3.1: Twenty problems which embody “3 — 1 = 2”
The answer to every one of the questions in Problem 78 is the same - at least, as a ‘pure number’. The goal is therefore not to “solve” each problem, but to distinguish between, and to reflect upon, the different ways in which the very simple mathematical structure “3 — 1 = 2” turns out to be the relevant “model” in each case.
Problem 78
(a) I was given three apples, and then ate two of them. How many were left?
(b) A barge-pole three metres long stands upright on the bottom of the canal, with one metre protruding above the surface. How deep is the water in the canal?
(c) Tanya said: “I have three more brothers than sisters”. How many more boys are there in Tanya’s family than girls?
(d) How many cuts do you have to make to saw a log into three pieces?
(e) A train was due to arrive one hour ago. We are told that it is three hours late. When can we expect it to arrive?
(f) A brick and a spade weigh the same as three bricks. What is the weight of the spade?
(g) The distance between each successive pair of milestones is 1 mile. I walk from the first milestone to the third one. How far do I walk?
(h) The arithmetic mean (or average) of two numbers is 3. If half their difference is 1, what is the smaller number?
(i) The distance from our house to the train station is 3 km. The distance from our house to Mihnukhin’s house along the same road is 1 km. What is the distance from the station to Mihnukhin’s house?
(j) In one hundred years’ time we will celebrate the tercentenary of our university. How many centuries ago was it founded?
(k) In still water I can swim 3 km in three hours. In the same time a log drifts 1 km downstream in the river. How many kilometres would I be able to swim in the same time travelling upstream in the same river?
(l) December 2 nd fell on a Sunday. How many working days preceded the first Tuesday of that month? 4
4 This question is historically correct. In 1946, in the Soviet Union, when these problems were formulated, Saturday was a working day.
(m) I walk with a speed of 3 km per hour. My friend is some distance ahead of me, and is walking in the same direction pushing his broken down motorbike at 1 km per hour. At what rate is the distance between us diminishing?
(n) A trench 3 km long was dug in a week by three crews of diggers, all working at the same rate as each other. How many such crews would be needed to dig a trench 1 km shorter in the same time?
(o) Moscow and Gorky are cities in adjacent time zones. What is the time in Moscow when it is 3 pm in Gorky? 5
(p) An old ‘rule-of-thumb’ for anti-aircraft gunners stated that: To hit a plane from a stationary anti-aircraft gun, one should aim at a point exactly three plane’s lengths ahead of the moving plane. Now suppose that the gun was actually moving in the same direction as the plane with one third of the plane’s speed. At what point should the gunner aim his fire?
(q) My brother is three times as old as I am. How many times my present age was his age when I was born?
(r) I add 1 to a number and the result is a multiple of 3. What would the remainder be if I were to divide the original number by 3?
(s) It takes 1 minute for a train 1 km long to completely pass a telegraph pole by the track side. At the same speed the train passes right through a tunnel in 3 minutes. What is the length of the tunnel?
(t) Three trams operate on a two-track route, with trams travelling in one direction on one track and returning on the other track. Each tram remains a fixed distance of 3 km behind the tram in front. At a particular moment one tram is exactly 1 km away from the tram on the opposite track. How far is the third tram from its nearest neighbour?


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Study Questions
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What does the term information systems ethics mean?
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What is a code of ethics? What is one advantage and one disadvantage of a code of ethics?
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What does the term intellectual property mean? Give an example.
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What protections are provided by a copyright? How do you obtain one?
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What is fair use?
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What protections are provided by a patent? How do you obtain one?
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What does a trademark protect? How do you obtain one?
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What does the term personally identifiable information mean?
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What protections are provided by HIPAA, COPPA, and FERPA?
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How would you explain the concept of NORA?
Exercises
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Provide one example of how information technology has created an ethical dilemma that would not have existed before the advent of information technology.
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Find an example of a code of ethics or acceptable use policy related to information technology and highlight five points that you think are important.
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Find an example of work done under a CC license.
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Do some original research on the effort to combat patent trolls. Write a two-page paper that discusses this legislation.
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Give an example of how NORA could be used to identify an individual.
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How are intellectual property protections different across the world? Pick two countries and do some original research, then compare the patent and copyright protections offered in those countries to those in the US. Write a two- to three-page paper describing the differences.


27.1: The Chemical Bond in the Hydrogen Molecular Cation
This system has only one electron, but since its geometry is not spherical (figure \(\PageIndex{1}\)), the TISEq cannot be solved analytically as for the hydrogen atom.
The electron is at point \(P\), while the two protons are at position \(A\) and \(B\) at a fixed distance \(R\). Using the Born-Oppenheimer approximation we can write the one-electron molecular Hamiltonian in a.u. as:
\[ \hat{H} = \hat{H}_e+\dfrac{1}{R} = \left( -\dfrac{1}{2}\nabla^2-\dfrac{1}{\mathbf{r}_A}-\dfrac{1}{\mathbf{r}_B} \right)+\dfrac{1}{R} \label{28.1.1} \]
As a first approximation to the variational wave function, we can build the one-electron molecular orbital (MO) by linearly combine two \(1s\) hydrogenic orbitals centered at \(A\) and \(B\), respectively:
\[ \varphi = c_1 a + c_2 b, \label{28.1.2} \]
with:
\[ \begin{equation} \begin{aligned} a &= 1s_A = \left( \psi_{100} \right)_A\\ b &= 1s_B = \left( \psi_{100} \right)_B. \end{aligned} \end{equation}\label{28.1.3} \]
Using Equation 27.3.2 and considering that the nuclei are identical, we can define the integrals \(H_{aa}=H_{bb}, H_{ab}=H_{ba}\) and \(S_{ab}=S\) (while \(S_{aa}=1\) because the hydrogen atom orbitals are normalized). The secular equation, Equation 27.3.4 can then be written:
\[ \begin{vmatrix} H_{aa}-E & H_{ab}-ES \\\ H_{ab}-ES & H_{aa}-E \end{vmatrix}=0 \label{28.1.4} \]
The expansion of the determinant results into:
\[ \begin{equation} \begin{aligned} (H_{aa}-E)^2 &=(H_{ab}-ES)^2 \\ H_{aa}-E &= \pm (H_{ab}-ES), \\ \end{aligned} \end{equation}\label{28.1.5} \]
with roots:
\[ \begin{equation} \begin{aligned} E_{+} &= \dfrac{H_{aa}+H_{ab}}{1+S} = H_{aa}+\dfrac{H_{ba}-SH_{aa}}{1+S}, \\ E_{-} &= \dfrac{H_{aa}-H_{ab}}{1-S} = H_{aa}-\dfrac{H_{ba}-SH_{aa}}{1-S}, \end{aligned} \end{equation}\label{28.1.6} \]
the first corresponding to the ground state, the second to the first excited state. Solving for the best value for the coefficients of the linear combination for the ground state \(E_{+}\), we obtain:
\[ c_1=c_2=\dfrac{1}{\sqrt{2+2S}}, \label{28.1.7} \]
which gives the bonding MO:
\[ \varphi_{+}=\dfrac{a+b}{\sqrt{2+2S}}. \label{28.1.8} \]
Proceeding similarly for the excited state, we obtain:
\[ c_1=\dfrac{1}{\sqrt{2-2S}}\;\quad c_2=-\dfrac{1}{\sqrt{2-2S}}, \label{28.1.9} \]
which gives the antibonding MO:
\[ \varphi_{-}=\dfrac{b-a}{\sqrt{2-2S}}. \label{28.1.10} \]
These results can be summarized in the molecular orbital diagram of figure \(\PageIndex{2}\) We notice that the splitting of the doubly degenerate atomic level under the interaction is non-symmetric for \(S\neq0\), the antibonding level being more repulsive and the bonding less attractive than the symmetric case occurring for \(S = 0\).
Calculating the values for the integrals and repeating these calculations for different internuclear distances, \(R\), results in the plot of figure \(\PageIndex{3}\) As we see from the plots, the ground state solution is negative for a vast portion of the plot. The energy is negative because the electronic energy calculated with the bonding orbital is lower than the nuclear repulsion. In other words, the creation of the molecular orbital stabilizes the molecular configuration versus the isolated fragments (one hydrogen atom and one proton).


9.4: Intergroup Behavior and Performance
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4. What are barriers to intergroup cooperation, and how do you take action to minimize such impediments and understand how to get the most out of the collective actions of groups in organizations in order to enhance industrial competitiveness?
