Problem 14 - Calculus: One and Several Variables

Question

Given that $$f(x)=\left\\{\begin{array}{cl}1-x, & x \leq 1 \\\2 x-1, & x>1\end{array} \text { and } g(x)=\left\\{\begin{aligned}0, & x ,: 2 \\\\-1, & x,: 2\end{aligned}\right.\right.$$ find \(f+g , f-g , f \cdot g .\) HINT: Break up the domains of the two functions in the same manner.

Step-by-Step Solution

Answer

The sum \(f + g\) is \(1 - x\) for \(x < 1\), \(2x - 1\) for \(1 \leq x < 2\), and \(2(x - 1)\) for \(x \geq 2\). The difference \(f - g\) is \(1 - x\) for \(x < 1\), \(2x - 1\) for \(1 \leq x < 2\), and \(2x\) for \(x \geq 2\). The product \(f \cdot g\) is 0 for \(x < 2\), and \(-2x + 1\) for \(x \geq 2\).

Step 1: Analyzing the given functions

Firstly, we will examine the piecewise functions f(x) and g(x). f(x) is defined as \(1 - x\) when \(x \leq 1\) and \(2x - 1\) when \(x > 1\). Similarly, g(x) is defined as 0 when \(x < 2\) and as -1 when \(x \geq 2\).

Step 2: Forming the functions for \(f + g\), \(f - g\) and \(f \cdot g\)

Following the given hint, we will break up the domains of the two functions in the same manner. Therefore, we will have three cases for each of \(f + g\), \(f - g\) and \(f \cdot g\): when \(x < 1\), when \(1 \leq x < 2\), and when \(x \geq 2\).

Step 3: Evaluating \(f + g\)

We calculate \(f + g\) for each of the three cases.\n\nCase 1: \(x < 1\): Here, \(f\) is \(1 - x\) and \(g\) is 0. Hence, \(f + g = (1 - x) + 0 = 1 - x\).\n\nCase 2: \(1 \leq x < 2\): Here, \(f\) is \(2x - 1\) and \(g\) is 0. So, \(f + g = (2x - 1) + 0 = 2x - 1\).\n\nCase 3: \(x \geq 2\): In this case, \(f\) is \(2x - 1\) and \(g\) is -1. So, \(f + g = (2x - 1) - 1 = 2x - 2 = 2(x - 1)\).

Step 4: Evaluating \(f - g\)

Next we'll calculate \(f - g\) for each of the three cases.\n\nCase 1: \(x < 1\): Here, \(f\) is \(1 - x\) and \(g\) is 0. Hence, \(f - g = (1 - x) - 0 = 1 - x\).\n\nCase 2: \(1 \leq x < 2\): Here, \(f\) is \(2x - 1\) and \(g\) is 0. Hence, \(f - g = (2x - 1) - 0 = 2x - 1\).\n\nCase 3: \(x \geq 2\): Here, \(f\) is \(2x - 1\) and \(g\) is -1. Hence, \(f - g = (2x - 1) - (-1) = 2x - 1 + 1 = 2x\).

Step 5: Evaluating \(f \cdot g\)

Finally, we'll calculate \(f \cdot g\) for each of the cases.\n\nCase 1: \(x < 1\): Here, \(f\) is \(1 - x\) and \(g\) is 0. Thus, \(f \cdot g = (1 - x) \cdot 0 = 0\).\n\nCase 2: \(1 \leq x < 2\): Here, \(f\) is \(2x - 1\) and \(g\) is 0. Thus, \(f \cdot g = (2x - 1) \cdot 0 = 0\).\n\nCase 3: \(x \geq 2\): Here, \(f\) is \(2x - 1\) and \(g\) is -1. Thus, \(f \cdot g = (2x - 1) \cdot (-1) = -2x + 1\).

Key Concepts

Function Operations

Functions are mathematical relationships between a set of inputs and outputs. In this exercise, we explore how to perform operations on piecewise functions, which are functions defined by multiple sub-functions, each applied to a specific interval of the domain. To find the sum, difference, and product of two functions, you apply operations between the functions' outputs. You need to do this while considering the intervals over which each sub-function is defined. For our given functions, the operations we perform are:

• Addition: Compute \( f(x) + g(x) \) for each interval.
• Subtraction: Compute \( f(x) - g(x) \) for each interval.
• Multiplication: Compute \( f(x) \times g(x) \) for each interval.

By handling each piece of the function separately and respecting its domain, we ensure the results accurately reflect the behavior of the entire function.

Domain Analysis

The domain of a function is the set of all possible input values (\(x\)) for which the function is defined. For piecewise functions, understanding how these domains change across different intervals is crucial. Our given functions, \(f(x)\) and \(g(x)\), each have specific intervals defined by their piecewise nature:

• For \(f(x)\), the domain is split as \(x \leq 1\) with output \(1 - x\), and \(x > 1\) with output \(2x - 1\).
• For \(g(x)\), the domain is defined as \(x < 2\) with output 0, and \(x \geq 2\) with output -1.

Breaking up the domain of each function into intervals ensures that any operations on \(f\) and \(g\) respect these boundaries. This process of domain analysis is essential to ensure the correctness of function operations.

Elementary Calculus

Piecewise functions and their operations touch upon fundamental ideas in calculus. Calculus often requires meticulous handling of function domains and operations, especially when dealing with discontinuous functions or different rules over intervals.In traditional calculus, you might be tasked with finding derivatives or integrals of such functions. While this exercise doesn't directly dive into differentiation or integration, it establishes the foundation needed for such tasks by reinforcing an understanding of how piecewise functions and their operations work. You start by familiarizing yourself with the functions' behavior over different intervals. For the given problem, setting boundaries at critical points in the domain, such as \( x = 1 \) and \( x = 2 \), prepares you for more advanced calculus concepts where continuous and differentiable functions are examined over similar intervals. Overall, these operations help strengthen your grasp of elementary calculus through practical application of domain and function behavior.