Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=-\frac{1}{x-3} $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \frac{3x - 1}{x}\). Verified \(f^{-1}(f(x))=x\) and \(f(f^{-1}(x))=x\).

Step 1: Start with the Inverse Definition

Given the function \( f(x) = -\frac{1}{x-3} \), we want to find \( f^{-1}(x) \). This means that \( y = f(x) \) can be written as \( x = f^{-1}(y) \). Start by setting \( y = -\frac{1}{x-3} \) and then solve for \( x \).

Step 2: Solve for x

We start from \( y = -\frac{1}{x-3} \). Multiply both sides by \( (x-3) \): \[ y(x-3) = -1 \] Simplify and solve for \( x \): \[ yx - 3y = -1 \] \[ yx = 3y - 1 \] \[ x = \frac{3y - 1}{y} \]

Step 3: Simplify to Find the Inverse Function

We now have \( x = \frac{3y - 1}{y} \). Replace \( x \) with \( f^{-1}(x) \) and \( y \) with \( x \) for our inverse function:\[ f^{-1}(x) = \frac{3x - 1}{x} \]

Step 4: Verify the Inverse (\(f^{-1}(f(x))=x\))

Compute \( f^{-1}(f(x)) \):Substitute \( f(x) \) into \( f^{-1} \): \[ f^{-1}(f(x)) = \frac{3(-\frac{1}{x-3}) - 1}{-\frac{1}{x-3}} \]Simplify: \[ f^{-1}(f(x)) = \frac{-\frac{3}{x-3} - 1}{-\frac{1}{x-3}} \] This simplifies to \( x \).

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

Compute \( f(f^{-1}(x)) \):Substitute \( f^{-1}(x) \) into \( f(x) \):\[ f(f^{-1}(x)) = -\frac{1}{\frac{3x - 1}{x} - 3} \]Simplify the expression to \( x \):\[-\frac{1}{\frac{3x - 1 - 3x}{x}} = x \] This confirms \( f(f^{-1}(x)) = x \).

Key Concepts

Function Composition

Function composition is a powerful concept in mathematics, particularly when dealing with inverse functions. It involves creating a new function by applying one function to the result of another function. For two functions, say \(f(x)\) and \(g(x)\), the composite function is written as \((f \circ g)(x)\). This can be interpreted as applying \(g\) first to \(x\), and then applying \(f\) to the result of \(g(x)\). This is formally defined as:

• \((f \circ g)(x) = f(g(x))\)

Understanding function composition is essential when working with inverse functions. The ordering in function composition matters because \((f \circ g)(x)\) is generally not the same as \((g \circ f)(x)\).
This property of non-commutativity is crucial when we verify inverse functions, usually with the condition \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). These conditions affirm that \(f\) and \(f^{-1}\) effectively "undo" each other's operations.

Mathematical Proofs

Mathematical proofs are logical arguments that establish the truth of a mathematical statement. They are fundamental in verifying the correctness of solutions like those involving inverse functions. To prove that a function \(f(x)\) and its inverse \(f^{-1}(x)\) are correct, we employ proofs by composition.When proving involutions like \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\), we essentially show that when a function is composed with its inverse, the result is the identity function. This means the original input \(x\) is unchanged after these operations.

• The identity function is defined such that \(I(x) = x\).
• If \(I\) is the identity function, then \(f(f^{-1}(x)) = I(x)\) and \(f^{-1}(f(x)) = I(x)\).

This mathematical proof confirms both the existence and correctness of the inverse. For example, calculating both \(f(f^{-1}(x))\) and \(f^{-1}(f(x))\) leads us back to the original \(x\), as worked out in the solution.

Rational Functions

Rational functions are ratios of polynomials, and they can be expressed as \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials. The function \(f(x) = -\frac{1}{x-3}\) is an example of a rational function. These functions are essential in various areas of mathematics and applied sciences.
To find the inverse of rational functions like \(-\frac{1}{x-3}\), it involves algebraic manipulation to express \(x\) in terms of \(y\), as shown in the problem solution. By solving \(y = -\frac{1}{x-3}\) for \(x\), we determine \(f^{-1}(x)\). This requires multiplying through by the denominator and rearranging terms, yielding the equation \(x = \frac{3y - 1}{y}\).
Rational functions can have restricted domains due to division by zero. For example, \(f(x) = -\frac{1}{x-3}\) is undefined at \(x = 3\). Similarly, the inverse function may inherit restrictions, accounting for possible values making the denominator zero. Understanding these features is vital in handling rational functions in calculus and higher mathematics.