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Write the rule of the function in the form \(\left.f(x)=P e^{k x} . \text { (See the discussion and box after Example } 11 .\right)\) $$f(x)=-2.2\left(.75^{x}\right)$$

Short Answer

Expert verified
Question: Write the given function \(f(x) = -2.2(.75^x)\) in the form \(f(x) = Pe^{kx}\), and identify the values of P and k. Answer: The function can be written as \(f(x) = -2.2e^{x*ln(.75)}\), where \(P = -2.2\) and \(k = ln(.75)\).

Step by step solution

01

Rewrite the function as \(f(x) = Pe^{kx}\)

In order to write the given function in the form \(f(x) = Pe^{kx}\), we need to express the term \(.75^x\) as an exponential function with base \(e\). To do this, we take the natural logarithm (ln) of both sides: We have: $$ .75^x = e^{ln(.75^x)} $$ Now, using logarithmic properties, we can rewrite the equation as: $$ e^{x*ln(.75)} $$ Now, substitute this into the given function, and get: $$ f(x) = -2.2 \times e^{x*ln(.75)} $$
02

Identify P and k

Our function is now in the form \(f(x)=Pe^{kx}\). Comparing the equation with the required format, we have: - \(P = -2.2\) - \(k = ln(.75)\) Our final function in the given format is: $$ f(x) = -2.2e^{x*ln(.75)} $$

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Key Concepts

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

Natural Logarithm
The natural logarithm is a mathematical function commonly denoted as ln, which is the inverse of the exponential function when the base is Euler's number, e (approximately equal to 2.71828). In other words, if we have an equation e^y = x, then the natural logarithm of x is y: ln(x) = y. This function is crucial when dealing with exponential expressions since it allows us to rewrite equations in a form that is easier to manage and understand.

When working with exponential functions that do not have the base e, like .75^x in the given exercise, we use the natural logarithm to convert it. This is accomplished by expressing any other base in terms of e, which is achieved by the property e^{ln(a)} = a. For the exercise, the transformation was made by taking the natural logarithm of 0.75, resulting in the exponent k = ln(.75). This makes the natural logarithm an indispensable tool in simplifying and analyzing the behavior of exponentially varying quantities.
Exponential Growth and Decay
Exponential growth and decay models describe how quantities increase or decrease at a rate that is proportional to their current value. These models are represented by functions of the form f(x) = Pe^{kx}, where:
  • P represents the initial amount or size of the quantity.
  • k is the growth or decay rate. If k > 0, the function represents exponential growth, and if k < 0, it indicates exponential decay.
  • x usually stands for time.
For example, in the context of the exercise, the negative sign of the initial amount P = -2.2 and the fact that k = ln(.75) is negative (because 0.75 is less than 1), suggests the function is modeling exponential decay. This indicates that the quantity represented by the function is decreasing over time.

Exponential functions are widely used in various fields such as biology, for population dynamics, in finance for compound interest, and in physics for radioactive decay. Understanding the nature of these functions helps predict the behavior of systems over time.
Function Transformation
Function transformation involves making algebraic modifications to a function's formula in order to shift, stretch, compress, or reflect its graph. There are various types of transformations:
  • Horizontal shifts occur when we add or subtract a constant from the x variable.
  • Vertical shifts happen when we add or subtract a constant from the whole function.
  • Reflections are when we multiply the function by -1, which flips it across an axis.
  • Stretching or compressing can be performed by multiplying the function by a constant factor.
In relation to the textbook exercise, we see a transformation from the standard form of a decaying exponential function to a form that fits a specific context. The given function f(x) = -2.2×(.75)^x does not obviously show its nature as an exponential function. By rewriting it to match the form f(x) = Pe^{kx}, we apply a transformation that helps to reveal the underlying characteristics of the function, namely its initial value and its rate of decay. This transformed form is more suitable for analysis and applications.

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Most popular questions from this chapter

Find the logarithm, without using a calculator. $$\log \frac{\sqrt{10}}{1000}$$

The population of Mexico was 100.4 million in 2000 and is expected to grow at the rate of \(1.4 \%\) per year. (a) Find the rule of the function \(f\) that gives Mexico's population (in millions) in year \(x,\) with \(x=0\) corresponding to 2000. (b) Estimate Mexico's population in 2010 . (c) When will the population reach 125 million people?

Beef consumption in the United States (in billions of pounds) in year \(x\) can be approximated by the function $$ f(x)=-154.41+39.38 \ln x \quad(x \geq 90) $$ where \(x=90\) corresponds to \(1990 .\) (a) How much beef was consumed in 1999 and in \(2002 ?\) (b) According to this model when will beef consumption reach 35 billion pounds per year?

Determine whether an exponential, power, or logarithmic model (or none or several of these) is appropriate for the data by determining which (if any) of the following sets of points are approximately linear: $$\\{(x, \ln y)\\}, \quad\\{(\ln x, \ln y)\\}, \quad\\{(\ln x, y)\\}$$ where the given data set consists of the points \(\\{(x, y)\\}\) $$\begin{array}{|l|c|c|c|c|c|c|} \hline x & 1 & 3 & 5 & 7 & 9 & 11 \\ \hline y & 2 & 25 & 81 & 175 & 310 & 497 \\ \hline \end{array}$$

Deal with functions of the form \(f(x)=P e^{k x}\) where \(k\) is the continuous exponential growth rate (see Example 6 ). The probability \(P\) percent of having an accident while driving a car is related to the alcohol level of the driver's blood by the formula \(P=e^{k t},\) where \(k\) is a constant. Accident statistics show that the probability of an accident is \(25 \%\) when the blood alcohol level is \(t=.15\). (a) Find \(k .\) IUse \(P=25,\) not .25 .1 (b) At what blood alcohol level is the probability of having an accident \(50 \% ?\)

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