Problem 6 - Discrete Mathematics with Applications

Question

Let \(f(x)=\lfloor x\rfloor\) and \( g(x)=\lceil x\rceil,\) where \(x \in \mathbb{R} .\) Compute each. $$(f \circ g)(-2.3)$$

Step-by-Step Solution

Answer

The short answer to the question is (f ⚬ g)(-2.3) = -2.

Step 1: Evaluate g(-2.3)

Let's start by evaluating the function g at the given value, -2.3. Since g(x) = ceil(x), we will be rounding the input value up to its nearest integer. Thus, \(g(-2.3) = \lceil -2.3 \rceil = -2\).

Step 2: Evaluate f(g(-2.3))

Now that we have found g(-2.3) = -2, we will input this result into the function f. Since f(x) = floor(x), and our input value is already an integer, the result will remain the same: \(f(g(-2.3)) = f(-2) = \lfloor -2 \rfloor = -2\).

Step 3: Final Answer

The composition of the two functions at the given value is (f ⚬ g)(-2.3) = -2.

Key Concepts

Function Composition

Function composition is when you apply one function to the result of another function. In mathematical terms, if you have two functions, say \( f(x) \) and \( g(x) \), then their composition is denoted as \((f \circ g)(x)\). This represents \(f(g(x))\), meaning you first apply \( g \) to \( x \), and then apply \( f \) to the outcome of \( g(x) \).

Here's how to tackle it:

• Evaluate the inside function first, \(g(x)\) in this case.
• Once you have the result, use it as the input for the outside function \(f\).

For example, in our exercise:

1. First, find \(g(-2.3)\).
2. Use this result and find \(f(g(-2.3))\).
By following these steps, you're systematically simplifying the problem to find the composition of two functions.

Real Numbers

Real numbers include all the numbers on the number line. This means anything from negative numbers, like -3, to fractions, like 1/2, and even irrational numbers like \(\sqrt{2}\).

Here's a quick breakdown:

• Integers: Whole numbers, including negative and positive numbers, like -3, 0, 1.
• Rational numbers: Numbers that can be expressed as a fraction, like 3/4.
• Irrational numbers: Numbers that cannot be written as a simple fraction, such as \(\pi\).

In the exercise, we use \(x \in \mathbb{R}\), indicating that \(x\) could be any real number. We particularly looked at how the floor and ceiling functions operate on these real numbers.

Integer Functions

Integer functions, like floor and ceiling functions, help transform real numbers into integers. They are useful in many areas of mathematics and computer algorithms.

• Floor Function \(\lfloor x \rfloor\): This rounds \(x\) down to the nearest integer. For example, \(\lfloor 3.7 \rfloor = 3\) and \(\lfloor -2.3 \rfloor = -3\).
• Ceiling Function \(\lceil x \rceil\): This rounds \(x\) up to the next integer. For example, \(\lceil 3.7 \rceil = 4\) and \(\lceil -2.3 \rceil = -2\).

In our problem, these functions were used to determine the outcome of something seemingly complex, like \((f \circ g)(-2.3)\). By understanding how to apply these integer functions, it's easy to see how real numbers are converted into manageable integers.