on \([b, 3b]\), we first note that \(w'(x) = a(be^{-bx}) = abe^{-bx}\). Since \(a\) and \(b\) are both positive and so is \(e^{-bx}\) for every value of \(x\), we see that \(w'(x) \gt 0\) for every value of \(x\). Because \(w\) is always increasing (specifically on \([b,3b]\)), \(w\) has its absolute minimum at the left endpoint (\(g(b) = a(1-e^{-b^2})\)) and its absolute maximum at the right endpoint (\(g(3b) = a(1-e^{-3b^2})\)).
4. Finally, for \(s(x) = \sin(kx)\) on \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\), we know \(s'(x) = k\cos(kx)\), and \(\cos(kx) = 0\) when \(kx = \frac{\pi}{2} \pm n\pi\) for some integer \(n\), so \(x = \frac{\pi}{2k} \pm n\pi = \frac{\pi \pm 2n\pi}{2k}\). These \(x\)-values are the critical numbers of \(s\). Note that when \(n = -1, 0, 1\), the corresponding critical numbers are \[\begin{aligned}\end{aligned}\] Comparing the values of \(-\frac{\pi}{2}\), \(\frac{\pi}{2k}\), and \(\frac{3\pi}{2k}\) to \(\frac{\pi}{3}\) and \(\frac{5\pi}{6}\), we see that only \(\frac{\pi}{2k}\) lies in the given interval \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\), and thus \(x = \frac{\pi}{2k}\) is the only critical number at which we need to evaluate \(s\).
  Evaluating \(s\) at the endpoints and relevant critical number, we find:
  \(s\left( \frac{\pi}{3k} \right) = \sin\left( k \cdot \frac{\pi}{3k} \right) = \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}\);
  \(s\left( \frac{5\pi}{6k} \right) = \sin\left( k \cdot \frac{5\pi}{6k} \right) = \sin\left( \frac{5\pi}{6} \right) = \frac{1}{2}\); and
  \(s\left( \frac{\pi}{2k} \right) = \sin\left( k \cdot \frac{\pi}{2k} \right) = \sin\left( \frac{\pi}{2} \right) = 1\).
  Hence the absolute max of \(s\) on \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\) is \(s\left( \frac{\pi}{2k} \right) = 1\) and the absolute min of \(s\) is \(s\left( \frac{5\pi}{6k} \right) = \frac{1}{2}\).
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. \(f'(x) \le 0\) for all \(x\) in \([a,b]\)
2. \(g\) has a critical number at \(c\) such that \(a \lt c\lt b\) and \(g'(x) \gt 0\) for \(x \lt c\) and \(g'(x) \lt 0\) for \(x \gt c\)
3. \(h(a) = h(b)\) and \(h''(x) \lt 0\) for all \(x\) in \([a,b]\)
4. \(p(a) \gt 0\), \(p(b) \lt 0\), and for the critical number \(c\) such that \(a \lt c \lt b\), \(p'(x) \lt 0\) for \(x \lt c\) and \(p'(x) \gt 0\) for \(x \gt c\)
Kuratidza mhinduro
1. The fact that \(f'(x) \leq 0\) for all \(x\) in \([a,b]\) means that \(f\) cannot increase in this interval. So either \(f\) is constant or \(f\) decreases somewhere in this interval. We can conclude that the global maximum value of \(f\) occurs at \(x=a\) and the global minimum value at \(x=b\).
2. The information we are given shows that \(g\) is increasing on \((a,c)\) and decreasing on \((c,b)\). So \(g\) has a global maximum value at \(x=c\). We know that \(g\) must have a global minimum at either \(x=a\) or \(x=b\). The information given does not tell us how much \(g\) increases on \((a,c)\) and decreases on \((c,b)\), so we cannot determine the endpoint at which \(g\) attains its global minimum value.
3. In order for \(h(a)=h(b)\) and \(h''(x) \lt 0\) for all \(x\) in \([a,b]\), the graph of \(h\) must look something like a parabola that opens down on the interval \([a,b]\). Consequently, \(h\) has its global minimum at \(x=a\) and \(x=b\), and its global maximum somewhere in \((a,b)\), but we do not know the exact point at which \(h\) attains its global maximum.
