Problem 31 - Thomas Calculus in SI Units

Question

Gives a formula for a function \(y=f(x) .\) In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1} .\) As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x\). $$f(x)=\frac{x+3}{x-2}$$

Step-by-Step Solution

Answer

The inverse is \(f^{-1}(x)=\frac{2x+3}{x-1}\); domain: \(x \neq 1\), range: \(y \neq 2\).

Step 1: Understand the Function

The given function is \( f(x) = \frac{x+3}{x-2} \). This is a rational function where the numerator is \(x+3\) and the denominator is \(x-2\). Notice that the function is undefined at \(x=2\) because division by zero is not possible.

Step 2: Express the Inverse Function

To find the inverse of the function \( f(x) = \frac{x+3}{x-2} \), we start by setting \( y = \frac{x+3}{x-2} \) and then solve for \( x \). This gives us:\[ y(x-2) = x+3 \]Rearranging terms, we have:\[ yx - 2y = x + 3 \]Isolating terms with \(x\), we get:\[ yx - x = 2y + 3 \]Factor out the \(x\) from the left side:\[ x(y - 1) = 2y + 3 \]Finally, isolate \(x\):\[ x = \frac{2y + 3}{y - 1} \]Thus, the inverse function is \( f^{-1}(x) = \frac{2x + 3}{x - 1} \).

Step 3: Determine the Domain and Range of the Inverse

For the inverse function \( f^{-1}(x) = \frac{2x + 3}{x - 1} \), the domain is all real numbers except \( x = 1 \), as the function is undefined there. Since the original function \( f(x) = \frac{x+3}{x-2} \) has a vertical asymptote at \( x = 2 \), the range of \( f^{-1} \) will be all real numbers except \( y = 2 \). So, the domain is \( x \in \mathbb{R}, x eq 1 \) and the range is \( y \in \mathbb{R}, y eq 2 \).

Step 4: Verify the Inverse by Composition

To verify that \( f(x) = \frac{x+3}{x-2} \) and \( f^{-1}(x) = \frac{2x+3}{x-1} \) are indeed inverse functions, we need to show:1. \( f(f^{-1}(x)) = x \)2. \( f^{-1}(f(x)) = x \)First, let's calculate \( f(f^{-1}(x)) = f\left(\frac{2x+3}{x-1}\right) \):\[ f\left(\frac{2x+3}{x-1}\right) = \frac{\left(\frac{2x+3}{x-1}\right) + 3}{\left(\frac{2x+3}{x-1}\right) - 2} = \frac{\frac{2x + 3 + 3(x - 1)}{x - 1}}{\frac{2x + 3 - 2(x - 1)}{x - 1}} = \frac{\frac{2x + 3 + 3x - 3}{x - 1}}{\frac{2x + 3 - 2x + 2}{x - 1}} = \frac{x(x-1)}{x(x-1)} = x \]Now, for \( f^{-1}(f(x)) = f^{-1}\left(\frac{x+3}{x-2}\right) \):\[ f^{-1}\left(\frac{x+3}{x-2}\right) = \frac{2\left(\frac{x+3}{x-2}\right) + 3}{\frac{x+3}{x-2} - 1} = \frac{\frac{2(x+3) + 3(x-2)}{x-2}}{\frac{x+3 - (x-2)}{x-2}} = \frac{x(x-2)}{x(x-2)} = x \]Both compositions result in \( x \), confirming the inverse calculation is correct.

Key Concepts

Rational Functions

Rational functions are a key concept in mathematics. They are the ratio of two polynomial functions, each represented as a fraction where the numerator and the denominator are polynomials. For example, consider the function \( f(x) = \frac{x+3}{x-2} \). In this rational function, \( x+3 \) is the numerator and \( x-2 \) is the denominator.

One important aspect of rational functions is that they can predict where the function will be undefined. This occurs at any value of \( x \) that makes the denominator zero because division by zero is mathematically undefined. In the case of our example function, \( f(x) \) becomes undefined when \( x = 2 \).

Another significant feature of rational functions is their graph behavior, which often includes asymptotes. Vertical asymptotes occur where the function becomes undefined. In our example, since the denominator is zero at \( x = 2 \), there is a vertical asymptote at this point.

Domain and Range

Understanding the domain and range of a function is crucial for graphing and solving problems involving rational functions. The domain of a function refers to all the possible input values (\( x \)-values) that the function can accept without becoming undefined. For a rational function like \( f(x) = \frac{x+3}{x-2} \), the domain is all real numbers except where the denominator is zero. Thus, the domain is \( x \in \mathbb{R}, x eq 2 \).

The range of a function, on the other hand, refers to all the possible output values (\( y \)-values) it can produce. The range can sometimes be more complex to determine, especially for inverse functions. For the inverse function \( f^{-1}(x) = \frac{2x + 3}{x - 1} \), the range is all real numbers except where \( y = 2 \). This is because the original function \( f(x) \) approaches 2 as \( x \) approaches infinity, suggesting a horizontal asymptote at \( y = 2 \).

To properly identify the domain and range, it is often helpful to graph the function, look for asymptotes, and test points close to those asymptotes.

Function Composition

Function composition is a technique used to verify the correctness of inverse functions. When you compose a function with its inverse, the result should return you to the starting point, which is the variable \( x \). If this holds true, then one function is indeed the inverse of the other.

In example exercises, such as verifying the inverse of \( f(x) = \frac{x+3}{x-2} \) and \( f^{-1}(x) = \frac{2x+3}{x-1} \), you perform the composition by substituting one function into the other:

• Perform \( f(f^{-1}(x)) \) by substituting \( f^{-1}(x) \) into \( f(x) \).
• Perform \( f^{-1}(f(x)) \) by substituting \( f(x) \) into \( f^{-1}(x) \).
• If both operations simplify to \( x \), the functions are confirmed as inverses.

This process not only verifies the mathematical work on finding inverses, but also deepens your understanding of function behavior. It is a fundamental technique in algebra, reinforcing concepts of equivalence and operation within functions.