Problem 73 - Precalculus Mathematics for Calculus

Question

Find the inverse function of \(f\). $$f(x)=\log _{2}(x-1)$$

Step-by-Step Solution

Answer

The inverse function of \( f(x) = \log_{2}(x-1) \) is \( f^{-1}(x) = 2^x + 1 \).

Step 1: Understand the Problem

The function \( f(x) = \log_2(x-1) \) needs to be reversed to express \( x \) in terms of \( f(x) \). The goal is to find \( f^{-1}(x) \).

Step 2: Substitute \( f(x) \) with \( y \)

To simplify the inverse finding process, substitute \( f(x) \) with \( y \). So, we have:\[y = \log_2(x-1)\]

Step 3: Switch \( x \) and \( y \)

To find the inverse, switch the \( x \) and \( y \) variables:\[x = \log_2(y-1)\]

Step 4: Convert Logarithmic Equation to Exponential Form

Convert the logarithmic equation into exponential form to solve for \( y \):\[y-1 = 2^x\]

Step 5: Solve for \( y \)

Now solve for \( y \) in terms of \( x \):\[ y = 2^x + 1 \]

Step 6: Finalize the Inverse Function

The inverse function is obtained:\[f^{-1}(x) = 2^x + 1\]

Key Concepts

Logarithmic Functions

Logarithmic functions are essential in solving inverse problems, like converting logarithms to exponentials and vice versa. A logarithm asks, "To what power must the base be raised, to produce a given number?" For example, in the function \( f(x) = \log_2(x-1) \), the base is 2. This specific logarithmic function can be interpreted as asking "What power do we raise 2 to, in order to get \( x-1 \)?"

Logarithmic functions have several properties:

• The logarithm of a number is the exponent to which the base must be raised to obtain that number.
• They accompany exponential functions and are typically used to "undo" or reverse exponentiations.
• The graph of a logarithmic function is the inverse of its corresponding exponential graph.
• When the base is greater than one, the function is increasing. If the base is a fraction, the function decreases.

Understanding logarithms helps in finding the inverse of a function, as seen in our exercise where we transformed the logarithmic equation into an exponential one to find the inverse.

Exponential Functions

Exponential functions are the counterparts of logarithmic functions. In exponential functions, the base is a constant, and the exponent is the variable. This is typically expressed as \( y = a^x \), where \( a \) is the base. They are vital in calculations involving growth and decay, such as in finance, science, and engineering.

An exponential function grows rapidly. With our given inverse, \( f^{-1}(x) = 2^x + 1 \), this indicates that as \( x \) increases, the function value grows exponentially.

• Exponential functions generally have a horizontal asymptote, typically along the x-axis.
• They illustrate rapid increase or decrease with increasing or decreasing \( x \) values respectively.
• The base \( b \) should be greater than zero and not equal to one since any number raised to zero is one, which isn't useful in these contexts.

This concept of exponential growth is crucial in many mathematical models and helps understand how fast a quantity can grow over time.

Function Notation

Function notation is a way to represent functions in mathematics, showing the relationship between input and output. Typically, functions are expressed as \( f(x) \), where \( f \) denotes the function, and \( x \) is the variable or input. For example, in the equation \( f(x) = \log_2(x-1) \), \( f \) is the function, \( x \) is the input, and \( \log_2(x-1) \) is the output.

This notation is useful because:

• It allows for easy substitution of different inputs to see resulting outputs.
• It provides clarity on which variable is the independent one.
• It helps in composing functions and finding inverses.

In our exercise, function notation plays a crucial role in expressing and finding the inverse. Here, by swapping \( f(x) \) with \( y \), finding the inverse is simplified. The final result \( f^{-1}(x) = 2^x + 1 \) indicates the original input has been expressed as an output, demonstrating the efficiency and importance of function notation in algebraic manipulation and communication.