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Use a graphing utility to graph \(f(x)=\left(1+\frac{0.5}{x}\right)^{x} \quad\) and \(\quad g(x)=e^{0.5}\) in the same viewing window. What is the relationship between \(f\) and \(g\) as \(x\) increases and decreases without bound?

Short Answer

Expert verified
As \(x\) increases and goes to positive infinity, \(f(x)\) approaches \(g(x)=e^{0.5}\). However, as \(x\) decreases without bound (or goes to negative infinity), \(f(x)\) diverges or moves away from \(g(x)\).

Step by step solution

01

Graph the Function \(f(x)\)

To graph the function \(f(x)=\left(1+\frac{0.5}{x}\right)^{x}\), a graphing calculator or software that can plot mathematical functions needs to be used. Input the function into the graphing tool and plot the graph.
02

Graph the Function \(g(x)\)

The next function to graph is \(g(x)=e^{0.5}\). As before, input the function \(g(x)\) into the graphing utility and plot the function. As this function is a constant, its graph will be a horizontal line in the same viewing window as the function \(f(x)\).
03

Observing The Relationships Between Two Functions

The exercise requires investigating the relation between \(f\) and \(g\) as \(x\) increases and decreases without bound. By observing the graphs, it is seen that as \(x\) becomes larger and goes to positive infinity, function \(f(x)\) appears to approach the line represented by \(g(x)\). Furthermore, as \(x\) goes to negative infinity, the function \(f(x)\) appears to diverge or move away from \(g(x)\). This can be interpreted as a limit where \(\lim_{x\to\infty} f(x) = e^{0.5}\), and as \(x\) goes to negative infinity, \(f(x)\) does not approach a finite number.

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Key Concepts

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

Understanding the Exponential Function
Exponential functions are a class of mathematical functions that allow us to model growth or decay processes. The general form of an exponential function is given by \[ f(x) = a^x \]where the base \(a\) is a constant. When a positive constant base is raised to varying powers, it produces a curve that has distinct characteristics:
- If \(a > 1\), the function models exponential growth.- If \(0 < a < 1\), the function models exponential decay.In the context of the given exercise, the function \(g(x) = e^{0.5}\) is a specific type of exponential function where the base is Euler's number \(e\), approximately 2.71828. This constant is crucial in calculus as it describes continuous growth processes. Even though \(g(x)\) appears as a constant in this exercise (specifically due to no variable component in its power), it represents a horizontal line at \(e^{0.5}\).
The function \(f(x) = \left(1+\frac{0.5}{x}\right)^{x}\) takes a slightly different form but behaves similarly to an exponential function, approaching \(g(x)\) as \(x\) increases. The expression \(\left(1 + \frac{0.5}{x}\right)^x\) is a key part in understanding limits and approximation of exponentials in calculus.
Graphing Functions
Graphing functions is a visual way to understand their behavior and relationships. In the provided exercise, two functions are graphed together to observe their interaction.
1. **Graphing \(f(x)\):** - Use a graphing calculator or software to plot \(f(x) = \left(1+\frac{0.5}{x}\right)^x\). - As \(x\) approaches infinity, observe how \(f(x)\) behaves.2. **Graphing \(g(x)\):** - Plot \(g(x) = e^{0.5}\), which is a constant function. Therefore, its graph is a horizontal line. - Notice the interaction between \(f(x)\) and \(g(x)\) as \(x\) changes.
Graphing these functions assists in visually identifying specific patterns, particularly how a function behaves as \(x\) grows larger or smaller. This method is a valuable tool in calculus as it provides insights into the function’s limits and possible points of intersection with other functions.
The Concept of Behavior at Infinity
When we talk about behavior at infinity in calculus, we are interested in understanding what happens to a function as the input value, \(x\), becomes extremely large or small.
- **As \(x\) increases to positive infinity:** - The function \(f(x) = \left(1+\frac{0.5}{x}\right)^{x}\) demonstrates a tendency to approach the value \(e^{0.5}\). - The limit can be mathematically expressed as \(\lim_{x\to\infty} f(x) = e^{0.5}\).- **As \(x\) decreases to negative infinity:** - Unlike the positive direction, \(f(x)\) behaves differently and does not settle near a particular value. - This phenomenon indicates that \(f(x)\) diverges, and is not constrained by \(g(x) = e^{0.5}\) when \(x\) is negative.
Understanding these behaviors helps in predicting and analyzing how functions compare to fixed values as they extend to the extreme ends of their domains. This concept is fundamental in limits and continuity, providing essential insights for both theoretical and practical applications in calculus.

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Most popular questions from this chapter

Home Mortgage \(A \$ 120,000\) home mortgage for 30 years at \(7 \frac{1}{2} \%\) has a monthly payment of \(\$ 839.06\) Part of the monthly payment covers the interest charge on the unpaid balance, and the remainder of the payment reduces the principal. The amount paid toward the interest is $$u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}$$ and the amount paid toward the reduction of the principal is $$v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}$$ In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time (in years). (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 30 years of mortgage payments.) (b) In the early years of the mortgage, is the greater part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71) .\) What can you conclude?

In Exercises \(97-102,\) determine whether the statement is true or false given that \(f(x)=\ln x .\) Justify your answer. $$\sqrt{f(x)}=\frac{1}{2} f(x)$$

Use a graphing utility to graph the functions \(y_{1}=\ln x-\ln (x-3)\) and \(y_{2}=\ln \frac{x}{x-3}\) in the same viewing window. Does the graphing utility show the functions with the same domain? If so, should it? Explain your reasoning.

Data Analysis The table shows the time \(t\) (in seconds) required for a car to attain a speed of \(s\) miles per hour from a standing start. $$\begin{array}{|c|c|} \hline \text { Speed, S } & \text { Time, t } \\ \hline 30 & 3.4 \\ 40 & 5.0 \\ 50 & 7.0 \\ 60 & 9.3 \\ 70 & 12.0 \\ 80 & 15.8 \\ 90 & 20.0 \\ \hline \end{array}$$ Two models for these data are as follows. \(t_{1}=40.757+0.556 s-15.817 \ln s\) \(t_{2}=1.2259+0.0023 s^{2}\) (a) Use the regression feature of a graphing utility to find a linear model \(t_{3}\) and an exponential model \(t_{4}\) for the data. (b) Use the graphing utility to graph the data and each model in the same viewing window. (c) Create a table comparing the data with estimates obtained from each model. (d) Use the results of part (c) to find the sum of the absolute values of the differences between the data and the estimated values given by each model. Based on the four sums, which model do you think best fits the data? Explain.

Home Mortgage The total interest \(u\) paid on a home mortgage of \(P\) dollars at interest rate \(r\) for \(t\) years is $$u=P\left[\frac{r t}{1-\left(\frac{1}{1+r / 12}\right)^{12 t}}-1\right]$$ Consider a \(\$ 120,000\) home mortgage at \(7 \frac{1}{2} \%\) (a) Use a graphing utility to graph the total interest function. (b) Approximate the length of the mortgage for which the total interest paid is the same as the size of the mortgage. Is it possible that some people are paying twice as much in interest charges as the size of the mortgage?

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