4. The given information forces the graph of \(p\) to decrease on \((a,c)\) then increase on \((c,b)\). Thus, we must have \(p(c) \lt p(b)\) and so \(p\) has its global minimum value at \(x=c\). Since \(p(a) \gt p(b)\), this makes \(p(a)\) the global maximum value of \(p\) on \([a,b]\).
Let \(s(t) = 3\sin(2(t-\frac{\pi}{6})) + 5\). Find the exact absolute maximum and minimum of \(s\) on the provided intervals by testing the endpoints and finding and evaluating all relevant critical numbers of \(s\).
1. \([\frac{\pi}{6}, \frac{7\pi}{6}]\)
2. \([0, \frac{\pi}{2}]\)
3. \([0, 2\pi]\)
4. \([\frac{\pi}{3}, \frac{5\pi}{6}]\)
Kuratidza mhinduro
1. For \(s(t) = 3\sin(2(t-\frac{\pi}{6})) + 5\), \(s'(t) = 3\cos(2(t-\frac{\pi}{6})) \cdot 2 = 6\cos(2(t-\frac{\pi}{6}))\). It follows that \(s'(t) = 0\) if and only if \(\cos(2(t-\frac{\pi}{6})) = 0\), and thus \(2(t-\frac{\pi}{6}) = \frac{\pi}{2} \pm k\pi\) for some integer value of \(k\). Solving this most recent equation for \(t\), we have \(t-\frac{\pi}{6} = \frac{\pi}{4} \pm \frac{k\pi}{2}\), so \[\begin{aligned}\end{aligned}\]. Now we consider the given interval \([\frac{\pi}{6}, \frac{7\pi}{6}]\), which we can think of as \([\frac{2\pi}{12}, \frac{14\pi}{12}]\). The value \(t = \frac{5\pi}{12}\) lies in this interval, as does \(t = \frac{5\pi}{12} + \frac{\pi}{2} = \frac{11\pi}{12}\). Thus, we have two critical numbers in the interval at which to evaluate \(s\), as well as at the endpoints.
  Doing so, we find:
  \(s(\frac{5\pi}{12}) = 8\);
  \(s(\frac{11\pi}{12}) = 2\);
  \(s(\frac{\pi}{6}) = 5\); and
  \(s(\frac{7\pi}{6}) = 5\).
  The absolute max of \(s\) on \([\frac{\pi}{6}, \frac{7\pi}{6}]\) is thus \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(\frac{11\pi}{12}) = 2\).
2. From our work in (a), we know that \(s\) has critical numbers at \(t = \frac{5\pi}{12} \pm \frac{k\pi}{2}\). On the interval \([0, \frac{\pi}{2}]\), only \(t = \frac{5\pi}{12}\) lies in the interval. Thus we have three values at which to evaluate \(s\).
  Doing so:
  \(s(\frac{5\pi}{12}) = 8\);
  \(s(0) = 3\sin(2(0-\frac{\pi}{6})) + 5 = 3\sin(-\frac{\pi}{3}) + 5 = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\); and
  \(s(\frac{\pi}{2}) = 3\sin(2(\frac{\pi}{2}-\frac{\pi}{6})) + 5 = 3\sin(\frac{2\pi}{3}) + 5 = 5 + \frac{3\sqrt{3}}{2} \approx 7.598\).
  The absolute max of \(s\) on \([0, \frac{\pi}{2}]\) is thus \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(0) = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\).
3. On the interval from \(t = 0\) to \(t = 2\pi\), \(s\) completes nearly two full periods, and thus achieves each of its maximum and minimum values. The absolute maximum is \(8\) as determined in (a), and the absolute minimum is \(2\). These occur, for example, at \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(\frac{11\pi}{12}) = 2\). There are other points at which the function achieves these values on the given interval.
4. On the interval \([\frac{\pi}{3}, \frac{5\pi}{6}] = [\frac{4\pi}{12}, \frac{10\pi}{12}]\), only the critical number \(t = \frac{5\pi}{12}\) lies in the interval. Thus we have three values at which to evaluate \(s\).
