Problem 63 - A Graphical Approach to College Algebra

Question

Use \(f(x)\) and \(g(x)\) to find each composition. Identify its domain. (Use a calculator if necessary to find the domain.) (a) \((f \circ g)(x) \quad\) (b) \((g \circ f)(x) \quad\) (c) \((f \circ f)(x)\). $$f(x)=\frac{1}{x+1}, \quad g(x)=5 x$$

Step-by-Step Solution

Answer

(f \circ g)(x) = \frac{1}{5x+1}; domain: x \neq -\frac{1}{5}. (g \circ f)(x) = \frac{5}{x+1}; domain: x \neq -1. (f \circ f)(x) = \frac{x+1}{x+2}; domain: x \neq -2.

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

To find \((f \circ g)(x)\), we evaluate the function \(f\) at \(g(x)\). Since \(g(x) = 5x\), we substitute \(5x\) into \(f(x) = \frac{1}{x+1}\). Thus, \((f \circ g)(x) = f(g(x)) = f(5x) = \frac{1}{5x + 1}\).

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

The expression \(\frac{1}{5x + 1}\) is undefined when the denominator is zero. Therefore, solve \(5x + 1 = 0\) to find that \(x = -\frac{1}{5}\). The domain of \((f \circ g)(x)\) is all real numbers except \(x = -\frac{1}{5}\).

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

For \((g \circ f)(x)\), we evaluate \(g\) at \(f(x)\). Substitute \(f(x) = \frac{1}{x+1}\) into \(g(x) = 5x\), resulting in \((g \circ f)(x) = g(f(x)) = 5 \left( \frac{1}{x+1} \right) = \frac{5}{x+1}\).

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

The expression \(\frac{5}{x+1}\) is undefined when its denominator is zero. Solving \(x + 1 = 0\) reveals \(x = -1\). Thus, the domain of \((g \circ f)(x)\) is all real numbers except \(x = -1\).

Step 5: Compute \((f \circ f)(x)\)

To calculate \((f \circ f)(x)\), evaluate \(f\) at \(f(x)\). Substitute \(f(x)=\frac{1}{x+1}\) into another \(f(x)\): thus \((f \circ f)(x) = f(f(x)) = \frac{1}{\left(\frac{1}{x+1}\right) + 1}\). Simplify it to \(\frac{x+1}{x+2}\).

Step 6: Determine the domain of \((f \circ f)(x)\)

The function \(\frac{x+1}{x+2}\) is undefined for any \(x\) making the denominator zero. Solve \(x+2=0\) to find \(x = -2\). Consequently, the domain of \((f \circ f)(x)\) includes all real numbers except \(x = -2\).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (usually represented by \(x\)) for which the function is defined. In simpler terms, it tells us which numbers we can plug into our function without causing any issues like division by zero or taking the square root of a negative number.
For rational functions, these are functions that are expressed as a ratio of polynomials, the domain is determined by finding where the denominator is not equal to zero. This is because division by zero is undefined.

• If you have a function \(f(x) = \frac{1}{x+1}\), the denominator \(x+1\) must not be zero for the function to be defined. This means \(x eq -1\).
• Further, in function composition, say \((f \circ g)(x)\), the domain is additionally constrained by ensuring that both input into \(g(x)\) and output from \(g(x)\) applied to \(f(x)\) remain in their respective domains.

Understanding the domain is crucial because it helps you understand the limits within which the function operates properly without errors.

Rational Functions

Rational functions are a type of function represented by the quotient of two polynomials. They have the general form \(R(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\).

The behavior and characteristics of rational functions are heavily influenced by their numerator and denominator. These functions can have:

• Vertical Asymptotes: Occur at values of \(x\) where the denominator is zero, indicating the function approaches infinity.
• Holes: Points where both the numerator and the denominator equal zero, after simplification.

In our original problem, the function \(f(x) = \frac{1}{x+1}\) is an example of a rational function. The domain is affected by the zero of the denominator \(x + 1 = 0\) resulting in an undefined point at \(x = -1\). Understanding these aspects of rational functions allows us to predict and interpret the graph's behavior and its features.

Composite Functions

Composite functions involve creating a new function by applying one function to the result of another. This is denoted as \((f \circ g)(x)\), which means to apply \(f\) to \(g(x)\).

Creating composite functions requires careful attention to the domains of both functions involved:

• The input function, like \(g(x)\), must have a defined output for all values in its domain.
• The function into which the input is fed, like \(f(x)\), must be defined for each result coming from \(g(x)\).

For example, if we calculate \((f \circ g)(x) = \frac{1}{5x + 1}\), it's important to find all \(x\) where \(5x + 1 = 0\) to identify parts outside the domain.
Understanding composite functions helps integrate different actions into complex calculations, identifying potential problems like undefined values early in the process.