Problem 44 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Replace the Function Notation

First, replace the function notation \(f(x)\) with \(y\). So, we have: \[ y = \sqrt{2x - 1} \] This helps in working with the equation directly when finding the inverse.

Step 2: Swap x and y

Next, swap \(x\) and \(y\) in the equation. This forms the basis for finding the inverse: \[ x = \sqrt{2y - 1} \] This step represents switching the roles of inputs and outputs in the function.

Step 3: Solve for y

Isolate \(y\) by first squaring both sides to remove the square root: \[ x^2 = 2y - 1 \]Then, add 1 to both sides: \[ x^2 + 1 = 2y \]Finally, divide by 2: \[ y = \frac{x^2 + 1}{2} \] This gives us the expression for \(y\), which is our inverse function.

Step 4: Write the Inverse Function

Replace \(y\) with \(f^{-1}(x)\) to denote the inverse function: \[ f^{-1}(x) = \frac{x^2 + 1}{2} \] This step signifies that we have found the inverse function of \(f\).

Key Concepts

Function Notation

Function notation is a way to symbolize and work with functions in mathematics using symbols like \(f(x)\). When we write \(f(x) = \sqrt{2x - 1}\), we are defining a function named \(f\) with variable \(x\). This means that for any value of \(x\), the function \(f\) produces the value \(\sqrt{2x - 1}\).

• Purpose: It provides a clear and concise method to communicate how a function operates.
• Relabeling: In the process of finding an inverse, we often replace \(f(x)\) with a variable like \(y\), simplifying the equation to work with directly.

Once \(y\) replaces \(f(x)\), the equation becomes \(y = \sqrt{2x - 1}\), noticeably simplifying further manipulation when finding the inverse. This change is purely for simplification and does not alter the function's operation.

Solving Equations

Solving equations involves finding the values of the variables that satisfy a given equation. In the context of finding inverse functions, solving equations entails isolating one variable to express one in terms of another, often by exchanging \(x\) and \(y\).
In our exercise, after switching \(x\) and \(y\) in the equation \(x = \sqrt{2y - 1}\), we aim to solve for \(y\). This process typically involves several steps:

• Removing the square root: By squaring both sides, the equation becomes \(x^2 = 2y - 1\). This 'undoes' the square root operation, allowing us to work with simpler algebraic terms.
• Isolate \(y\): Next, we add 1 to both sides, resulting in \(x^2 + 1 = 2y\). Then, divide by 2 to solve for \(y\), which finally gives us \(y = \frac{x^2 + 1}{2}\).

This process of solving for \(y\) in terms of \(x\) is essential in determining the inverse function from the original.

Square Roots

The square root function \(\sqrt{x}\) identifies a value which, when multiplied by itself, results in \(x\). In function problems, square roots often present challenges given their restrictive nature.

• Properties: The square root is defined for non-negative numbers only, making the domain an important consideration when dealing with root functions.
• Inverses: When dealing with inverse functions, square roots can be 'undone' by squaring; this is fundamental to solving equations involving square roots.

In our exercise, when \(f(x) = \sqrt{2x - 1}\), we managed square roots when solving for the inverse. Squaring \(x = \sqrt{2y - 1}\) gave us \(x^2 = 2y - 1\). This step is vital to eliminate the square root and progress in solving for the inverse. Understanding how square roots function and their inverse operations is crucial for mastering these types of function problems.