We are now ready to move on to an examination of intergroup behavior. That is, what happens when one group in an organization must interact with another? Clearly, in any corporation, a high degree of intergroup interaction is vital to organizational success. Even in small companies, the production group must interact with the sales group, and both must accommodate the finance and accounting groups. Without smooth intergroup relations, organizational effectiveness and industrial competitiveness are virtually impossible.
Determinants of Intergroup Performance
To understand how groups interact with one another, it is important to identify the primary variables that characterize intergroup behavior.27 We can do this by suggesting a model of intergroup performance. This model is outlined in Exhibit 9.10 . As shown, intergroup behavior occurs when two groups intersect. Each group has its own characteristics and uniqueness, but both operate within the larger confines of organizational policies, culture, reward systems, and so forth. Within this context, performance is largely influenced by three types of interaction requirements: interdependence requirements, information flow requirements, and integration requirements. The quality of intergroup performance is affected by the extent to which all parties to the interaction can meet these requirements.
Interdependence Requirements. Interdependence requirements relate to the frequency and quality of interactions among groups; high-quality interaction is required for successful task accomplishment. To successfully achieve corporate objectives, organizations must achieve enough intergroup interaction to coordinate resource allocation and utilization. The amount of interaction required is determined by the extent and nature of the groups’ interdependence. Group interdependence takes three primary forms (see Exhibit 9.11 ):
- Pooled interdependence. Pooled interdependence occurs when various groups are largely independent of one another, even though each contributes to and is supported by the larger organization. For example, although the physics and music departments may not interact frequently, both contribute to the larger goals of the university, and both use university resources. In a factory setting, pooled interdependence can be seen in a company with two distinct manufacturing divisions; e.g., one for consumer products and one for industrial products. Although produced separately, both kinds of products come together in the shipping department, and both represent products of the same company.
- Sequential interdependence. Sequential interdependence exists when the outputs of one unit or group become the inputs for another. For example, the manufacturing department in a company is clearly dependent on the purchasing department for the success of its own operation, whereas the purchasing department is much less dependent on manufacturing.
- Reciprocal interdependence. Reciprocal interdependence occurs when two or more groups depend on one another for inputs. For example, without product engineering, the marketing department would have nothing to sell. On the other hand, without consumer information from marketing, product engineering might not know what to manufacture. The two units are highly dependent on each other, thereby requiring a high degree of interaction.
In summary, the type of interdependence determines in large part the degree of interrelationship that develops among two or more groups. High interdependence typically requires high intergroup interaction, whereas low interdependence typically requires relatively low intergroup interaction.
Information Flow Requirements. The second requirement for successful intergroup performance is optimal information flow . To be successful, groups need the appropriate amount of information. Information flow is influenced to a large degree by the extent of task uncertainty . When groups are working on highly uncertain tasks (e.g., a new product, an experiment, or an old product in a new environment), the need for communication increases. When task uncertainty is low, less information is typically needed.
Task uncertainty, in turn, is influenced by two factors. The first, task clarity , is the extent to which the requirements and responsibilities of the group are clearly understood. The use of standard operating procedures in organizations is an example of a group requirement. The other consideration is task environment , those factors inside and outside the organization that can affect the group’s performance. The task environment has two aspects: the number of groups that must be dealt with and the relative stability of the environment. Obviously, the more groups that must interact and the more dynamic the environment, the greater the task uncertainty. In a dynamic environment, groups tend to expand their information-gathering efforts to detect and cope with environmental changes. Hence, the greater the task uncertainty, the greater the need for comprehensive information flow systems.
Integration Requirements. The final requirement for successful intergroup performance is integration. Integration requirements focus on the extent of collaboration, cooperation, or structural relationships among groups needed to ensure success. Typically, various departments within an organization have different goals and time orientations. A technical research department, for example, often sees its goals in scientific terms and has a long-term perspective. A marketing department in the same company, focusing its goals on market considerations on the other hand, would typically have a short-term orientation. The production department, concerned with technical goals, would probably attempt to maintain a moderate time orientation in order to take advantage of the economies of scale associated with longer production runs.
A successful organization finds ways to integrate groups so that they coordinate their efforts on behalf of corporate objectives. The trick is to achieve some commonly acceptable coordinating mechanism—not a state in which all units have the same goals and time orientations. It would prove disastrous, for example, if the research unit looked for short-term results or the marketing department ignored short-term shifts in the marketplace. Through integration, various units can accommodate one another’s needs while maintaining their individuality. In this way, the strengths of all groups are used in addressing organizational problems.
When we put these various requirements and their antecedents together, we can see why achieving intergroup coordination and performance is no easy task. Table 9.3 shows the defining characteristics of four typical units of an organization: research, development, sales, and manufacturing. The interdependence, task uncertainty, and time and goal orientation of each unit are shown. Consider the complexities managers face in attempting to lead such an organization efficiently and effectively. Indeed, business magazines are filled with examples of corporate failures that can be traced to poor coordination of such units. These examples point to an endless array of potential sources of conflict that can reduce the capacity of a company to compete successfully in an ever-changing environment.
Managing Intergroup Behavior and Performance
When we analyze the challenge of managing intergroup behavior and performance, the key issue facing managers is the issue of coordination. In most situations, various units or departments in the organization all have talent needed to ensure task accomplishment. Yet, each unit has its own culture, goals, norms, and so forth. Hence, the challenge for managers is harnessing and coordinating this talent in such a way that group harmony is maintained while organizational objectives are achieved.
There are several techniques for managing intergroup relations and performance. These techniques include using rules and procedures, member exchange, linking roles, task forces, and decoupling. Let us briefly consider each as it relates to intergroup coordination and performance.
Rules and Procedures. A common way to manage intergroup relations is for senior management to establish rules and procedures governing the interactions of two or more departments or units. For example, if units consistently fail to communicate with one another, which leads to poor coordination, the company may institute a new policy requiring all groups to post certain types of information at regular time intervals or to inform other department heads of proposed new activities or changes. By simply increasing communication flow, group coordination should increase.
Member Exchange. In some circumstances, it is desirable for the organization to temporarily transfer a member from one group to another. Such exchanges offer the employee an opportunity to better understand the problems and procedures of the other group. Upon returning to his original group, the employee can share information about the other group. In addition, the transferred employee often develops better interpersonal contacts with the other department, thereby enhancing communication and coordination. An example of this can be seen when a company transfers a production engineer into the quality assurance department. As a result, the employee sees firsthand the problems of the quality control group and can take the knowledge back to production engineering.
Linking Roles. A linking role is a position or unit within the organization that is charged with overseeing and coordinating the activities of two or more groups. A good example is a product manager who is responsible for coordinating manufacturing, sales, quality control, and product research for a certain product line (see Exhibit 9.12 ). In essence, these linking role positions are designed to enhance communication among the various functional units and ensure that the right products are designed, manufactured, and marketed. We will say more about the product manager’s role in Understanding and Managing Work Teams.
Task Forces. A task force serves much the same purpose as a linking role except that the role is temporary instead of permanent. In a task force, individuals from several units are brought together to solve a specific problem, usually in a short period of time. It is felt that each unit has expertise to contribute and that by coordinating these efforts, a better solution can be achieved. A typical task force arrangement can be seen in Exhibit 9.14 . For instance, a company facing a major financial cutback may create a task force consisting of members from across the company to identify ways to resolve the crisis. Or a company may create a task force to consider a joint venture offer from a foreign company. In both cases, the problem is that immediate and diverse skills are required to reach an optimal solution.
Decoupling. Finally, there are situations in which two or more closely related groups simply don’t work together effectively. In such cases, decoupling may be the answer. 28 Decoupling involves separating two groups—physically or administratively—in such a way that the required tasks of the organization are fulfilled while the interaction between the two groups is minimized. For instance, hardware and software engineers ideally should work closely together on the design of a new computing system. Yet sometimes these people see problems and solutions quite differently, which may lead to overt hostility and uncooperative behavior. One solution would be to separate the two groups physically and then have one group (e.g., the hardware engineers) outline product specifications. Software engineers then could work more on their own to design software to meet these specifications. Obviously, some coordination would be required. Even so, such an approach could retain the services of two valued groups of engineers who see problems quite differently— a wise compromise strategy for the high-tech company.
In essence, several strategies are available to assist managers in coordinating the diverse talents of interdependent groups in ways that help achieve organizational goals. The choice of an appropriate technique depends upon the unique situation facing the manager. One such approach to managing intergroup coordination was practiced at General Motors Company as it approached the design and manufacture of the Saturn.
expanding around the globe
Engagement on Global Work Teams: IBM
Since 2008, IBM has increased its focus on becoming a globally integrated enterprise. Employing over 200,000 people from different countries and backgrounds, there are major challenges that IBM faces when managing its work teams on such a large global scale.