  Doing so:
  \(s(\frac{5\pi}{12}) = 8\);
  \(s(\frac{\pi}{3}) = 3\sin(2(\frac{\pi}{3}-\frac{\pi}{6})) + 5 = 3\sin(\frac{\pi}{3}) + 5 = \frac{3\sqrt{3}}{2} + 5 \approx 7.598\); and
  \(s(\frac{5\pi}{6}) = 3\sin(2(\frac{5\pi}{6}-\frac{\pi}{6})) + 5 = 3\sin(\frac{4\pi}{3}) + 5 = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\).
  The absolute max of \(s\) on \([\frac{\pi}{3}, \frac{5\pi}{6}]\) is thus \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(\frac{5\pi}{6}) \approx 2.402\).
Based on the given information about each function, decide whether the function has 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 \(f''(x) \lt 0\) for every \(x\).
2. \(g\) is a function with two critical numbers \(a\) and \(b\) (where \(a \lt b\)), and \(g'(x) \lt 0\) for \(x \lt a\), \(g'(x) \lt 0\) for \(a \lt x \lt b\), and \(g'(x) \gt 0\) for \(x \gt b\).
3. \(h\) is a function with two critical numbers \(a\) and \(b\) (where \(a \lt b\)), and \(h'(x) \lt 0\) for \(x \lt a\), \(h'(x) \gt 0\) for \(a \lt x \lt b\), and \(h'(x) \lt 0\) for \(x \gt b\). In addition, \(\lim_{x \to \infty} h(x) = 0\) and \(\lim_{x \to -\infty} h(x) = 0\).
4. \(p\) is a function differentiable everywhere except at \(x = a\) and \(p''(x) \gt 0\) for \(x \lt a\) and \(p''(x) \lt 0\) for \(x \gt a\).
Kuratidza mhinduro
1. It's not possible to say without more information: for instance, the functions \(f(x) = e^{-x}\), \(f(x) = e^{-x}\), and \(f(x) = x^2\) are each always concave up, but only one of them has a global minimum. We can say for sure that by being always concave up, \(f\) cannot have a global maximum, and it also cannot have both a global maximum and a global minimum. It might have a global minimum, or it might have neither.
2. By being twice differentiable, we know that \(g'(x) = 0\) any time \(x\) is a critical number of \(g\). Thus, given the signs of \(g'(x)\) on the stated intervals, \(g\) is decreasing before \(x = a\) and decreasing on \(a \lt x \lt b\), with a horizontal tangent line at \(x = a\). In addition, \(g\) is increasing for \(x \gt b\). Thus, we know that \(g\) has a global minimum at \(x = b\).
3. Since \(h\) is always decreasing for \(x \lt a\), always increasing for \(a \lt x \lt b\), and always decreasing for \(x \gt b\), we know that \(h\) has a local minimum at \(x = a\) and a local maximum at \(x = b\). Moreover, these local extremes are both global extremes since \(h(x) to 0\) as \(x \to \pm \infty\). In particular, since \(h(x)\) decreases to \(0\) for \(x \gt b\) and increases on \(a \lt x \lt b\), it follows that \(h(b)\) is the largest value of \(h(x)\) on \((a, \infty)\), and that \(h(b) \gt 0\). Similarly, since \(h(x)\) decreases on \((-\infty, a)\) and approaches \(0\) as \(x \to -\infty\), and \(h(x)\) increases on \(a \lt x \lt b\), it must be the case that \(h(a)\) is the least value of \(h(x)\) on \((-\infty, b)\) and that \(h(b) \lt 0\). Hence, \(h(x)\) is always negative on \(x \lt a\) and \(h(x)\) is always positive on \(x \gt b\). From all of these observations, it follows that \(h\) has a global minimum at \(x = a\) and a global maximum at \(x = b\).
4. First, observe that \(p\) is defined for all real numbers by assumption, so \(p(a)\) has some finite value. However, \(p\) is not differentiable at \(x = a\) and \(p\) is concave up to the left of \(x = a\) and concave down to the right of \(x = a\). This indicates that \(p\) has a relative maximum at \(x = a\). Not enough information is provided, however, to determine if this relative maximum is a global maximum, nor if \(p\) happens to have a global minimum at some other point.
