Problem 1 - Applied Calculus

Question

In Problems \(1-3,\) find the following: (a) \(\quad f(g(x))\) (b) \(g(f(x))\) (c) \(f(f(x))\) $$f(x)=5 x-1 \text { and } g(x)=3 x+2$$

Step-by-Step Solution

Answer

(a) \(f(g(x))=15x+9\); (b) \(g(f(x))=15x-1\); (c) \(f(f(x))=25x-6\).

Step 1: Find \(f(g(x))\)

To find \(f(g(x))\), we first need to substitute\(g(x)\) into \(f(x)\). Let's start by finding \(g(x)\): \(g(x) = 3x + 2\). Next, substitute \(g(x)\) into \(f(x)\): \[ f(g(x)) = f(3x + 2) = 5(3x + 2) - 1.\] Now, distribute and simplify:\[ f(g(x)) = 15x + 10 - 1 = 15x + 9. \] Thus, \(f(g(x))=15x+9\).

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

To determine \(g(f(x))\), substitute \(f(x)\) into \(g(x)\). Begin by writing \(f(x)\): \(f(x) = 5x - 1\). Now, substitute \(f(x)\) into \(g(x)\):\[ g(f(x)) = g(5x - 1) = 3(5x - 1) + 2. \]Distribute and simplify:\[ g(f(x)) = 15x - 3 + 2 = 15x - 1. \] Therefore, \(g(f(x))=15x-1\).

Step 3: Find \(f(f(x))\)

To find \(f(f(x))\), substitute \(f(x)\) back into itself.Start with \(f(x) = 5x - 1\) and substitute:\[ f(f(x)) = f(5x - 1) = 5(5x - 1) - 1. \]Now, distribute and simplify:\[ f(f(x)) = 25x - 5 - 1 = 25x - 6. \] Thus, \(f(f(x))=25x-6\).

Key Concepts

function operations

Function operations are similar to the arithmetic operations you use with numbers. However, here you operate with entire functions rather than just individual numbers. These operations typically include addition, subtraction, multiplication, and division of functions. You simply apply the operation to the functions as wholes, much like you would add or multiply numbers.
For example, if you have two functions, say,

• \( f(x) = 5x - 1 \)
• \( g(x) = 3x + 2 \)

you can perform operations such as:

• **Addition:** \((f + g)(x) = f(x) + g(x) = (5x - 1) + (3x + 2) = 8x + 1\)
• **Subtraction:** \((f - g)(x) = f(x) - g(x) = (5x - 1) - (3x + 2) = 2x - 3\)
• **Multiplication:** \((f \cdot g)(x) = f(x) \cdot g(x) = (5x - 1)(3x + 2)\)
• **Division:** \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{5x - 1}{3x + 2}\)

Each of these operations combines the functions in different ways, forming new functions.

composition of functions

Composition of functions is a more intricate operation than basic function operations. It involves applying one function to the results of another function. This operation essentially stacks functions on top of each other, creating complex mappings from inputs to outputs.
For example, let's consider the previously mentioned functions

• \( f(x) = 5x - 1 \)
• \( g(x) = 3x + 2 \)

The composition \( f(g(x)) \) involves substituting the entire function \( g(x) \) into \( f(x) \).
This is calculated as follows:

• Substitute \( g(x) = 3x + 2 \) into \( f(x) = 5x - 1 \)
• We get \( f(g(x)) = 5(3x + 2) - 1 = 15x + 9 \)

Similarly, for \( g(f(x)) \), substitute \( f(x) = 5x - 1 \) into \( g(x) = 3x + 2 \):

• \( g(f(x)) = 3(5x - 1) + 2 = 15x - 1 \)

Lastly, \( f(f(x)) \) means substituting \( f(x) \) back into itself:

• \( f(f(x)) = 5(5x - 1) - 1 = 25x - 6 \).

Composition is powerful because it allows us to build more complex functions from simple ones, enabling deeper analysis and greater control in mathematical modeling.

step-by-step solution

When tackling problems involving function operations and compositions, a step-by-step solution is crucial for clarity and understanding. Let's break down how such an approach serves to demystify the process:
To solve for compositions like \( f(g(x)) \), \( g(f(x)) \), and \( f(f(x)) \), follow these steps:

• **Step 1:** Identify each function provided. Here, \( f(x) = 5x - 1 \) and \( g(x) = 3x + 2 \).
• **Step 2:** Replace the variable in the first function with the entire second function. For \( f(g(x)) \), replace \( x \) in \( f(x) \) with \( g(x) \), i.e., \( 3x + 2 \).
• **Step 3:** Simplify the resulting expression by distributing and combining like terms. This transforms the nested functions into a new, single expression.
• **Step 4:** Repeat these substitutions for each required composition, ensuring accuracy at each step.

By methodically working through each step, students gain a clearer understanding of how compositions work and how they change function behavior. This structured approach not only helps solve specific problems but also equips learners with the skills to tackle similar challenges independently.