One of those main components is time zone management. Instead of being highly rigorous in work hours and causing employees to be available on teams at all hours of the night or day, IBM decided they would implement a results-oriented work environment (ROWE). This strategy allows employees to work where they live on virtual teams and base their hours on their own schedules. ROWE allows employees to work at the hours that they feel naturally most productive.
Another key component to managing a global work team is clear communication. IBM structures its leadership of the work teams with leaders that consist of four or five senior executives from multiple geographies. They must work side by side to understand one another’s cultural differences, as well as provide input on their overall team objectives that enable business growth in that country. They are able to accommodate local differences, learn from one another’s differences, and come to common objectives because of their remote location differences for a better outcome. Additionally, these leaders are better equipped to understand local nuances because of their deeper understanding of the global and cultural nuances of their team members.
IBM continues to focus on growing “global IBMers” by offering opportunities for global leadership experiences as well as offering opportunities to acquire new skills. The company focuses on three key actions:
- Grow locally and globally via a consistent methodology. Align business strategies with national priorities and societal goals, build local expertise, and expand market relevance.
- Develop leadership. Provide more employees with opportunities to enhance their skills, and offer more varied global experiences early in careers.
- Enable the global integrated enterprise (GIE) vision. Accelerate enterprise-wide collaboration and an organizational culture based on shared values.
These key actions are clearly communicated by IBM and are demonstrated by leadership to help engage employees behind the methods. Utilizing the best technology to improve collaboration can garner the most productivity and empower employees. If leadership is engaged, employees will also engage with their work and workplace, which helps drive team cohesiveness overall.
Questions:
- What challenges does IBM face due to the size and global reach of its employee base?
- Name at least three strategies that managers and leaders can employ to help keep employees engaged, even when working in remote work teams.
Sources: K. White, “Enabling Growth through Global Enablement Teams,” IBM Thought Leadership , accessed January 3, 2019, https://ibm.com/services/us/gbs/thou...l-leaders.html ; “Enabling Growth through Global Enablement teams,” IBM Thought Leadership , accessed January 3, 2019, https://ibm.com/services/us/gbs/bus/...workforce.html ; T. Neely, “Global Teams that Work,” Harvard Business Review , October 2015, hbr.org/2015/10/global-teams-that- work; D. DeRosa, “3 Companies With High-Performing Teams,” Onpoint Consulting , October 3, 2017, www.onpointconsultingllc.com...-virtual-teams.
concept check
- Are well-functioning teams or groups in complex tasks more productive and leave workers more satisfied than in traditional arrangements?
- What is the importance of the ability to effectively manage both the task requirements and the process or maintenance aspects of the group for them to function well?


11.2: Two angles of a triangle
In this section we will prove a weaker form of Theorem 7.4.1 which holds in any neutral plane.
Let \(\triangle ABC\) be a nondegenerate triangle in the neutral plane. Then
\(|\measuredangle CAB| + |\measuredangle ABC| < \pi.\)
Note that according to Theorem 3.3.1 , the angles \(ABC, BCA\), and \(CAB\) have the same sign. Therefore, in the Euclidean plane the theorem follows immediately from Theorem 7.4.1 .
- Proof
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Let \(X\) be the reflection of \(C\) across the midpoint \(M\) of \([AB]\). By Proposition 7.2.1 \(\measuredangle BAX = \measuredangle ABC\) and therefore
\[\measuredangle CAX \equiv \measuredangle CAB + \measuredangle ABC.\]
Since \([BM]\) and \([MX]\) do not intersect \((CA)\), the points \(B\), \(M\), and \(X\) lie on the same side of \((CA)\). Therefore the angles \(CAB\) and \(CAX\) have the same sign. By Theorem 3.3.1 , the angles \(CAB, ABC\) have the same sign; that is all angles in 11.2.1 have the same sign.
Note that \(\measuredangle CAX \not\equiv \pi\), otherwise \(X\) would lie on \((AC)\). Therefore the identity 11.2.1 implies that
\(|\measuredangle CAB|+|\measuredangle ABC|=|\measuredangle CAX| < \pi.\)
Assume \(A,B,C\), and \(D\) are points in a neutral plane such that
\(2 \cdot \measuredangle ABC + 2 \cdot \measuredangle BCD \equiv 0.\)
Show that \((AB)\parallel (CD)\).
Note that one cannot apply the transversal property ( Theorem 7.3.1 )
- Hint
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Arguing by contradiction, assume \(2 \cdot (\measuredangle ABC + \measuredangle BCD) \equiv 0\), but \((AB) \nparallel (CD)\). Let \(Z\) be the point of intersection of \((AB)\) and \((CD)\).
Note that \(2 \cdot \measuredangle ABC \equiv 2 \cdot \measuredangle ZBC\), and \(2 \cdot \measuredangle BCD \equiv 2 \cdot \measuredangle BCZ\).
Apply Proposition \(\PageIndex{1}\) to \(\triangle ZBC\) and try to arrive at a contradiction.
Prove the side-angle-angle congruence condition in the neutral geometry.
In other words, let \(ABC\) and \(A'B'C'\) be two triangles in a neutral plane; suppose that \(\triangle A'B'C'\) is nondegenerate. Show that \(\triangle ABC \cong \triangle A'B'C'\) if
\(AB = A'B'\), \(\measuredangle ABC = \pm \measuredangle A'B'C'\) and \(\measuredangle BCA = \pm \measuredangle B'C'A'.\)
- Hint
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Let \(C'' \in [B'C')\) be the point such that \(B'C'' = BC\).
Note that by SAS, \(\triangle ABC \cong \triangle A'B'C''\). Conclude that \(\measuredangle B'C'A' = \measuredangle B'C''A'\).
Therefore, it is sufficient to show that \(C'' = C'\). If \(C' \ne C''\) apply Proposition \(\PageIndex{1}\) to \(\triangle A'C'C''\) and try to arrive at a contradiction.
Note that in the Euclidean plane, the above exercise follows from ASA and the theorem on the sum of angles of a triangle ( Theorem 7.4.1 ). However, Theorem 7.4.1 cannot be used here, since its proof uses Axiom V. Later ( Theorem 13.1.1 ) we will show that Theorem 7.4.1 does not hold in a neutral plane.
Assume that the point \(D\) lies between the vertices \(A\) and \(B\) of \(\triangle ABC\) in a neutral plane. Show that
\(CD < CA\) or \(CD < CB\).
- Hint
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Use Exercise 5.2.2 and Proposition \(\PageIndex{1}\).
Alternatively, use the same argument as in the solution of Exercise 5.6.1 .


15.0: Introduction
The mineral fluorite, CaF 2 Figure \(\PageIndex{1}\): , is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. Deposits of fluorite are formed through a process called hydrothermal precipitation in which calcium and fluoride ions dissolved in groundwater combine to produce insoluble CaF 2 in response to some change in solution conditions. For example, a decrease in temperature may trigger fluorite precipitation if its solubility is exceeded at the lower temperature. Because fluoride ion is a weak base, its solubility is also affected by solution pH, and so geologic or other processes that change groundwater pH will also affect the precipitation of fluorite. This chapter extends the equilibrium discussion of other chapters by addressing some additional reaction classes (including precipitation) and systems involving coupled equilibrium reactions.


16.2: Workplace Safety Factors
- Identify factors that contribute to workplace safety
According to OSHA, most successful safety and health programs are based on a common set of key elements that include management leadership, worker participation, and a systematic approach to finding and fixing hazards. OSHA’s Safe and Sound page provides resources to support the development of a Safe + Sound Program and promotes an annual Safe + Sound week (August 12–19 in 2019). Resources include a variety of why and how materials including videos/webinars, tip sheets, course offerings and recommended practices.
There are three core elements of effective safety and health programs:
- Management Leadership
- Worker Participation
- Find & Fix
Let’s take a closer look at each of these elements.
Management Leadership
Executive commitment to implementing a program and using it to drive continuous improvements in safety and health.
Associated action items:
- Developing and communicating a safety and health policy statement.
- Providing the resources needed to implement and operate the program.
- Factoring safety and health into operational planning and decisions.
- Recognizing or rewarding safety and health contributions and achievements.
- Leading by example, by practicing safe behaviors and making safety part of daily conversations.
- Additional perspective and specifics in OSHA’s Be Safe + Sound at Work: Management Leadership Guide
Worker Participation
Effective safety and health programs tap into workers’ collective experience, knowledge, and insight in order to find solutions to workplace safety and health challenges.