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. \(p(x) = x^3 - a^2x\), \([0,a]\) (\(a \gt 0\))
2. \(r(x) = axe^{-bx}\), \([\frac{1}{2b}, \frac{2}{b}]\) (\(a \gt 0, b \gt 1\))
3. \(w(x) = a(1-e^{-bx})\), \([b, 3b]\) (\(a, b \gt 0\))
4. \(s(x) = \sin(kx)\), \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\) (\(k \gt 0\))
Kuratidza mhinduro
1. Given \(p(x) = x^3 - a^2x\) on \([0,a]\), we first find the critical number(s) of \(p\) that lie in \([0,a]\). Since \(p'(x) = 3x^2 - a^2 = 3\left(x^2 - \frac{a^2}{3} \right)\), we know \(p\) has critical numbers at \(x = \pm \frac{a}{\sqrt{3}}\). The negative critical number lies outside the interval, while the positive one lies within the interval since \(0 \lt \frac{1}{\sqrt{3}} \lt 1\). Evaluating \(p\) at the endpoints and the critical number within the interval, we find that \(p(0) = 0\), \(p(a) = a^3 - a^2 \cdot a = 0\), and \(p\left( \frac{a}{\sqrt{3}} \right) = \left( \frac{a}{\sqrt{3}} \right)^3 - a^2 \cdot \left( \frac{a}{\sqrt{3}} \right) = \frac{a^3}{3\sqrt{3}} - \frac{a^3}{\sqrt{3}} = -\frac{2a^3}{3\sqrt{3}}\). Hence, the absolute maximum of \(p\) on \([0,a]\) is \(p(0) = p(a) = 0\) and the absolute minimum is \(p\left( \frac{a}{\sqrt{3}} \right) = -\frac{2a^3}{3\sqrt{3}}\).
2. Given \(r(x) = axe^{-bx}\) on \(\left[\frac{1}{2b}, \frac{2}{b}\right]\), we first find the critical numbers of \(r\) that lie in the given interval. Note that by the product rule, \(r'(x) = ax e^{-bx} (-b) + e^{-bx} a\). Factoring, \(r'(x) = ae^{-bx}(-bx + 1)\), and thus the only critical number of \(r\) occurs where \(x = \frac{1}{b}\). Note that \(\frac{1}{2b} \lt \frac{1}{b} \lt b\) since \(b \gt 1\), and thus the critical number lies within the interval.
  Evaluating \(r\) at the endpoints and the critical number, we find
  \(r\left( \frac{1}{2b} \right) = a \cdot \frac{1}{2b} \cdot e^{-1/2} \approx 0.607 \frac{a}{2b} = 0.3035 \frac{a}{b}\)
  \(r\left( \frac{2}{b} \right) = a \cdot \frac{2}{b} \cdot e^{-2} \approx 0.135 \frac{2a}{b} = 0.270 \frac{a}{b}\)
  \(r\left( \frac{1}{b} \right) = a \cdot \frac{1}{b} \cdot e^{-1} \approx 0.368 \frac{a}{b}\)
  The absolute max of \(r\) is therefore \(r\left( \frac{1}{b} \right) \approx 0.368 \frac{a}{b}\) and the absolute min is \(r\left( \frac{2}{b} \right) \approx 0.270 \frac{a}{b}\).
3. For \(w(x) = a(1-e^{-bx})\) on \([b, 3b]\), we first note that \(w'(x) = a(be^{-bx}) = abe^{-bx}\). Since \(a\) and \(b\) are both positive and so is \(e^{-bx}\) for every value of \(x\), we see that \(w'(x) \gt 0\) for every value of \(x\). Because \(w\) is always increasing (specifically on \([b,3b]\)), \(w\) has its absolute minimum at the left endpoint (\(g(b) = a(1-e^{-b^2})\)) and its absolute maximum at the right endpoint (\(g(3b) = a(1-e^{-3b^2})\)).
4. Finally, for \(s(x) = \sin(kx)\) on \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\), we know \(s'(x) = k\cos(kx)\), and \(\cos(kx) = 0\) when \(kx = \frac{\pi}{2} \pm n\pi\) for some integer \(n\), so \(x = \frac{\pi}{2k} \pm n\pi = \frac{\pi \pm 2n\pi}{2k}\). These \(x\)-values are the critical numbers of \(s\). Note that when \(n = -1, 0, 1\), the corresponding critical numbers are \[\begin{aligned}\end{aligned}\] Comparing the values of \(-\frac{\pi}{2}\), \(\frac{\pi}{2k}\), and \(\frac{3\pi}{2k}\) to \(\frac{\pi}{3}\) and \(\frac{5\pi}{6}\), we see that only \(\frac{\pi}{2k}\) lies in the given interval \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\), and thus \(x = \frac{\pi}{2k}\) is the only critical number at which we need to evaluate \(s\).
