Problem 48 - Precalculus with Limits

Question

In Exercises 41-48, find (a) \(f \circ g\), and (b) \(g \circ f\). Find the domain of each function and each composite function. \(f(x) = \frac{3}{x^2 - 1}\), \(g(x) = x+1\)

Step-by-Step Solution

Answer

The functions \(f \circ g\) and \(g \circ f\) are \(\frac{3}{(x + 1)^2 - 1}\) and \(\frac{3}{x^2 - 1} + 1\) respectively. The domain of \(f\) is all real numbers except \(\pm 1\), of \(g\) is all real numbers, of \(f \circ g\) is all real numbers except \(0\), and of \(g \circ f\) is all real numbers except \(\pm 1\).

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

Compute \(f \circ g\) by substituting \(g(x)\) into \(f(x)\). This means you put \(x+1\) inside \(f(x)\) where there is \(x\): \[f(g(x)) = f(x + 1) = \frac{3}{(x + 1)^2 - 1}.\]

Step 2: Compute \(g \circ f\)

Compute \(g \circ f\) by substituting \(f(x)\) into \(g(x)\). This means you put \(\frac{3}{x^2 - 1}\) inside \(g(x)\) where there is \(x\): \[g(f(x)) = g \left( \frac{3}{x^2 - 1} \right) = \frac{3}{x^2 - 1} + 1.\]

Step 3: Find the domain of \(f\)

To find the domain of a function you have to find all values of \(x\) that makes the function defined. \(f(x) = \frac{3}{x^2 - 1}\) is defined when the denominator \((x^{2} - 1)\) is not equal to zero. Solving \((x^{2} - 1) \neq 0\) gives \(x \neq \pm 1\). So, the domain of \(f\) is \(x != \pm 1\).

Step 4: Find the domain of \(g\)

In a similar fashion, for \(g(x) = x+1\), the function is defined for all real values of \(x\). So, the domain of \(g\) is all real numbers.

Step 5: Find the domain of \(f \circ g\)

To find the domain for the composite function \(f(g(x))\), the inside function \(g(x)\) must be in the domain of \(f(x)\) and \(g(x)\) itself. The function \(g\), from Step 4, is defined for all real numbers and \(f(x)\) is defined when \(x \neq \pm 1\). Evaluating \(g(x) \neq \pm 1\), we get \(x \neq 0\). So, the domain of the composite function \(f(g(x))\) is all real numbers except \(0\).

Step 6: Find the domain of \(g \circ f\)

To find the domain of \(g \circ f\) or \(g(f(x))\), the inside function \(f(x)\) must be in the domain of \(g(x)\) and \(f(x)\) itself. From Step 3, implies that \(f(x)\) is defined when \(x \neq \pm 1\) and, from Step 4, \(g(x)\) is defined for all real numbers. Thus, the domain of \(g \circ f\) is \(x \neq \pm 1\).

Key Concepts

Domain of a Function

Understanding the domain of a function is essential as it tells us the allowable input values, meaning the numbers you can safely plug into the function without breaking any mathematical rules. For the function

• \( f(x) = \frac{3}{x^2 - 1} \): To find the domain, identify the values that make the denominator zero, because division by zero is undefined. Here, the denominator is \(x^2 - 1\), which equals zero for \(x = \pm 1\). Therefore, the domain of this function is all real numbers except \(x = \pm 1\).
• \( g(x) = x+1 \): This function has no denominator to worry about or square roots that could become negative, so it is defined for all real values of \(x\). Thus, the domain is all real numbers.

Finding the domain ensures that the functions are working properly within their constraints and helps avoid mathematical errors.

Composite Function

A composite function occurs when one function is applied to the result of another function. It's like feeding the output of one function directly into another. In our case, we worked with two composite functions:

• \( f \circ g \) or \( f(g(x)) \): Here, we place the function \(g(x) = x + 1\) into \(f(x) = \frac{3}{x^2 - 1}\). This results in the expression \( f(x + 1) = \frac{3}{(x + 1)^2 - 1} \).
• \( g \circ f \) or \( g(f(x)) \): This time, we substitute \( f(x) = \frac{3}{x^2 - 1} \) into \( g(x) = x + 1 \). Hence, the expression becomes \( g \left( \frac{3}{x^2 - 1} \right) = \frac{3}{x^2 - 1} + 1 \).

The composite functions are crucial as they allow us to build more complex operations from simpler ones, providing a richer framework for solving problems.

Function Operations

Function operations include various ways in which two or more functions can interact or be combined. Such operations are vital in exploring mathematical concepts deeply because they allow us to manipulate and understand different behavioral patterns of functions. Here's how this works:

• **Addition and Subtraction**: You can add or subtract the outputs of functions, creating new functions that describe new relationships. For example, \( (f + g)(x) = f(x) + g(x) \).
• **Multiplication and Division**: Functions can also be multiplied or divided, although with division, it's essential to ensure that the denominator is not zero, just like individual fractions.
• **Composition**: As seen earlier, you can compose functions, meaning applying one function to the results of another, creating a chain of operations.

Understanding these operations helps in creating models and solving a wide range of mathematical problems efficiently.