Opportunities for worker participation include:
- Developing the initial program design.
- Reporting incidents (including near misses) so they can be investigated.
- Analyzing hazards associated with routine and nonroutine jobs, tasks, and processes.
- Defining and documenting safe work practices.
- Conducting site inspections and incident investigations.
- Training current coworkers and new hires.
- Evaluating program performance and identifying ways to improve it.
- Additional perspective and specifics in OSHA’s Be Safe + Sound at Work: Worker Participation
Find & Fix
Developing a systematic process for identifying and controlling (i.e., finding and fixing) workplace hazards is at the core of every effective safety and health program.
A systematic approach includes:
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- Involving workers, who often have the best understanding of the conditions that create hazards and insights into how they can be controlled.
- Reviewing all available information about hazards that might be present.
- Conducting inspections to identify new or emerging hazards.
- Investigating incidents to identify root causes and potential solutions.
- Evaluating options using the “hierarchy of controls.”
- Considering how to protect workers during emergencies and nonroutine activities.
- Checking that existing controls are intact and remain effective.
- Additional perspective and specifics in OSHA’s Be Safe + Sound at Work: Find & Fix Hazards Resource Guide
Contributors and Attributions
- Authored by : StartupStockPhotos. Provided by : Pixabay. Located at : pixabay.com/photos/startup-meeting-brainstorming-594090/. License : CC0: No Rights Reserved . License Terms : Pixabay License
- Authored by : geralt. Provided by : Pixabay. Located at : pixabay.com/illustrations/feedback-confirming-businessmen-2990424/. License : CC0: No Rights Reserved . License Terms : Pixabay License


2.E: Functions of Several Variables (Exercises)
2.1: Functions of Two or Three Variables
A
2.1.1. \(f (x, y) = x^ 2 + y^ 2 −1\)
2.1.2. \(f (x, y) = \frac{1}{ x^ 2 + y^ 2}\)
2.1.3. \(f (x, y) = \sqrt{ x^ 2 + y^ 2 −4}\)
2.1.4. \(f (x, y) = \frac{x^ 2 +1}{ y}\)
2.1.5. \(f (x, y, z) = \sin (x yz)\)
2.1.6. \(f (x, y, z) = \sqrt{ (x−1)(yz −1)}\)
For Exercises 7-18, evaluate the given limit.
2.1.7. \(\lim\limits_{(x,y)→(0,0)} \cos (x y)\)
2.1.8. \(\lim\limits_{(x,y)→(0,0)} e^{ x y}\)
2.1.9. \(\lim\limits_{(x,y)→(0,0)} \frac{x^ 2 − y^ 2}{ x^ 2 + y^ 2}\)
2.1.10. \(\lim\limits_{(x,y)→(0,0)} \frac{x y^2}{ x^ 2 + y^ 4}\)
2.1.11. \(\lim\limits_{(x,y)→(1,−1)} \frac{x^ 2 −2x y+ y^ 2}{ x− y}\)
2.1.12. \(\lim\limits_{(x,y)→(0,0)} \frac{x y^2}{ x^ 2 + y^ 2}\)
2.1.13. \(\lim\limits_{(x,y)→(1,1)} \frac{x^ 2 − y^ 2}{ x− y}\)
2.1.14. \(\lim\limits_{(x,y)→(0,0)} \frac{x^ 2 −2x y+ y^ 2}{ x− y}\)
2.1.15. \(\lim\limits_{(x,y)→(0,0)} \frac{y^ 4 \sin (x y)}{ x^ 2 + y^ 2}\)
2.1.16. \(\lim\limits_{(x,y)→(0,0)} (x^ 2 + y^ 2 )\cos \left ( \frac{1}{ x y}\right ) \)
2.1.17. \(\lim\limits_{(x,y)→(0,0)} \frac{x}{ y}\)
2.1.18. \(\lim\limits_{(x,y)→(0,0)} \cos \left ( \frac{1}{ x y}\right ) \)
B
2.1.19. Show that \(f (x, y) = \frac{1}{ 2πσ^2} e^{ −(x^ 2+y^ 2 )/2σ^ 2}\), for \(σ > 0\), is constant on the circle of radius \(r > 0\) centered at the origin. This function is called a Gaussian blur , and is used as a filter in image processing software to produce a “blurred” effect.
2.1.20. Suppose that \(f (x, y) ≤ f (y, x) \text{ for all }(x, y)\) in \(\mathbb{R}^ 2\). Show that \(f (x, y) = f (y, x) \text{ for all }(x, y)\) in \(\mathbb{R}^ 2\).
2.1.21. Use the substitution \(r = \sqrt{ x^ 2 + y^ 2}\) to show that
\[\lim\limits_{(x,y)→(0,0)} \frac{\sin \sqrt{ x^ 2 + y^ 2}}{ \sqrt{ x^ 2 + y^ 2}} = 1 .\]
( Hint: You will need to use L’Hôpital’s Rule for single-variable limits. )
C
2.1.22. Prove Theorem 2.1(a) in the case of addition. ( Hint: Use Definition 2.1 .)
2.1.23. Prove Theorem 2.1(b).
2.2: Partial Derivatives
A
For Exercises 1-16, find \(\frac{∂f}{ ∂x} \text{ and }\frac{∂f}{ ∂y}\).
2.2.1. \(f (x, y) = x^ 2 + y^ 2\)
2.2.2. \(f (x, y) = \cos (x+ y)\)
2.2.3. \(f (x, y) = \sqrt{ x^ 2 + y+4}\)
2.2.4. \(f (x, y) = \frac{x+1}{ y+1}\)
2.2.5. \(f (x, y) = e^{ x y} + x y\)
2.2.6. \(f (x, y) = x^ 2 − y^ 2 +6x y+4x−8y+2\)
2.2.7. \(f (x, y) = x^ 4\)
2.2.8. \(f (x, y) = x+2y\)
2.2.9. \(f (x, y) = \sqrt{ x^ 2 + y^ 2}\)
2.2.10. \(f (x, y) = \sin (x+ y)\)
2.2.11. \(f (x, y) = \sqrt[3]{ x^ 2 + y+4}\)
2.2.12. \(f (x, y) = \frac{x y+1}{ x+ y}\)
2.2.13. \(f (x, y) = e^{ −(x^ 2+y^ 2 )}\)
2.2.14. \(f (x, y) = \ln (x y)\)
2.2.15. \(f (x, y) = \sin (x y)\)
2.2.16. \(f (x, y) = \tan (x+ y)\)
For Exercises 17-26, find \(\frac{∂^ 2 f}{ ∂x^ 2} ,\, \frac{∂^ 2 f}{ ∂y^ 2} \text{ and }\frac{∂^ 2 f}{ ∂y∂x}\) (use Exercises 1-8, 14, 15).
2.2.17. \(f (x, y) = x^ 2 + y^ 2\)
2.2.18. \(f (x, y) = \cos (x+ y)\)
2.2.19. \(f (x, y) = \sqrt{ x^ 2 + y+4}\)
2.2.20. \(f (x, y) = \frac{x+1}{ y+1}\)
2.2.21. \(f (x, y) = e^{ x y} + x y\)
2.2.22. \(f (x, y) = x^ 2 − y^ 2 +6x y+4x−8y+2\)
2.2.23. \(f (x, y) = x^ 4\)
2.2.24 . \(f (x, y) = x+2y\)
2.2.25. \(f (x, y) = \ln (x y)\)
2.2.26. \(f (x, y) = \sin (x y)\)
B
2.2.27. Show that the function \(f (x, y) = \sin (x+ y)+\cos (x− y)\) satisfies the wave equation
\[\frac{∂^ 2 f}{ ∂x^ 2} − \frac{∂^ 2 f}{ ∂y^ 2} = 0 .\]
The wave equation is an example of a partial differential equation .
2.2.28 Let \(u \text{ and }v\) be twice-differentiable functions of a single variable, and let \(c \neq 0\) be a constant. Show that \(f (x, y) = u(x+ c y)+v(x− c y)\) is a solution of the general one-dimensional wave equation
\[\frac{∂^ 2 f}{ ∂x^ 2} − \frac{1}{ c^ 2} \frac{∂^ 2 f}{ ∂y^ 2} = 0\]
2.3: Tangent Plane to a Surface
A
For Exercises 1-6, find the equation of the tangent plane to the surface \(z = f (x, y)\) at the point \(P\).
