/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Table 2.4 gives \(P=f(t),\) the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Table 2.4 gives \(P=f(t),\) the number of households, in millions, in the US with cable television \(t\) years since \(1998^{4}\) (a) Does \(f^{\prime}(4)\) appear to be positive or negative? What does this tell you about the percent of households with cable television? (b) Estimate \(f^{\prime}(2) .\) Estimate \(f^{\prime}(10) .\) Explain what each is telling you, in terms of cable television. $$\begin{array}{c|c|c|c|c|c|c|c}\hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\\\\hline P & 64.65 & 66.25 & 66.732 & 65.727 & 65.141 & 64.874 & 60.958 \\\\\hline\end{array}$$

Short Answer

Expert verified
(a) Negative. Cable households decreased at t=4. (b) f'(2) ≈ 0.5205; f'(10) ≈ -1.04575. Increasing at t=2, decreasing at t=10.

Step by step solution

01

Evaluate sign of f'(4)

To determine if \(f^{\prime}(4)\) is positive or negative, compare the values of \(f(4)\) and \(f(6)\). \(P\) decreases from \(66.732\) at \(t=4\) to \(65.727\) at \(t=6\), indicating \(f^{\prime}(4)\) is negative. This implies the number of households with cable television is decreasing at \(t=4\).
02

Estimate f'(2)

To estimate \(f^{\prime}(2)\), compute the slope of the tangent line at \(t=2\) using points \((t=0, f(0)=64.65)\) and \((t=4, f(4)=66.732)\).\[f^{\prime}(2) \approx \frac{f(4) - f(0)}{4 - 0} = \frac{66.732 - 64.65}{4} = \frac{2.082}{4} = 0.5205\]This tells us the number of households with cable was increasing at a rate of about 0.5205 million per year at \(t=2\).
03

Estimate f'(10)

To estimate \(f^{\prime}(10)\), compute the slope of the tangent line at \(t=10\) using points \((t=8, f(8)=65.141)\) and \((t=12, f(12)=60.958)\).\[f^{\prime}(10) \approx \frac{f(12) - f(8)}{12 - 8} = \frac{60.958 - 65.141}{4} = \frac{-4.183}{4} = -1.04575\]This indicates the number of households with cable was decreasing at a rate of approximately 1.04575 million per year at \(t=10\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative Interpretation
In calculus, the derivative of a function provides crucial insight into how the function behaves at any given point. When we talk about the derivative, we are essentially referring to the rate at which one quantity changes with respect to another. This is a foundational concept in understanding real-world changes and patterns in data.

For example, in the context of the exercise, the derivative of the function \(f(t)\) represents the rate of change of the number of households with cable television with respect to time. A positive derivative indicates an increase in the number of households over time, while a negative derivative signals a decrease.

Understanding whether the derivative is positive or negative at a certain point can help infer trends. Such as, \(f^{\prime}(4)\) being negative tells us that, around the fourth year, the number of households with cable began to decline. These interpretations can guide businesses and policymakers in decision-making related to cable television distribution strategies.
Slope Estimation
Estimating the slope of a function over different intervals can provide insights into how quickly or slowly a change is occurring. In applied calculus, this method is often used to predict trends and analyze scenarios over specific periods.

In the given problem, slope estimation was used to find \(f^{\prime}(2)\) and \(f^{\prime}(10)\). To estimate these, we looked at the difference in function values over the chosen intervals and calculated the average rate of change. For \(f^{\prime}(2)\), the calculation involved comparing from \(t=0\) to \(t=4\). The estimate of 0.5205 suggested that during this time, the number of households was increasing annually by approximately 0.5205 million. Conversely, \(f^{\prime}(10)\) showed a decrease of about 1.04575 million households per year between years 8 and 12.

Such estimations are vital in the context of business to foresee how trends might evolve and to plan accordingly. It simplifies complex data into manageable and interpretable information, which aids in strategic planning and forecasting.
Rates of Change
Rates of change are at the heart of calculus and provide essential information about trends and shifts in real-world phenomena. A rate of change can help us understand not just the existence of change, but its direction and magnitude.

In the context of cable television households, the rates of change tell us how quickly the market is expanding or contracting. When we see \(f^{\prime}(2)\) as an increasing rate of about 0.5205 million households per year, it indicates a period of growth. On the other hand, the negative rate of \(f^{\prime}(10)\) by approximately 1.04575 million households per year marks a period of decline.

Understanding these rates allows stakeholders to set goals and develop strategies. For instance, if the decline is due to emerging technologies, companies might need to innovate to retain customers. Rates of change help in identifying such pivotal shifts in trends, ensuring that timely adaptations can be made to evolving market demands.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Yesterday's temperature at \(t\) hours past midnight was \(f(t)^{\circ} \mathrm{C} .\) At noon the temperature was \(20^{\circ} \mathrm{C} .\) The first derivative, \(f^{\prime}(t),\) decreased all morning, reaching a low of \(2^{\circ} \mathrm{C} /\) hour at noon, then increased for the rest of the day. Which one of the following must be correct? (a) The temperature fell in the morning and rose in the afternoon. (b) At 1 pm the temperature was \(18^{\circ} \mathrm{C}\) (c) At 1 pm the temperature was \(22^{\circ} \mathrm{C}\) (d) The temperature was lower at noon than at any other time. (e) The temperature rose all day.

The table gives the number of passenger cars, \(C=f(t)\) in millions, \(^{24}\) in the US in the year \(t\) (a) Do \(f^{\prime}(t)\) and \(f^{\prime \prime}(t)\) appear to be positive or negative during the period \(1975-1990 ?\) (b) Do \(f^{\prime}(t)\) and \(f^{\prime \prime}(t)\) appear to be positive or negative during the period \(1990-2000 ?\) (c) Estimate \(f^{\prime}(2005) .\) Using units, interpret your answer in terms of passenger cars. $$\begin{array}{c|c|c|c|c|c|c|c}\hline t & 1975 & 1980 & 1985 & 1990 & 1995 & 2000 & 2005 \\ \hline C & 106.7 & 121.6 & 127.9 & 133.7 & 128.4 & 133.6 & 136.6 \\\\\hline\end{array}$$

In \(2011,\) the Greenland Ice Sheet was melting at a rate between 82 and 224 cubic km per year. \(^{15}\) (a) What derivative does this tell us about? Define the function and give units for each variable. (b) What numerical statement can you make about the derivative? Give units.

The cost, \(C=f(w),\) in dollars of buying a chemical is a function of the weight bought, \(w,\) in pounds. (a) In the statement \(f(12)=5,\) what are the units of the 12? What are the units of the 5? Explain what this is saying about the cost of buying the chemical. (b) Do you expect the derivative \(f^{\prime}\) to be positive or negative? Why? (c) In the statement \(f^{\prime}(12)=0.4,\) what are the units of the \(12 ?\) What are the units of the \(0.4 ?\) Explain what this is saying about the cost of buying the chemical.

The average weight, \(W,\) in pounds, of an adult is a function, \(W=f(c),\) of the average number of Calories per day, \(c,\) consumed. (a) Interpret the statements \(f(1800)=155\) and \(f^{\prime}(2000)=0\) in terms of diet and weight. (b) What are the units of \(f^{\prime}(c)=d W / d c ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.