Problem 56 - College Algebra with Modeling and Visualization

Question

Exercises \(53-56:\) Use the given \(f(x)\) and \(g(x)\) to evaluate each expression. \(f(x)=\frac{1}{x-4}, g(x)=5\) (a) \((f \circ g)(3)\) (b) \((g \circ f)(8)\)

Step-by-Step Solution

Answer

(f \circ g)(3) = 1; (g \circ f)(8) = 5.

Step 1: Understanding Function Composition

Function composition \( (f \circ g)(x) \) means you first apply \( g(x) \) and then apply \( f(x) \) to the result. Conversely, \( (g \circ f)(x) \) means you first apply \( f(x) \) and then apply \( g(x) \) to the result.

Step 2: Evaluate (f \circ g)(3)

Since \( g(x) = 5 \) is a constant function, it assigns the value 5 to any input, including 3. Therefore, \( g(3) = 5 \). Now, apply \( f \) to this result: \( f(5) = \frac{1}{5 - 4} = \frac{1}{1} = 1 \). Thus, \( (f \circ g)(3) = 1 \).

Step 3: Evaluate (g \circ f)(8)

First evaluate \( f(8) \): \( f(8) = \frac{1}{8 - 4} = \frac{1}{4} \). Then, since \( g(x) = 5 \) is a constant, \( g(f(8)) = g(\frac{1}{4}) = 5 \). So, \( (g \circ f)(8) = 5 \).

Key Concepts

Constant Function

In mathematics, a constant function is a simple yet significant concept that assigns the same value to every input. No matter what number you plug into the function, the output remains unchanged. For example, if we have a constant function such as \( g(x) = 5 \), it means every input to the function results in the output of 5, whether \( x = 3 \), \( x = 10 \), or \( x = -7 \).

This stability makes constant functions easy to handle because they don’t depend on the input value. This is why we could smoothly evaluate \((f \circ g)(3)\) in the original exercise. Regardless of the initial input (in this case, 3), \(g(x)\) always yields 5. Then, solving becomes straightforward by further applying any subsequent functions to this constant value.

Understanding and using constant functions is helpful in both simple calculations and complex mathematical modeling, offering predictability and reliability.

Evaluate Expressions

Evaluating expressions is the stepping stone of solving any mathematical problem. It involves replacing the variables with their given values and performing the operations indicated to get a final answer.

Let's see how we evaluate the expression \((f \circ g)(3)\) in the context of the original exercise.

• First, we identify that \(g(x) = 5\) is a constant function, so \(g(3) = 5\).
• Next, we apply this result as the input to the function \(f(x) = \frac{1}{x-4}\). So, we evaluate \( f(5) = \frac{1}{5-4} = 1 \).

As you can see, evaluating expressions step-by-step ensures clarity and accuracy.

This process reinforces understanding of mathematical operations and helps in recognizing patterns, allowing for more streamlined solutions to future problems.

Inverse Functions

Inverse functions are unique functions that "undo" each other. If you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), applying them in succession will return you to your initial value. Mathematically, this is expressed as \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

Although inverse functions were not directly utilized in the given exercise, understanding their concept is crucial as it lays the foundation for more complex topics in calculus and algebra. For instance:

• If you know \( f(x) \) involves multiplying by a number, then \( f^{-1}(x) \) would involve dividing by that same number.
• The graph of an inverse function is essentially the reflection of the original graph across the line \( y = x \).

Grasping inverse functions can tremendously aid in solving equations and understanding transformations. They offer an excellent example of how functions interact, providing deeper insights into the symmetric nature of mathematical operations.