2.3.1. \(f (x, y) = x^ 2 + y^ 3 ,\, P = (1,1,2)\)
2.3.2. \(f (x, y) = x y,\, P = (1,−1,−1) \)
2.3.3. \(f (x, y) = x^ 2 y,\, P = (−1,1,1)\)
2.3.4. \(f (x, y) = xe^ y ,\, P = (1,0,1)\)
2.3.5. \(f (x, y) = x+2y,\, P = (2,1,4)\)
2.3.6. \(f (x, y) = \sqrt{ x^ 2 + y^ 2},\, P = (3,4,5)\)
For Exercises 7-10, find the equation of the tangent plane to the given surface at the point \(P\).
2.3.7. \(\frac{x^ 2}{ 4} + \frac{y^ 2}{ 9} + \frac{z^ 2}{ 16} = 1,\, P = \left ( 1,2, \frac{2 \sqrt{ 11}}{ 3} \right ) \)
2.3.8. \(x^ 2 + y^ 2 + z^ 2 = 9,\, P = (0,0,3)\)
2.3.9. \(x^ 2 + y^ 2 − z^ 2 = 0,\, P = (3,4,5)\)
2.3.10. \(x^ 2 + y^ 2 = 4,\, P = ( \sqrt{ 3},1,0)\)
2.4: Directional Derivatives and the Gradient
A
For Exercises 1-10, compute the gradient \(∇f\).
2.4.1. \(f (x, y) = x^ 2 + y^ 2 −1\)
2.4.2. \(f (x, y) = \frac{1}{ x^ 2 + y^ 2}\)
2.4.3. \(f (x, y) = \sqrt{ x^ 2 + y^ 2 +4}\)
2.4.4. \(f (x, y) = x^ 2 e^ y\)
2.4.5. \(f (x, y) = \ln (x y)\)
2.4.6. \(f (x, y) = 2x+5y\)
2.4.7. \(f (x, y, z) = \sin (x yz)\)
2.4.8. \(f (x, y, z) = x^ 2 e^{ yz}\)
2.4.9. \(f (x, y, z) = x^ 2 + y^ 2 + z^ 2\)
2.4.10. \(f (x, y, z) = \sqrt{ x^ 2 + y^ 2 + z^ 2}\)
For Exercises 11-14, find the directional derivative of \(f\) at the point \(P\) in the direction of \(v = \left ( \frac{1}{ \sqrt{ 2}} , \frac{1}{ \sqrt{ 2}} \right ) \).
2.4.11. \(f (x, y) = x^ 2 + y^ 2 −1,\, P = (1,1)\)
2.4.12. \(f (x, y) = \frac{1}{ x^ 2 + y^ 2} ,\, P = (1,1)\)
2.4.13. \(f (x, y) = \sqrt{ x^ 2 + y^ 2 +4},\, P = (1,1)\)
2.4.14. \(f (x, y) = x^ 2 e^ y ,\, P = (1,1)\)
For Exercises 15-16, find the directional derivative of \(f\) at the point \(P\) in the direction of \(v = \left ( \frac{1}{ \sqrt{ 3}} , \frac{1}{ \sqrt{ 3}} , \frac{1}{ \sqrt{ 3}} \right ) \).
2.4.15. \(f (x, y, z) = \sin (x yz),\, P = (1,1,1)\)
2.4.16. \(f (x, y, z) = x^ 2 e^{ yz} ,\, P = (1,1,1)\)
2.4.17. Repeat Example 2.16 at the point \((2,3)\).
2.4.18. Repeat Example 2.17 at the point \((3,1,2)\).
B
For Exercises 19-26, let \(f (x, y) \text{ and }g(x, y)\) be continuously differentiable real-valued functions, let \(c\) be a constant, and let \(v\) be a unit vector in \(\mathbb{R}^ 2\). Show that:
2.4.19. \(∇(c f ) = c∇f \)
2.4.20. \(∇(f + g) = ∇f + ∇g\)
2.4.21. \(∇(f g) = f ∇g + g∇f\)
2.4.22. \(∇(f /g) = \frac{g∇f − f ∇g}{ g^ 2}\text{ if }g(x, y) \neq 0\)
2.4.23. \(D_{−v} f = −D_v f\)
2.4.24. \(D_v(c f ) = c D_v f\)
2.4.25. \(D_v(f + g) = D_v f + D_v g\)
2.4.26. \(D_v(f g) = f D_v g + g D_v f\)
2.4.27. The function \(r(x, y) = \sqrt{ x^ 2 + y^ 2}\) is the length of the position vector \(\textbf{r} = x\textbf{i} + y\textbf{j}\) for each point \((x, y)\) in \(\mathbb{R}^ 2\). Show that \(∇r = \frac{1}{ r} \textbf{r}\) when \((x, y) \neq (0,0)\), and that \(∇(r^ 2 ) = 2\textbf{r}\).
2.5: Maxima and Minima
A
For Exercises 1-10, find all local maxima and minima of the function \(f (x, y)\).
2.5.1. \(f (x, y) = x^ 3 −3x+ y^ 2\)
2.5.2. \(f (x, y) = x^ 3 −12x+ y^ 2 +8y\)
2.5.3. \(f (x, y) = x^ 3 −3x+ y^ 3 −3y\)
2.5.4. \(f (x, y) = x^ 3 +3x^ 2 + y^ 3 −3y^ 2\)
2.5.5. \(f (x, y) = 2x^ 3 +6x y+3y^ 2\)
2.5.6. \(f (x, y) = 2x^ 3 −6x y+ y^ 2\)
2.5.7. \(f (x, y) = \sqrt{ x^ 2 + y^ 2}\)
2.5.8. \(f (x, y) = x+2y\)
2.5.9. \(f (x, y) = 4x^ 2 −4x y+2y^ 2 +10x−6y\)
2.5.10. \(f (x, y) = −4x^ 2 +4x y−2y^ 2 +16x−12y\)
B
2.5.11. For a rectangular solid of volume 1000 cubic meters, find the dimensions that will minimize the surface area. ( Hint: Use the volume condition to write the surface area as a function of just two variables. )
2.5.12. Prove that if \((a,b)\) is a local maximum or local minimum point for a smooth function \(f (x, y)\), then the tangent plane to the surface \(z = f (x, y)\) at the point \((a,b, f (a,b))\) is parallel to the \(x y\)-plane. ( Hint: Use Theorem 2.5 .)
C
2.5.13. Find three positive numbers \(x, y, z\) whose sum is 10 such that \(x^ 2 y^ 2 z\) is a maximum.
2.6: Unconstrained Optimization: Numerical Methods
C
2.6.1. Recall Example 2.21 from the previous section, where we showed that the point \((2,1)\) was a global minimum for the function \(f (x, y) = (x −2)^4 +(x −2y)^ 2\). Notice that our computer program can be modified fairly easily to use this function (just change the return values in the fx, fy, fxx, fyy and fxy function definitions to use the appropriate partial derivative). Either modify that program or write one of your own in a programming language of your choice to show that Newton’s algorithm does lead to the point \((2,1)\). First use the initial point \((0,3)\), then use the initial point \((3,2)\), and compare the results. Make sure that your program attempts to do 100 iterations of the algorithm. Did anything strange happen when your program ran? If so, how do you explain it? ( Hint: Something strange should happen. )
2.6.2. There is a version of Newton’s algorithm for solving a system of two equations
\[f_1(x, y) = 0 \quad \text{ and }\quad f_2(x, y) = 0 ,\]
where \(f_1(x, y) \text{ and }f_2(x, y)\) are smooth real-valued functions:
Pick an initial point \((x_0 , y_0)\). For \(n = 0,1,2,3,...,\) define:
\[x_{n+1} = x_n - \frac{\begin{vmatrix} f_1(x_n, y_n) & f_2(x_n, y_n) \\ \frac{∂f_1}{ ∂y} (x_n, y_n) & \frac{∂f_2}{ ∂y} (x_n, y_n) \\ \end{vmatrix}}{D(x_n, y_n)},\quad y_{n+1} = y_n + \frac{\begin{vmatrix} f_1(x_n, y_n) & f_2(x_n, y_n) \\ \frac{∂f_1}{ ∂x} (x_n, y_n) & \frac{∂f_2}{ ∂x} (x_n, y_n) \\ \end{vmatrix}}{D(x_n, y_n)},\text{ where} \]
\[D(x_n, y_n) = \frac{∂f_1}{ ∂x} (x_n, y_n) \frac{∂f_2}{ ∂y} (x_n, y_n)− \frac{∂f_1}{ ∂y} (x_n, y_n) \frac{∂f_2}{ ∂x} (x_n, y_n) .\]
Then the sequence of points \((x_n, y_n)_{n=1}^{\infty}\) converges to a solution. Write a computer program that uses this algorithm to find approximate solutions to the system of equations
\[f_1(x, y) = \sin (x y)− x− y = 0 \quad \text{ and }\quad f_2(x, y) = e^{ 2x} −2x+3y = 0 .\]
Show that you get two different solutions when using \((0,0) \text{ and }(1,1)\) for the initial point \((x_0 , y_0)\).