  Evaluating \(s\) at the endpoints and relevant critical number, we find:
  \(s\left( \frac{\pi}{3k} \right) = \sin\left( k \cdot \frac{\pi}{3k} \right) = \sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}\);
  \(s\left( \frac{5\pi}{6k} \right) = \sin\left( k \cdot \frac{5\pi}{6k} \right) = \sin\left( \frac{5\pi}{6} \right) = \frac{1}{2}\); and
  \(s\left( \frac{\pi}{2k} \right) = \sin\left( k \cdot \frac{\pi}{2k} \right) = \sin\left( \frac{\pi}{2} \right) = 1\).
  Hence the absolute max of \(s\) on \(\left[\frac{\pi}{3k}, \frac{5\pi}{6k}\right]\) is \(s\left( \frac{\pi}{2k} \right) = 1\) and the absolute min of \(s\) is \(s\left( \frac{5\pi}{6k} \right) = \frac{1}{2}\).
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. \(f'(x) \le 0\) for all \(x\) in \([a,b]\)
2. \(g\) has a critical number at \(c\) such that \(a \lt c\lt b\) and \(g'(x) \gt 0\) for \(x \lt c\) and \(g'(x) \lt 0\) for \(x \gt c\)
3. \(h(a) = h(b)\) and \(h''(x) \lt 0\) for all \(x\) in \([a,b]\)
4. \(p(a) \gt 0\), \(p(b) \lt 0\), and for the critical number \(c\) such that \(a \lt c \lt b\), \(p'(x) \lt 0\) for \(x \lt c\) and \(p'(x) \gt 0\) for \(x \gt c\)
Kuratidza mhinduro
1. The fact that \(f'(x) \leq 0\) for all \(x\) in \([a,b]\) means that \(f\) cannot increase in this interval. So either \(f\) is constant or \(f\) decreases somewhere in this interval. We can conclude that the global maximum value of \(f\) occurs at \(x=a\) and the global minimum value at \(x=b\).
2. The information we are given shows that \(g\) is increasing on \((a,c)\) and decreasing on \((c,b)\). So \(g\) has a global maximum value at \(x=c\). We know that \(g\) must have a global minimum at either \(x=a\) or \(x=b\). The information given does not tell us how much \(g\) increases on \((a,c)\) and decreases on \((c,b)\), so we cannot determine the endpoint at which \(g\) attains its global minimum value.
3. In order for \(h(a)=h(b)\) and \(h''(x) \lt 0\) for all \(x\) in \([a,b]\), the graph of \(h\) must look something like a parabola that opens down on the interval \([a,b]\). Consequently, \(h\) has its global minimum at \(x=a\) and \(x=b\), and its global maximum somewhere in \((a,b)\), but we do not know the exact point at which \(h\) attains its global maximum.
4. The given information forces the graph of \(p\) to decrease on \((a,c)\) then increase on \((c,b)\). Thus, we must have \(p(c) \lt p(b)\) and so \(p\) has its global minimum value at \(x=c\). Since \(p(a) \gt p(b)\), this makes \(p(a)\) the global maximum value of \(p\) on \([a,b]\).
Let \(s(t) = 3\sin(2(t-\frac{\pi}{6})) + 5\). Find the exact absolute maximum and minimum of \(s\) on the provided intervals by testing the endpoints and finding and evaluating all relevant critical numbers of \(s\).
