Question

Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=\frac{x-1}{x+1} $$

Short answer

The inverse is \( f^{-1}(x) = \frac{-x-1}{x-1} \), and it satisfies both verification properties.

Step-by-step solution

Understand the problem

We are asked to find the inverse of the function \( f(x) = \frac{x-1}{x+1} \) and verify two properties of inverse functions: \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \). This involves finding an expression for \( f^{-1}(x) \) and substituting it back into these properties for verification.

Set up the equation to find the inverse

Start by setting \( y = f(x) = \frac{x-1}{x+1} \). We need to solve for \( x \) in terms of \( y \), which will give us \( f^{-1}(y) \).

Cross-multiply to eliminate the fraction

Rearrange the equation by cross-multiplying to get rid of the fraction: \[ y(x + 1) = x - 1 \] Expand this to find: \[ yx + y = x - 1 \]

Rearrange terms to solve for x

Next, rearrange all terms involving \( x \) on one side and constant terms on the other: \[ yx - x = -y - 1 \] This simplifies to: \[ x(y - 1) = -y - 1 \].

Solve for x

Solve for \( x \) by dividing both sides through by \( (y - 1) \): \[ x = \frac{-y - 1}{y - 1} \]. This gives us the inverse function: \( f^{-1}(y) = \frac{-y - 1}{y - 1} \).

Verify \( f^{-1}(f(x)) = x \)

Substitute \( f(x) \) into \( f^{-1}(x) \):\[ f^{-1}\left(\frac{x-1}{x+1}\right) = \frac{-\left(\frac{x-1}{x+1}\right) - 1}{\left(\frac{x-1}{x+1}\right) - 1} \]Simplify the expression:\[ \frac{-(x-1) - (x+1)}{x-1 - (x+1)} = x \]Which simplifies correctly to \( x \), verifying \( f^{-1}(f(x)) = x \).

Verify \( f(f^{-1}(x)) = x \)

Substitute \( f^{-1}(x) \) into \( f(x) \): \[ f\left(\frac{-x-1}{x-1}\right) = \frac{\left(\frac{-x-1}{x-1}\right) - 1}{\left(\frac{-x-1}{x-1}\right) + 1} \]Simplify this expression:\[ \frac{-(x+1) - (x-1)}{-(x+1) + (x-1)} = x \]Which simplifies correctly to \( x \), verifying \( f(f^{-1}(x)) = x \).

Key concepts

Understanding Calculus and Inverse Functions

Calculus is a field of mathematics focused on studying rates of change and the accumulation of quantities. One of the key concepts in calculus is the idea of a function and its inverse. A function maps an input to an output. An inverse function reverses this process, mapping the output back to the original input. Finding an inverse function involves switching the roles of the dependent and independent variables and solving for the original variable. Consider the function given in the exercise, \[ f(x) = \frac{x-1}{x+1} \] To find the inverse, you interchange the variables to have \( y = \frac{x-1}{x+1} \) and solve for \( x \) in terms of \( y \). You'll navigate through algebraic manipulation to isolate \( x \), eventually coming to the inverse function formula.

Verifying Inverse Functions

Verifying that a function is an inverse involves demonstrating that the function and its inverse undo each other's operations. Hence, you perform function verification through two key properties:

• \( f^{-1}(f(x)) = x \)
• \( f(f^{-1}(x)) = x \)

These properties ensure that applying the function followed by its inverse returns you to the starting point, and vice versa. In this exercise, substitute the original function into its inverse, then verify that the simplification returns \( x \). Similarly, substitute the inverse into the original function. Only when both substitutions simplify to \( x \), you confirm the inverse relationship.

Mathematical Proof Through Simplification

A mathematical proof is a logical argument that verifies the truth of a mathematical statement. In this exercise, proving that we have found the correct inverse function involves simplifying equations. The original function \( f(x) = \frac{x-1}{x+1} \) and the inverse \( f^{-1}(x) = \frac{-x-1}{x-1} \) when substituted into one another should simplify to \( x \). This is done through algebraic simplification: breaking down the equation by expanding it, simplifying terms, and ensuring no incorrect assumptions arise. Each simplification step confirms that the inverse function indeed reverses the transformation made by the original function, thus completing the proof.