2.7: Constrained Optimization: Lagrange Multipliers
A
2.7.1. Find the constrained maxima and minima of \(f (x, y) = 2x+ y\) given that \(x^ 2 + y^ 2 = 4\).
2.7.2. Find the constrained maxima and minima of \(f (x, y) = x y\) given that \(x^ 2 +3y^ 2 = 6\).
2.7.3. Find the points on the circle \(x^ 2+ y^ 2 = 100\) which are closest to and farthest from the point \((2,3)\).
B
2.7.4. Find the constrained maxima and minima of \(f (x, y, z) = x + y^ 2 +2z\) given that \(4x^ 2 +9y^ 2 − 36z^ 2 = 36\).
2.7.5. Find the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid
\[\frac{x^ 2}{ a^ 2} + \frac{y^ 2}{ b^ 2} + \frac{z^ 2}{ c^ 2} = 1 .\]


4.7: Domain and Range of a Function
The domain of a function is all possible values of x that can be used as input to the function, which will result in a real number as the output. The range of a function is the set of all possible output values of a function.
Find the domain and range of the following function:
\(f(x) = 5x + 3 \)
Solution
Any real number, negative, positive or zero can be replaced with x in the given function. Therefore, the domain of the function \(f(x) = 5x + 3 \) is all real numbers, or as written in interval notation, is: \(D:(−\infty , \infty )\). Because the function \(f(x) = 5x + 3\) is a polynomial of degree 1, it is a straight line (without any breaks or holes).
The range of any polynomial of degree 1 is all real numbers or written in interval notation, is: \(R:(−\infty , \infty )\).
Find the domain and range of the following function:
\(g(x) = 2\sqrt{ x − 4}\)
Solution
Pay attention to the square root part of this function. The radicand (what’s inside the square root) must be nonnegative. Set the radicand greater than or equal to zero to find the domain:
\(\begin{aligned} x − 4 &\geq 0 && \text{Set the radicand greater than or equal to 0 }\\ x &\geq 4 &&\text{ Solve the inequality } \\ D&:[4, \infty ) &&\text{Write the solution in interval notation }\end{aligned}\)
Therefore, the domain of the function \(g(x) = 2\sqrt{ x − 4}\) is all real numbers in the interval from \([4, \infty )\), which is written \(D:[4, \infty )\).
To find the range of \(g(x) = 2\sqrt{ x − 4}\), let’s observe the behavior of the function for different values of x that are in the domain.
Let \(x = 4\), \(g(4) = 2\sqrt{ 4 − 4}\), so \(g(4) = 0\).
Let \(x = 5\), \(g(5) = 2\sqrt{ 5 − 4}\), so \(g(5) = 2\).
Let \(x = 8\), \(g(8) = 2\sqrt{ 8 − 4}\), so \(g(8) = 4\).
Any non-negative value chosen for x will result in a non-negative value for \(g(x)\). The function values for the range (the output from the function \(g(x)\)) are non-negative numbers, written as \(R:[0, \infty )\).
Find the domain and range of the following function:
\(h(x) = −2x^2 + 4x − 9\)
Solution
Any real number, negative, positive or zero can replace x in the given function.
Therefore, the domain of the function \(h(x) = 2x^2 + 4x − 9\) is all real numbers, or as written in interval notation, is: \(D:(−\infty , \infty )\).
Because the function \(h(x) = 2x^2 + 4x − 9\) is a quadratic of degree 2, when graphed, it is a parabola (without any breaks or holes). Identify two things about this parabola:
- Which way does it open, up or down? and
- Where is the vertex?
The sign of the coefficient of the leading term of the quadratic function (\(2x^2\) ) shows which way the parabola opens. The coefficient is 2, and since it’s positive, the quadratic function opens upward.
Now find the vertex. The y-value of the vertex ordered pair will show where the range begins.
The vertex is \(\left(−\dfrac{b}{2a} , f\left( −\dfrac{b}{2a} \right)\right)\), with \(a = 2\) and \(b = 4\).
The vertex is \(\left(− \dfrac{4 }{2∗2} , f \left(− \dfrac{4 }{2∗2}\right)\right)\)
The vertex is \((− 1, f(− 1))\), which is \((− 1, 2 ∗ (−1)^2 − 9))\) or \((− 1, −11)\)
The range will start at −11, and continue to increase, since the parabola opens upward. \(R:[-11, \infty)\)
Find the domain and range of the following function:
\(j(x) = \vert z − 6 \vert − 3\)
Solution
This function contains an absolute value. Any value can be chosen for \(z\), so the domain of the function is all real numbers, or as written in interval notation, is: \(D:(−\infty , \infty )\)
To find the range, examine inside the absolute value symbols. This quantity, \(\vert z−6 \vert\) will always be either 0 or a positive number, for any values of z. First, find what makes the expression z−6 equal to zero, which is the number 6.
\(\begin{aligned} j(x) &= \vert z − 6 \vert − 3 &&\text{ Original function } \\ j(x) &= \vert 6 − 6 \vert − 3 && \text{Replace z with 6 } \\j(x) &= \vert 0 \vert − 3 && \text{Simplify } \\ j(x) &= −3 && j(x) \text{ is } −3 \end{aligned}\)
Therefore, the range of the function \(j(x) = \vert z − 6 \vert − 3\) is −3 or higher, or as written in interval notation, is: \(R:[-3, \infty)\)
Some types of functions are more difficult to work with. Here are some examples of functions where the domain can be found but the range would be too difficult to find, and out of the scope of this course:
Find the domain of the following functions:
\(f(x) = \dfrac{x − 4 }{x^2 − 2x − 15 }\)
Solution
With any rational function (a quotient of polynomials), be aware of division by 0. Set the denominator polynomial equal to 0 and solve.
\(\begin{aligned} x^2 − 2x − 15 &= 0 &&\text{Set the denominator function equal to } 0 \\ (x − 5)(x + 3) &= 0 &&\text{Factor the quadratic equation } \\ x − 5 &= 0 && \text{Set the first binomial factor equal to zero } \\ x &= 5 &&\text{Solve the first binomial factor } \\ x + 3 &= 0 &&\text{Set the second binomial factor equal to zero } \\ x &= −3 &&\text{Solve the second binomial factor} \end{aligned}\)
There are two solutions to the quadratic equation, 5 and −3.
These values must be excluded from the domain, because if \(x\) is either 5 or −3, the denominator will equal zero.
Division by zero is undefined. The domain of the function \(f(x) = x − 4 x^2 − 2x − 15\) is \((−\infty , −3) \cup (−3, −5) \cup (−5, \infty )\).
Find the domain of the following function:
\(g(x) = \dfrac{x }{x^2 − 9}\)
Solution
Once again this is a rational function, and the concern is to avoid division by 0. Set the denominator function equal to 0 and solve.
\(\begin{aligned} x^{2}-9&=0 && \text { Set the denominator function equal to } 0 \\ (x-3)(x+3)&=0 &&\text{Factor the quadratic equation} \\ x-3&=0 && \text{Set the first binomial factor equal to zero} \\ x&=3 &&\text{Solve the first binomial factor}\\ x+3&=0 && \text{Set the second binomial factor equal to zero} \\ x&=-3 && \text{Solve the second binomial factor} \end{aligned}\)
There are two solutions to the quadratic equation, 3 and −3. These values must be excluded from the domain, because if \(x\) is either 3 or −3, the denominator will equal zero. Division by zero is undefined. The domain of the function \(g(x) =\dfrac{ x}{ x^2 − 9 }\) is \((−\infty , −3) \cup (−3, 3) \cup (3, \infty )\).
Find the domain of the following function:
\(g(t) = \sqrt{6 + t − t^2}\)
Solution
The radicand of this square root function must be non-negative. Set the radicand greater than or equal to 0 and solve.
\(\begin{aligned} 6 + t − t^2 &\geq 0 &&\text{Set the radicand equal to }0 \\ −t^2 + t + 6 &\geq 0 &&\text{Rewrite the function with the leading term first } \\ (−t + 3)(t + 2) &= 0 && \text{Factor the quadratic equation } \\−t + 3 &= 0 && \text{Set the first binomial factor equal to zero } \\ t &= 3 && \text{Solve the first binomial factor } \\ t + 2 &= 0 &&\text{Set the second binomial factor equal to zero } \\ t &= −2 &&\text{Solve the second binomial factor} \end{aligned}\)
There are two values that will make the radicand of this square root function zero, 3 and −2.
