Problem 56 - Brief Applied Calculus

Question

For each pair of functions \(f(x)\) and \(g(x)\), find a. \(f(g(x))\) b. \(g(f(x))\) and c. \(f(f(x))\) $$ f(x)=x^{3}+x ; g(x)=\frac{x^{4}+1}{x^{4}-1} $$

Step-by-Step Solution

Answer

f(g(x)) = \left( \frac{x^4+1}{x^4-1} \right)^3 + \frac{x^4+1}{x^4-1}; g(f(x)) = \frac{(x^3 + x)^4 + 1}{(x^3 + x)^4 - 1}; f(f(x)) = (x^3 + x)^3 + (x^3 + x).

Step 1: Identify the Functions

We have two functions given: \( f(x) = x^3 + x \) and \( g(x) = \frac{x^4+1}{x^4-1} \). We will use these functions to find the compositions \( f(g(x)) \), \( g(f(x)) \), and \( f(f(x)) \).

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

To find \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \):\[f(g(x)) = f\left( \frac{x^4+1}{x^4-1} \right) = \left( \frac{x^4+1}{x^4-1} \right)^3 + \frac{x^4+1}{x^4-1}.\]

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

To find \( g(f(x)) \), substitute \( f(x) \) into \( g(x) \):\[g(f(x)) = g(x^3 + x) = \frac{(x^3 + x)^4 + 1}{(x^3 + x)^4 - 1}.\]

Step 4: Compute \( f(f(x)) \)

To find \( f(f(x)) \), substitute \( f(x) \) into itself:\[f(f(x)) = f(x^3 + x) = (x^3 + x)^3 + (x^3 + x).\] Expand if needed for detailed calculation.

Key Concepts

Understanding Calculus

Calculus is a branch of mathematics that deals with change and motion. It's a vast field that includes topics like differentiation and integration. At its core, calculus helps us understand rates of change—how things grow or shrink, speed up or slow down. In the context of function composition, calculus allows us to explore more complex functions. We start to see how functions interact when combined. For example, when you compose functions, you're essentially chaining them together. This helps you understand how the output of one function becomes the input for another. It's important to be comfortable with basic calculus concepts like limits and derivatives, as they form the foundation for more advanced topics. In equations like the ones in the exercise, we need to consider the continuity and differentiability of functions. These properties ensure we can apply calculus techniques effectively.

Diving Into Composite Functions

Composite functions involve applying one function to the results of another. This concept is fundamental in calculus, where chaining operations allows for more complex problem-solving. When you see expressions like \( f(g(x)) \), you're looking at a composite function.

• Function: The building blocks; each function takes an input and gives an output.
• Composition: The operation of taking an output from \( g(x) \) and using it as input for \( f(x) \).

In our exercise, we have \( f(x) = x^3 + x \) and \( g(x) = \frac{x^4+1}{x^4-1} \). To find \( f(g(x)) \), substitute \( g(x) \) into \( f(x) \). It transforms the problem from two separate functions into one compound function. This process involves recalling the order of operations and function rules, ensuring each step is accurate for further calculus applications.

Exploring Function Operations

Function operations are the nuts and bolts of calculus. They allow you to apply different types of operations on functions—addition, subtraction, multiplication, division, and composition. It’s a bit like mixing ingredients to create new recipes, resulting in more complex flavors (or in math, outcomes).

• Addition/Subtraction: Combining functions by adding or subtracting their outputs.
• Multiplication/Division: Producing new function outputs by multiplying or dividing existing ones.
• Composition: As seen in our example, nesting functions like Russian dolls.

In the example exercise, the focus was on compositions. Function composition won’t change the underlying functions but allows insights into how these mathematical forms can interact and intersect more dynamically. Understanding these interactions aids in predicting behaviors in complex systems, crucial for applications in physics, engineering, and beyond.