Problem 40 - Contemporary Precalculus

Question

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

Step-by-Step Solution

Answer

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 1: 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: $$ f(x) = -2.2 \times e^{x*ln(.75)} $$

Step 2: 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: $$ f(x) = -2.2e^{x*ln(.75)} $$

Key Concepts

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.