Since the radicand must be non-negative, test the regions between the found solutions.
If \(x < −2\), for example, −4, \(g(−4) = \sqrt{6 + (−4) − (−4)^2}\) is negative, which is not allowed for the radicand.
If \(x\) is between −2 and 3, for example, 0, \(g(0) = \sqrt{6 + (0) − (0)^2}\) is positive. This region between −2 and 3 will be in the domain of the function.
There is one more region to check, where \(x > 3\). Let \(x = 4\). \(g(4) = \sqrt{ 6 + (4) − (4)^2}\) is negative, which is not allowed for the radicand. The domain of the function \(g(t) = \sqrt{6 + t − t^2}\) is \([−2, 3]\)
Find the domain and range of the following functions:
- \(f(x) = x ^2 − 8x + 12\)
- \(g(x) = \sqrt{x + 10}\)
- \(h(x) = \vert − 2x + 1\vert\)
Find the domain of the following functions:
- \(f(x) = \dfrac{6x + 7 }{5x + 2}\)
- \(f(x) = \dfrac{2x }{2x^2 + 3x − 20}\)
- \(f(x) =\dfrac{ 4x + 11 }{x^2 + 6x + 9}\)
- \(f(x) = \dfrac{3x }{x^2 − 5x − 14}\)
- \(f(x) = \dfrac{2x + 1}{ 6x^2 − x − 2}\)
- \(f(x) = \dfrac{−6 }{25x^2 − 4}\)


1.11: Ceremonial Speaking- Identification, Magnification, Narration
Identification
Identification
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Words have incredible power.
They can make people’s hearts soar,
or they can make people’s hearts sore.
– Dr. Mardy Grothe
psychologist and author
The category of speeches that you are most likely to have to give is ceremonial speeches. Ceremonial speaking, originally called epideictic oratory, includes graduation speeches, wedding speeches, eulogy speeches, after-dinner speeches, award speeches, toast speeches, and tribute speeches. The biggest thing that these have in common is that they have a heavy emphasis on pathos-emotion. When listening to a ceremonial speech, an audience expects to feel something and to feel a part of something. To achieve that, a good ceremonial speech will include identification, narration, and magnification.
In many ways, special occasions are the punctuation marks of life.
They are the rituals that draw us together
in celebration of some person or event
Teri and Michael Gamble
communication scholars
This chapter will look at identification, magnification, and narration and how they function in ceremonial speeches to help you prepare to construct a speech of your own.
In all speeches, the audience should feel like you are considering their needs, this is especially true in a ceremonial speech. In a ceremonial speech, your audience in some way becomes part of the speech. To achieve this, you should use a lot of phrases like “we” and “our.” In the graduation speech by Matthew McConaughey, he uses “we,” and “you” words over 400 times–that is a lot of identification!
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He uses “you” nearly 250 times and “we” 170 times in a 45-minute speech
Fun fact: Matthew McConaughey is a professor of practice at the Moody College of Communication at the University of Texas at Austin. https://moody.utexas.edu/faculty/matthew-mcconaughey
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Here are a few ways that my students used identification in their speeches.
The Survivor Tree, an Elm, still stands as a witness of that day as a symbol of our resilience. The same kind of resilience all Americans have when faced with a tough obstacle to overcome. Tribute to OK City Memorial
My grandpa Jack is like the wind. We can’t see him but we can feel him. He is always with us .
Lily, Tribute to Grandpa JackToday, we gather here to honor a famous, and beloved man, Stanley Martin Lieber. W e all knew him as Stan Lee
He brought us together
He may be gone but we will not forget what for he did for us .
David Lester, Tribute to Stan Lee
Tip for Identification
When I do one-on-one speech consultations. I pull up two or three empty chairs and then I ask the person who I am coaching to tell me who is sitting in the chair and listening to their speech. I ask them to show me how that person is included in the speech. I read parts of their speech out loud with them thinking about people in those chairs and then we talk about how we can reword some things to make sure those folks feel included.
In addition to identification, ceremonial speeches should contain magnification. Magnification is where you take an attribute of a person, and you magnify it and make it seem larger than life. It is where you highlight a positive attribute of a person or group. This works best if you tap into values that the audience share.
These are the most common magnification themes:
It is easiest to understand how these work in a speech by way of example. Watch the following speeches to understand how magnification can be achieved.
Magnification is where you make an action larger than life. Notice how Oprah Winfrey elaborates on the action of Rosa Parks and makes it large and heroic.
I grew up in the South, and Rosa Parks was a hero to me long before I recognized and understood the power and impact that her life embodied. I remember my father telling me about this colored woman who had refused to give up her seat. And in my child’s mind, I thought, “She must be really big.” I thought she must be at least a hundred feet tall. I imagined her being stalwart and strong and carrying a shield to hold back the white folks.
…So I thank you again, Sister Rosa, for not only confronting the one white man who[se] seat you took, not only confronting the bus driver, not only for confronting the law, but for confronting history, a history that for 400 years said that you were not even worthy of a glance, certainly no consideration. I thank you for not moving.

Before Mohammad Ali passed, he struggled with tremors. In this eulogy speech, Former President Clinton describes watching him light the Olympic flame. The mere act of walking was a simple thing that was made great by the circumstances.
In addition to identification and magnification, ceremonial speeches should include stories.
This example is from a University of Arkansas student speech and he is telling the story of picking out his first dog. Notice how the small details help to draw us in and help us to see the event unfold. I left this sample in speech format so you could see how it was written on his manuscript to have a feeling for the rhythm of the speech. (/// slanted lines, signal a pause…the more lines, the longer the pause)
It was a nice day and I can still picture you running towards me
when I really try that is
As we approach the house,///
three little girls
accompanied by their father
open a shed door to the side of the property
I rode my Heelys /// down a small
//concrete sidewalk
Excited to see the little balls of fluff
My parents said to pick a good one
So I picked you /// because you bit my shoe.
Bryan Stevenson, author of Just Mercy: A Story of Justice and Redemption gave this acceptance speech when winning the Andrew Carnegie Medal for Excellence in Fiction and NonFiction. Notice how his use of narration makes us feel something about his grandmother and then we carry those feelings as he talks about racial justice.
I had a very close relationship with my grandmother. My grandmother was the daughter of people who were enslaved. Her parents were born into slavery in Virginia in the 1840s. She was born in the 1880s, and the only thing that my grandmother insisted that I know about her enslaved father is that he learned to read before emancipation, and that reading is a pathway to survival and success. So I learned to read. I put books and words in my head and in my heart, so that I could get to the places that she needed me to go.
I’m thinking about my grandmother tonight, because she had these qualities about her. She was like lots of African American matriarchs. She was the real force in our family. She was the end of every argument. She was also the beginning of a lot of arguments! She was tough, and she was strong but she was also kind and loving. When I was a little boy, she’d give me these hugs, she’d squeeze me so tightly I could barely breathe. And then she’d see me an hour later and she’d say, “Bryan, do you still feel me hugging you?” And if I said no, she would assault me again!
Did you notice the magnification (the greatness of a simple thing) in learning to read and in hugging?
Did you notice his use of identification by saying, “She was like lots of African American matriarchs?”
Speaking of identification, check out how he uses “we”
We need to change the narrative in this country about race, and poverty. We’re a country that has a difficult time dealing with our shame, our mistakes. We don’t do shame very well in America, and because of that we allow a lot of horrific things to go unaddressed.
Read the whole speech here:
https://www.publishersweekly.com/pw/by-topic/industry-news/awards-and-prizes/article/67546-is-this-the-greatest-book-award-acceptance-speech-ever.html
You can use identification, magnification, and narration in all your speeches, but these three elements are especially important in ceremonial type speeches. Maya Angelou said, “People will forget what you said, people will forget what you did, but people will never forget how you made them feel.”
Remember This!
I want to hear from you.
Do you have an activity to include?
Did you notice a typo that I should correct?
Are you planning to use this as a resource and do you want me to know about it?
Do you want to tell me something that really helped you?
Albanese, A. (2015). Is this the greatest book award acceptance speech ever? Publisher’s Weekly. https://www.publishersweekly.com/pw/by-topic/industry-news/awards-and-prizes/article/67546-is-this-the-greatest-book-award-acceptance-speech-ever.html
Clinton, B. ( 2020). Bill Clinton’s eulogy for Muhammad Ali. [Video]. YouTube. https://www.youtube.com/watch?v=feGjy5bKMb0 Standard YouTube License.
