Problem 83 - A Graphical Approach to College Algebra

Question

Assume that \(f(x)=a^{x},\) where \(a>1\) If \(f^{-1}\) exists, find an equation for \(y=f^{-1}(x),\) using the method described in Section \(5.1 .\) (You need not solve for \(y .)\)

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \log_a(x) \).

Step 1: Understand the Function

We start by recognizing that the function given is an exponential function of the form \( f(x) = a^x \) where \( a > 1 \). This function is one-to-one and hence has an inverse.

Step 2: Define the Inverse Function

Since \( f(x) = a^x \), the inverse function \( f^{-1}(x) \) will swap the roles of \( x \) and \( y \). Thus, we take \( y = a^x \) and swap to look for \( x \) in terms of \( y \).

Step 3: Write the Inverse Equation

Starting from \( y = a^x \), we need to express \( x \) as the subject. This requires using logarithms since they are the inverse of exponentiation. Taking the logarithm base \( a \) of both sides, we get \( \log_a(y) = x \).

Step 4: Express the Inverse Function

For the inverse function \( f^{-1}(x) \), we now have \( f^{-1}(x) = \log_a(x) \). Therefore, we have expressed the inverse of the function \( f(x) = a^x \) as \( f^{-1}(x) = \log_a(x) \).

Key Concepts

Exponential Functions

In mathematics, exponential functions are critical in understanding growth and decay processes. The general form of an exponential function is given by \( f(x) = a^x \), where \( a \) is a positive constant, and \( a > 1 \). This type of function features a constant base raised to a variable exponent. Exponential functions have several unique properties that make them powerful:

• They are continuous and defined for all real numbers.
• As \( x \) increases, \( f(x) \) grows rapidly without bound if \( a > 1 \).
• They have a horizontal asymptote at \( y = 0 \).

An exponential function's shape on a graph is a smooth curve that rises sharply, showcasing rapid growth.Exponential functions are unique because they are one-to-one, meaning they pass the horizontal line test. This characteristic makes finding an inverse possible. Understanding exponential functions is essential because they model real-world scenarios such as population growth, radioactive decay, and compound interest.

Logarithms

Logarithms are the inverse operations of exponentiation, just as subtraction is the inverse of addition. If you have an exponential equation \( y = a^x \), solving for \( x \) involves logarithms. The equation can be rewritten using logarithms as \( x = \log_a(y) \). Here, \( \log_a(y) \) represents the logarithm of \( y \) with base \( a \). Here are some properties of logarithms:

• \( \log_a(a^x) = x \)
• \( a^{\log_a(x)} = x \)

The primary role of logarithms is to simplify calculations involving exponential growth and decay by reducing multiplicative problems into additive ones. Logarithms also provide a convenient way to handle the inverses of exponential functions. For exponential functions like \( f(x) = a^x \), where \( a > 1 \), the inverse function is expressed as \( f^{-1}(x) = \log_a(x) \). This inverse relationship underlies the connection between exponential and logarithmic functions.

One-to-One Functions

One-to-one functions, also called injective functions, have a distinct output for each unique input. In simpler terms, no two different inputs produce the same output. This characteristic is essential because it ensures that each output is linked to a single input, which is necessary for the existence of an inverse function.Here is what defines one-to-one functions:

• If \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).
• The function passes the horizontal line test on its graph, meaning any horizontal line cuts the curve at most once.

Exponential functions such as \( f(x) = a^x \) (where \( a > 1 \)) are classic examples of one-to-one functions. This exclusivity paves the way for finding unique inverse functions.For example, with \( f(x) = a^x \), we can find its inverse, \( f^{-1}(x) = \log_a(x) \), because the function is one-to-one. Having a firm grasp of one-to-one functions helps in solving equations that involve finding inverse functions, ensuring clarity and accuracy in mathematical processes.