Problem 34 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{x+2}{x-1}, \quad g(x)=\frac{x-5}{x+4} $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = -\frac{x+1}{3}\), domain: \(x \neq -4, 5\); (b) \((g \circ f)(x) = \frac{-4x + 7}{5x - 2}\), domain: \(x \neq 1, \frac{2}{5}\).

Step 1: Compute \((f \circ g)(x)\)

To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). This gives us:\[ f(g(x)) = f\left(\frac{x-5}{x+4}\right) = \frac{\frac{x-5}{x+4} + 2}{\frac{x-5}{x+4} - 1} \]Now, simplify the expression:\[ \text{Numerator}: \frac{x-5}{x+4} + 2 = \frac{x-5 + 2(x+4)}{x+4} = \frac{x-5 + 2x + 8}{x+4} = \frac{3x + 3}{x+4} \]\[ \text{Denominator}: \frac{x-5}{x+4} - 1 = \frac{x-5 - (x+4)}{x+4} = \frac{x-5-x-4}{x+4} = \frac{-9}{x+4} \]Divide the two fractions:\[ f(g(x)) = \frac{\frac{3x + 3}{x+4}}{\frac{-9}{x+4}} = \frac{3x + 3}{-9} = -\frac{x+1}{3} \]

Step 2: Determine the Domain of \((f \circ g)(x)\)

For the domain of \(f \circ g\), we require that both \(g(x)\) and \(f(g(x))\) are defined. \(g(x)\) is undefined when the denominator is zero, i.e., \(x+4=0\), so \(x eq -4\).In \(f(g(x))\), the denominator is also undefined when \(x eq 5\) to avoid the original \(f(x)\) denominator from being zero after substitution. Therefore, the domain of \(f \circ g\) is all real numbers except \(x = -4\) and \(x = 5\).

Step 3: Compute \((g \circ f)(x)\)

To find \((g \circ f)(x)\), we need to substitute \(f(x)\) into \(g(x)\). This gives us:\[ g(f(x)) = g\left(\frac{x+2}{x-1}\right) = \frac{\frac{x+2}{x-1} - 5}{\frac{x+2}{x-1} + 4} \]Simplify the expression:\[ \text{Numerator}: \frac{x+2}{x-1} - 5 = \frac{x+2 - 5(x-1)}{x-1} = \frac{x+2 - 5x + 5}{x-1} = \frac{-4x + 7}{x-1} \]\[ \text{Denominator}: \frac{x+2}{x-1} + 4 = \frac{x+2 + 4(x-1)}{x-1} = \frac{x+2 + 4x - 4}{x-1} = \frac{5x - 2}{x-1} \]Divide the two fractions:\[ g(f(x)) = \frac{\frac{-4x + 7}{x-1}}{\frac{5x - 2}{x-1}} = \frac{-4x + 7}{5x - 2} \]

Step 4: Determine the Domain of \((g \circ f)(x)\)

The domain of \(g \circ f\) requires that \(f(x)\) and \(g(f(x))\) are both defined. \(f(x)\) is undefined when \(x=1\), hence \(x eq 1\). Additionally, the denominator of \(g(f(x))\) is zero when \(5x - 2 = 0\), or when \(x = \frac{2}{5}\). Therefore, the domain of \(g \circ f\) is all real numbers except \(x = 1\) and \(x = \frac{2}{5}\).

Key Concepts

Domain of a Function

The domain of a function is a fundamental concept in mathematics. It indicates the set of input values ("x"-values) for which the function is defined. When working with composed functions, like \((f \circ g)(x)\), we need to ensure both functions are defined for the resulting values.

To find the domain of \((f \circ g)(x)\), you must:

• Check when \(g(x)\) is defined. This is done by ensuring that the denominator of \(g(x)\) does not equal zero.
• Substitute \(g(x)\) into \(f(x)\) and find where \(f(g(x))\) is defined by considering its denominator.

In the given exercise, \(g(x)\) becomes undefined when \(x+4 = 0\) leading to \(x eq -4\), and \(f(g(x))\) must avoid \(x = 5\) as it leads to a zero denominator in \(f(x)\).

Similarly, for \((g \circ f)(x)\), determine the domain by checking where \(f(x)\) and \(g(f(x))\) are defined. This involves ensuring the denominator of \(f(x)\) is not zero at \(x = 1\), and solving \(5x - 2 = 0\) indicates that \(x eq \frac{2}{5}\).

Understanding these constraints ensures precise identification of the domain, crucial for correct function analysis.

Inverse Functions

Inverse functions reverse the operation of the original function. If a function \(f(x)\) takes an input \(x\) and gives an output \(y\), its inverse, denoted \(f^{-1}(x)\), reverses this process and takes \(y\) back to \(x\).

To find an inverse function:

• Replace "f(x)" with "y".
• Switch the places of "x" and "y".
• Solve for "y".
• Replace "y" with "f^{-1}(x)".

This method ensures that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\), indicating that applying the function and its inverse leads you back to your starting point.

In some cases, a function does not have an inverse because it isn't one-to-one, which means it doesn't pass the horizontal line test. For instance, a quadratic function like \(x^2\) without restriction cannot have an inverse as it fails this test. Understanding if a function has an inverse is crucial in problem-solving and mathematical modeling.

Rational Functions

Rational functions are quotients of polynomial functions, with the general form \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\).

These functions are significant due to their complex behaviors and interesting characteristics like vertical and horizontal asymptotes.

• **Vertical Asymptotes** occur where the denominator is zero; these are the "forbidden" points within the domain.
• **Horizontal Asymptotes** describe the behavior as "x" approaches infinity. They occur when the degree of the numerator and denominator are different.

The given functions in the exercise are rational:

• **\(f(x) = \frac{x+2}{x-1}\) shows vertical asymptotes at \(x = 1\), illustrating where the function is undefined.
• **\(g(x) = \frac{x-5}{x+4}\) has a vertical asymptote at \(x = -4\).

Analyzing rational functions involves simplifying expressions and understanding their asymptotic behaviors, which plays a key role in graph interpretation and calculus.