Gamble, T. & Gamble, M.W. (1998). Public Speaking in the Age of Diversity. Allyn and Bacon.
Justis, T. (2019). Photo of graduation speaker, Anna Francis Chandler.
Lester, D. Tribute to Stan Lee. University of Arkansas.
McConaughey, M. (2016). Matthew McConaughey University of Houston Speech. https://www.youtube.com/watch?v=BmCTQ_mkzHU
Rigsby, R. (2017). The most inspiring speech: The wisdom of a third-grade dropout will change your life. [Video]. YouTube. https://www.youtube.com/watch?v=Bg_Q7KYWG1g Standard YouTube License.
Samovar, L. A., & McDaniel, E. R. (2007). Public speaking in a multicultural society . Los Angeles, CA: Roxbury.
Saylor Academy (2012) . Stand up speak out: The practice and ethics of public speaking. Saylor Academy.
Snippe, E. (2016). 101 quotes to inspire speakers. Angelou, M. Quote. https://speakerhub.com/blog/101-quotes
Thibodeau, P.H and Boroditsky, L. (2011). Metaphors we think with: The role of metaphor in reasoning. Stanford University.
Winfrey, O. (2010). Oprah eulogy [Video]. YouTube. https://www.youtube.com/watch?v=5cfhtfNfIPE&t=1s Standard YouTube License.


2.5: Limits
In Section 1A, we saw how to go from a position graph to a velocity graph. However, the graphs we were dealing with were piecewise linear, which made it very easy to find the velocities, or the slopes. If the position graphs are not piecewise linear, it is more difficult to find the slope at a given point on the graph.
There is a very nice way of doing this for many functions, but if we’re not careful, it will require division by zero! What does being careful entail? It means knowing your limits!
Numerical Limits
Suppose you wanted to evaluate the function \(f(x) = \frac{x^3 - 8}{x - 2}\) at \(x = 2\). Plugging in \(x = 2\) into the \(f(x)\) formula gives
\(\frac{2^3 - 8}{2 - 2} = \frac{0}{0}\)
But there is a problem — we’ve divided by zero.
(image credit: Jaggery)
Hopefully you didn’t actually do the division by zero. What can we do instead? Let’s make a table to see what happens when we get close to putting in two, without actually doing it.
| x | f(x) |
|---|---|
| 1.5 | 9.25 |
| 1.9 | 11.41 |
| 1.99 | 11.9401 |
| 1.999 | 11.994 |
| 1.9999 | 11.9994 |
| 2.0001 | 12.0006 |
| 2.001 | 12.006 |
| 2.01 | 12.0601 |
| 2.1 | 12.61 |
| 2.5 | 15.25 |
If you look at the table, it looks like \(f(x)\) SHOULD be equal to \(12\) at \(x = 2\). We can’t plug in \(x = 2\) because of division by zero, but it really should be \(12\) if it has a value. Can we just say it’s 12 and call it a day? Well, not quite, since we want to distinguish between functions that are actually equal to 12, and ones that just should be. That’s where limits come in.
We say \(f(2)\) is undefined, but we can write \(\lim_{x \to 2} f(x) = 12\), which we read as “the limit of \(f\) of \(x\) as \(x\) approaches two is twelve”. That example is the idea behind a limit.
More technically, a limit is a value \(L\) the \(y\)-value of a function \(f(x)\) approaches as the \(x\)-value approaches a certain value \(a\), either from the right, the left or both. Notationally, \(\lim_{x \to a^-} f(x) = L\) is the left hand limit, \(\lim_{x \to a^+} f(x) = L\) is the right hand limit, and \(\lim_{x \to a} f(x) = L\) is the two-sided limit. Let’s look at some pictures to make this more intuitive.
Graphical Limits
Consider the following \(f(x)\).
To find the value of a function, recall that you look at how high that function is at a given \(x\) value. For example, \(f(1)\) is about \(2\), and \(f(3)\) is about \(10\).
What’s happening at \(x = 2\), or \(f(2)\)? The filled-in circle shows the function value. The white circle indicates that value is not part of the function. So \(f(2) = 8\).
However, there is something funny going on at \(x = 2\). Namely, the function seems to “jump from \(4\), stop at \(8\) momentarily, then finally jump to \(12\) and continue. This is called a discontinuity , and is usually a bad thing, or at least something that can be hard to deal with. This is where limits can be helpful. The notation \(\lim_{x \to 2^-} f(x)\) indicates the \(y\)-value of the function as \(x\) approaches the value of \(2\) from the left. Here is the picture:
In other words, \(\lim_{x \to 2^-} f(x)\) is the value that \(f(x)\) should be at \(2\), if you were approaching from the left. In this case, \(f(x)\) should be 4 if everything were right in the world. Therefore, \(\lim_{x \to 2^-} f(x) = 4\). The limit allows us to fill in what the function should be, even though it isn’t the case.
The right handed limit \(\lim_{x \to 2^+} f(x)\) is the same way, but it approaches from the right.
Here, the function approaches \(12\) as we approach \(2\) from the right, and therefore we write \(\lim_{x \to 2^+} = 12\).
A two-sided limit exists if \(\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)\). In other words, if the left and right limits are the same. The notation for this is \(\lim_{x \to a} f(x)\). Since the left and right limits are different, so we just write “Does Not Exist” or “DNE”. So \(\lim_{x \to 2} f(x) = \text{DNE}\).
Other Examples
For each value of \(a\), find \(f(a)\), \(\lim_{x \to a^-} f(x)\), \(\lim_{x \to a^+} f(x)\), \(\lim_{x \to a} f(x)\).
- \(a = -5\)
- \(a = -3\)
- \(a = -1\)
- \(a = 1\)
- \(a = 3\)
- \(a = 5\)
- \(a = \infty\)
- \(a = -\infty\)
- We are looking at the function at \(x = -5\). The actual value here is \(f(-5) = 3\). However, the value it should be if we were coming at \(-5\) from the left is \(-2\), so \(\lim_{x \to -5^-} f(x) = -2\). Coming from the right, it should be \(3\), so \(\lim_{x \to -5^+} f(x) = 3\). Finally, \(\lim_{x \to -5} f(x) = \text{DNE}\), since the left and right limits are not equal.
- We are looking at the function at \(x = -3\). We have \(f(-3) = -1\), since that’s where the black dot is. But the value SHOULD be \(2\), whether we approach from the right or left. So \(\lim_{x \to -3^-} f(x) = \lim_{x \to -3^+} f(x) = 2\). Since the left and right limits are equal, we have \(\lim_{x \to -3} f(x) = 2\).
- We have \(f(-1) = 5\), \(\lim_{x \to -1^-} f(x) = 4\), \(\lim_{x \to -1^+} f(x) = 1\), and \(\lim_{x \to -1} f(x) = \text{DNE}\).
- Everything is nice and happy — there are no discontinuities here. Therefore \(f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = \lim_{x \to 1} f(x) = 0\).
- We have what is called a vertical asymptote, and the function basically goes off infinitely far in both directions. Here, we say \(f(3) = \text{DNE}\), since there is no obvious value to make \(f(3)\) equal to. We say \(\lim_{x \to 3^-} f(x) = \infty\), since it goes up infinitely high (the \(\infty\) symbol means “infinity”). Likewise, \(\lim_{x \to 3^+} f(x) = -\infty\), since it goes down infinitely low. Therefore, we can see \(\lim_{x \to 3} f(x)\) does not exist.
- We have \(f(5) = 3\), \(\lim_{x \to 5^-} f(x) = 3\), \(\lim_{x \to 5^+} f(x) = -2\), and \(\lim_{x \to 5} f(x)= \text{DNE}\).
- We have \(x = \infty\). What we are asking here is what happens to \(f(x)\) as \(x\) gets really big. In other words, what happens to the function as \(x\) goes to infinity? Well, it looks like perhaps \(f(x)\) is just heading towards zero. So we say \(\lim_{x \to \infty} f(x) = 0\). There is no such thing as a right hand or two-sided limit in this case, nor does it make sense to talk about \(f(\infty)\). So we just leave it as \(\lim_{x \to \infty} f(x) = 0\). This is the same thing as a horizontal asymptote.
- Similarly we are looking at what happens to \(f(x)\) when \(x\) goes more and more negative. It looks like maybe \(f(x)\) is heading towards \(1\), so we write \(\lim_{x \to -\infty} f(x) = 1\).