Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(11-16\), find the inverse of the given function \(f\) and verify that \(f\left(f^{-1}(x)\right)=x\) for all \(x\) in the domain of \(f^{-1}\), and \(f^{-1}(f(x))=x\) for all \(x\) in the domain of \(f\). $$ f(x)=\log _{2}\left(\frac{x+1}{2 x}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Function

The given function is \( f(x) = \log_2\left( \frac{x+1}{2x} \right) \). It is a logarithmic function with base 2.

Step 2: Set up the Equation for Inversion

To find the inverse, we need to solve the equation \( y = \log_2\left( \frac{x+1}{2x} \right) \). This is equivalent to \( 2^y = \frac{x+1}{2x} \).

Step 3: Solve for \( x \)

From \( 2^y = \frac{x+1}{2x} \), multiply both sides by \( 2x \): \( 2x \, 2^y = x + 1 \). Rearrange terms to get: \( x(2 \, 2^y - 1) = 1 \). Solve for \( x \): \( x = \frac{1}{2 \, 2^y - 1} \).

Step 4: Expression for the Inverse Function

Replace \( y \) with \( x \) to express \( f^{-1}(x) \): \( f^{-1}(x) = \frac{1}{2^{x+1} - 1} \).

Step 5: Verify Inverse Relationship \( f\left(f^{-1}(x)\right) = x \)

Substitute \( f^{-1}(x) \) back into \( f \): \( f\left( \frac{1}{2^{x+1} - 1} \right) = \log_2 \left( \frac{\frac{1}{2^{x+1} - 1} + 1}{2 \cdot \frac{1}{2^{x+1} - 1}} \right) \). This simplifies to \( \log_2(x) = x \). Thus, \( f\left(f^{-1}(x)\right) = x \).

Step 6: Verify Inverse Relationship \( f^{-1}(f(x)) = x \)

Substitute \( f(x) \) back into \( f^{-1} \): \( f^{-1}(\log_2(\frac{x+1}{2x})) = \frac{1}{2^{\log_2(\frac{x+1}{2x})+1} - 1}\). Simplifying gives \( f^{-1}(f(x)) = x \).

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse operations of exponential functions. They are essential in mathematical expressions, allowing us to transform equations and solve problems related to growth and decay. In these functions, the logarithm answers the question: "What power must the base be raised to yield a given number?" For example, in the function \(f(x) = \log_2(x)\), we are seeking the power to which 2 must be raised to produce \(x\).

In our specific problem, we have a slightly more sophisticated function: \(f(x) = \log_2\left( \frac{x+1}{2x} \right)\). It requires us to understand the transformation from the fraction \(\frac{x+1}{2x}\) into a power of 2. This transformation allows us to manipulate complex functions more easily by leveraging the properties of logarithms, such as the ability to split products and quotients into sums and differences. This characteristic is especially useful when dealing with inverse functions and verifying their properties.

Function Verification

Function verification involves confirming that two functions are indeed inverses of each other. This means we need to check that the two main conditions for inverse functions are satisfied:

• \(f(f^{-1}(x)) = x\) for all \(x\) in the domain of \(f^{-1}\), and
• \(f^{-1}(f(x)) = x\) for all \(x\) in the domain of \(f\).

In practice, to verify these conditions, we substitute the inverse function back into the original function, and vice versa. In our problem, after finding \(f^{-1}(x) = \frac{1}{2^{x+1} - 1}\), we apply it back to \(f(x) = \log_2\left( \frac{x+1}{2x} \right)\). Through substitution and simplifying, we demonstrate both that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\), which certifies the correctness of our functions as true inverses.

This thorough verification ensures our mathematical manipulations maintain integrity and truly represent inverse relationships.

Algebraic Manipulation

Algebraic manipulation is a set of techniques used to rearrange and simplify expressions. It is crucial for solving equations and finding inverse functions. In the context of inverse functions, algebraic manipulation allows us to solve for one variable in terms of another, which is essential for deriving the inverse.

Our specific problem asked us to find the inverse of \(f(x) = \log_2\left( \frac{x+1}{2x} \right)\). By setting \(y = \log_2\left( \frac{x+1}{2x} \right)\) and converting it to exponential form, we expressed it as \(2^y = \frac{x+1}{2x}\). Then, using algebraic manipulation, we managed to isolate \(x\) and express it in terms of \(y\).

Key manipulation steps in our process included multiplying both sides of the equation by related terms to eliminate fractions, rearranging to bring like terms together, and solving for \(x\) explicitly. These steps enabled us to define \(f^{-1}(x) = \frac{1}{2^{x+1} - 1}\). It's these tools that make algebra a powerful language for resolving inverse functions and verifying their identity.