Problem 24 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-24: Determine the domain and range of \(f^{-1}\) for the given function without actually finding \(f^{-1}\). Hint: First find the domain and range of \(f\). $$ f(x)=\frac{2 x-7}{9 x+1} $$

Step-by-Step Solution

Answer

Domain of \( f^{-1} \): all reals except \( \frac{2}{9} \); Range of \( f^{-1} \): all reals except \( -\frac{1}{9} \).

Step 1: Determine the Domain of \( f \)

To find the domain of the function \( f(x) = \frac{2x-7}{9x+1} \), we need to set the denominator not equal to zero, as division by zero is undefined. Thus, solve:\[\begin{align*}9x + 1 & eq 0 \9x & eq -1 \x & eq -\frac{1}{9}\end{align*}\]Hence, the domain of \( f \) is all real numbers except \( x = -\frac{1}{9} \).

Step 2: Determine the Range of \( f \)

To find the range of \( f(x) = \frac{2x-7}{9x+1} \), we need to analyze the horizontal asymptote of the rational function. As \( x \to \infty \) or \( x \to -\infty \), the function approaches the ratio of the leading coefficients:\[ y = \frac{2}{9} \]Therefore, \( y eq \frac{2}{9} \) becomes the restriction for the range. Thus, the range of \( f \) is all real numbers except \( y = \frac{2}{9} \).

Step 3: Link the Domain and Range of \( f \) to \( f^{-1} \)

For an inverse function \( f^{-1} \), the domain of \( f^{-1} \) is the range of \( f \), and the range of \( f^{-1} \) is the domain of \( f \). Therefore:- The domain of \( f^{-1} \) is all real numbers except \( y = \frac{2}{9} \).- The range of \( f^{-1} \) is all real numbers except \( x = -\frac{1}{9} \).

Key Concepts

Domain

The domain of a function refers to all the input values (which we can think of as "x" values) for which the function is defined. In simpler terms, the domain is a set of possible numbers you can plug into a function without causing any mathematical errors. For rational functions, like the one in our exercise, the primary restriction comes from ensuring that the denominator is not zero.
To find the domain for our specific function, \( f(x) = \frac{2x-7}{9x+1} \), we set the denominator not equal to zero:

• 9x + 1 must not be equal to 0.
• Solving 9x + 1 = 0 gives \( x eq -\frac{1}{9} \).

Thus, the domain of the function \( f(x) \) is all real numbers except for \( x = -\frac{1}{9} \). This means that, for all other values of \( x \), the function will provide a valid output. This understanding is fundamental for exploring how the domain of a function affects its inverse.

Range

The range of a function is the set of all possible output values (or "y" values) that a function can produce. In the context of rational functions, identifying the range can often involve considering asymptotes. An asymptote is a line that the graph of a function approaches but never actually touches.
For our function \( f(x) = \frac{2x-7}{9x+1} \), as \( x \to \infty \) or \( x \to -\infty \), the function's behavior is dominated by the leading terms. In simple terms, the function approaches the ratio of the coefficients from the highest power terms:

• The horizontal asymptote is \( y = \frac{2}{9} \).
• Therefore, the function doesn’t reach \( y = \frac{2}{9} \).

Consequently, the range of this function is all real numbers except \( y = \frac{2}{9} \). For an inverse function, this range inverts to become its domain. Understanding this switch is key to mastering inverse functions.

Rational Functions

Rational functions are ratios of polynomials. Essentially, they are functions where one polynomial is divided by another. They often show interesting behaviors due to their denominators, particularly in terms of domain and range restrictions.
Consider the built-in characteristics of rational functions:

• Vertical asymptotes occur where the denominator equals zero (the function is undefined).
• Horizontal asymptotes are determined by the degrees and leading coefficients of the numerator and denominator.

In our problem with \( f(x) = \frac{2x-7}{9x+1} \), these attributes help discern the domain and range:

• Domain: Any point where the denominator doesn't result in zero; \( x eq -\frac{1}{9} \).
• Range: Avoids the value where the horizontal asymptote lies, \( y eq \frac{2}{9} \).

Rational functions thus present both opportunities and challenges in calculus, compelling a solid understanding of these key features to resolve both simple and complex problems. Comprehending the domain and range is not only about including or excluding numbers but also about recognizing the intricate balance created by the function's elements.