1. \([\frac{\pi}{6}, \frac{7\pi}{6}]\)
2. \([0, \frac{\pi}{2}]\)
3. \([0, 2\pi]\)
4. \([\frac{\pi}{3}, \frac{5\pi}{6}]\)
Kuratidza mhinduro
1. For \(s(t) = 3\sin(2(t-\frac{\pi}{6})) + 5\), \(s'(t) = 3\cos(2(t-\frac{\pi}{6})) \cdot 2 = 6\cos(2(t-\frac{\pi}{6}))\). It follows that \(s'(t) = 0\) if and only if \(\cos(2(t-\frac{\pi}{6})) = 0\), and thus \(2(t-\frac{\pi}{6}) = \frac{\pi}{2} \pm k\pi\) for some integer value of \(k\). Solving this most recent equation for \(t\), we have \(t-\frac{\pi}{6} = \frac{\pi}{4} \pm \frac{k\pi}{2}\), so \[\begin{aligned}\end{aligned}\]. Now we consider the given interval \([\frac{\pi}{6}, \frac{7\pi}{6}]\), which we can think of as \([\frac{2\pi}{12}, \frac{14\pi}{12}]\). The value \(t = \frac{5\pi}{12}\) lies in this interval, as does \(t = \frac{5\pi}{12} + \frac{\pi}{2} = \frac{11\pi}{12}\). Thus, we have two critical numbers in the interval at which to evaluate \(s\), as well as at the endpoints.
  Doing so, we find:
  \(s(\frac{5\pi}{12}) = 8\);
  \(s(\frac{11\pi}{12}) = 2\);
  \(s(\frac{\pi}{6}) = 5\); and
  \(s(\frac{7\pi}{6}) = 5\).
  The absolute max of \(s\) on \([\frac{\pi}{6}, \frac{7\pi}{6}]\) is thus \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(\frac{11\pi}{12}) = 2\).
2. From our work in (a), we know that \(s\) has critical numbers at \(t = \frac{5\pi}{12} \pm \frac{k\pi}{2}\). On the interval \([0, \frac{\pi}{2}]\), only \(t = \frac{5\pi}{12}\) lies in the interval. Thus we have three values at which to evaluate \(s\).
  Doing so:
  \(s(\frac{5\pi}{12}) = 8\);
  \(s(0) = 3\sin(2(0-\frac{\pi}{6})) + 5 = 3\sin(-\frac{\pi}{3}) + 5 = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\); and
  \(s(\frac{\pi}{2}) = 3\sin(2(\frac{\pi}{2}-\frac{\pi}{6})) + 5 = 3\sin(\frac{2\pi}{3}) + 5 = 5 + \frac{3\sqrt{3}}{2} \approx 7.598\).
  The absolute max of \(s\) on \([0, \frac{\pi}{2}]\) is thus \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(0) = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\).
3. On the interval from \(t = 0\) to \(t = 2\pi\), \(s\) completes nearly two full periods, and thus achieves each of its maximum and minimum values. The absolute maximum is \(8\) as determined in (a), and the absolute minimum is \(2\). These occur, for example, at \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(\frac{11\pi}{12}) = 2\). There are other points at which the function achieves these values on the given interval.
4. On the interval \([\frac{\pi}{3}, \frac{5\pi}{6}] = [\frac{4\pi}{12}, \frac{10\pi}{12}]\), only the critical number \(t = \frac{5\pi}{12}\) lies in the interval. Thus we have three values at which to evaluate \(s\).
  Doing so:
  \(s(\frac{5\pi}{12}) = 8\);
  \(s(\frac{\pi}{3}) = 3\sin(2(\frac{\pi}{3}-\frac{\pi}{6})) + 5 = 3\sin(\frac{\pi}{3}) + 5 = \frac{3\sqrt{3}}{2} + 5 \approx 7.598\); and
  \(s(\frac{5\pi}{6}) = 3\sin(2(\frac{5\pi}{6}-\frac{\pi}{6})) + 5 = 3\sin(\frac{4\pi}{3}) + 5 = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\).
  The absolute max of \(s\) on \([\frac{\pi}{3}, \frac{5\pi}{6}]\) is thus \(s(\frac{5\pi}{12}) = 8\), while the absolute min is \(s(\frac{5\pi}{6}) \approx 2.402\).

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Global optimization

What are the differences between finding relative extreme values and global extreme values of a function?
How is the process of finding the global maximum or minimum of a function over the function's entire domain different from determining the global maximum or minimum on a restricted domain?
For a function that is guaranteed to have both a global maximum and global minimum on a closed, bounded interval, what are the possible points at which these extreme values occur?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Tarisa yako

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

More in Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests