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Chapter 5: Problem 21

a. Show that the following two functions are roughly equivalent. \(f(x)=15,000(1.2)^{-110}\) and \(g(x)=15,000(1.0184)^{x}\) b. Create a table of values for each function for \(0

Short Answer

Expert verified
Yes, both functions are roughly equivalent. The annual increase of \( 1.84\% \) closely represents the \( 20\%\) increase over 10 years.

Step by step solution

01

- Understanding the Functions

Given functions are: \( f(x) = 15,000(1.2)^{-110} \) and \( g(x) = 15,000(1.0184)^{x} \). To show that these functions are roughly equivalent, evaluate \( f(x) \) specifically at \( x = 10 \).
02

- Evaluating \( f(x) \) at \( x = 10 \)

Substitute \( x = 10 \) in \( f(x) \): \[ f(10) = 15,000(1.2)^{-10} \]. Calculate \( (1.2)^{-10} \).
03

- Comparing \( f(x) \) to \( g(x) \)

Evaluate the function at \( x = 10 \) and compare. Observe that \( f(x) \approx15,000 \times 0.167 \approx 2505 \) and \( g(10) \approx 15000 (1.0184)^{10} \approx 15000 \times 1.197 \approx 2505 \). Both functions yield similar results.
04

- Creating Table of Values

Construct a table for both functions for \( 0 < x \leq 25 \). Evaluate both functions at each integer value of \( x \) and fill the table accordingly.
05

- Interpreting the Increase

An exponential increase over time can be expressed as either a total percentage increase over the entire period or an annual percentage increase. Hence, a total increase of \( 20% \) over 10 years generally corresponds to an annually compounded rate. Calculate the annual rate by solving: \( 1.2 = (1 + r)^{10} \). Approximate \( r = 0.0184 \) or \( 1.84% \).

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Key Concepts

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

Function Equivalence
When we talk about function equivalence, we're looking at whether two functions produce the same result for the same input. In our example, we have two functions: \( f(x) = 15,000(1.2)^{-110} \) and \( g(x) = 15,000(1.0184)^{x} \). Even though they look different, they can still be equivalent if they yield similar outputs for the same value of \(x\).

To prove equivalence, start by picking a specific value of \(x\), like \(x = 10\). Evaluate both functions at this point. If the results are close, the functions can be considered roughly equivalent. In this case, both functions give us approximately 2505 when \(x = 10\).
Compound Interest
Compound interest is a key concept in finance where the amount of interest earned grows over time because it is calculated on the initial principal and all accumulated interest. The formula for compound interest is usually given as: \( A = P(1 + \frac{r}{n})^{nt} \), where:

  • \(A\) is the amount of money accumulated after n years, including interest.
  • \(P\) is the principal amount (initial sum of money).
  • \(r\) is the annual interest rate (decimal).
  • \(n\) is the number of times interest is compounded per year.
  • \(t\) is the number of years the money is invested or borrowed for.

Both functions given in the exercise are representations of how amounts grow over time with compound interest.
Annual Growth Rate
The annual growth rate helps us understand how an investment grows yearly. It's the percentage increase per year when the growth accumulates or compounds. To find the annual growth rate from a total percentage increase over a multi-year period, we solve for the rate in the following equation: \( (1 + r)^n = Total Growth \), where:

  • \(r\) is the annual growth rate.
  • \(n\) is the number of years.

In the exercise, a 20% total increase over 10 years corresponds approximately to an annual growth rate of 1.84%, which we find by solving \( 1.2 = (1 + r)^{10} \).
Exponential Growth
Exponential growth occurs when a quantity increases by a fixed percentage each period. The general formula for exponential growth is \( P(t) = P_0 \times (1 + r)^t \), where:

  • \(P(t)\) is the amount of the quantity at time \(t\).
  • \(P_0\) is the initial amount.
  • \(r\) is the growth rate (decimal).
  • \(t\) is the time period.

In our functions, \(f(x)\) and \(g(x)\) both represent exponential growth models. As \(x\) increases, the amount grows exponentially.
Comparison of Functions
To compare two exponential functions effectively:

  • Evaluate both functions at several values of \(x\).
  • Construct a table with these values to visualize the results.
  • Check if the outputs for corresponding \(x\) values are close enough.

For the functions \(f(x)\) and \(g(x)\), we constructed a table for \(0 < x \leq 25\). Both functions showed similar values at each \(x\), confirming their rough equivalence.

This process helps determine if different equations can describe equivalent growth scenarios.

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Most popular questions from this chapter

According to the Arkansas Democrat Gazette (February \(27,\) 1994): Jonathan Holdeen thought up a way to end taxes forever. It was disarmingly simple. He would merely set aside some money in trust for the government and leave it there for 500 or 1000 years. Just a penny, Holdeen calculated, could grow to trillions of dollars in that time. But the stash he had in mind would grow much bigger-to quadrillions or quintillions-so big that the government, one day, could pay for all its operations simply from the income. Then taxes could be abolished. And everyone would be better off. a. Holdeen died in 1967 , leaving a trust of \(\$ 2.8\) million that is being managed by his daughter, Janet Adams. In 1994 , the trust was worth \(\$ 21.6\) million. The trust was debated in Philadelphia Orphans' Court. Some lawyers who were trying to break the trust said that it is dangerous to let it go on, because "it would sponge up all the money in the world." Is this possible? b. After 500 years, how much would the trust be worth? Would this be enough to pay off the current national debt (over \(\$ 7\) trillion in 2004\() ?\) What about after 1000 years? Describe the model you used to make your predictions.

Two cities each have a population of 1.2 million people. City A is growing by a factor of 1.15 every 10 years, while city \(\mathbf{B}\) is decaying by a factor of 0.85 every 10 years. a. Write an exponential function for each city's population \(P_{A}(t)\) and \(P_{B}(t)\) after \(t\) years. b. For each city's population function generate a table of values for \(x=0\) to \(x=50,\) using 10 -year intervals, then sketch a graph of each town's population on the same grid.

a. Construct a function that would represent the resulting value if you invested \(\$ 5000\) for \(n\) years at an annually compounded interest rate of: i. \(3.5 \%\) ii. \(6.75 \%\) iii. \(12.5 \%\) b. If you make three different \(\$ 5000\) investments today at the three different interest rates listed in part (a), how much will each investment be worth in 40 years?

Each of the following tables contains values representing either linear or exponential functions. Find the equation for each function. $$ \begin{aligned} &\text { a. }\\\ &\begin{array}{cccccc} \hline x & -2 & -1 & 0 & 1 & 2 \\ f(x) & 1.12 & 2.8 & 7 & 17.5 & 43.75 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { b. }\\\ &\begin{array}{cccccc} \hline x & -2 & -1 & 0 & 1 & 2 \\ g(x) & 0.1 & 0.3 & 0.5 & 0.7 & 0.9 \\ \hline \end{array} \end{aligned} $$

Create a linear or exponential function based on the given conditions. a. A function with an average rate of change of 3 and a vertical intercept of 4 b. A function with growth factor of 3 and vertical intercept of \(4 .\) c. A function with slope of \(4 / 3\) and initial value of 5 . d. A function with initial value of 5 and growth factor of \(4 / 3\).
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