Problem 39 - Calculus Volume 1

Question

For the following exercises, for each pair of functions, find a. \(f+g\) b. \(f-g\) c. \(f \cdot g\) d. \(f / g .\) Determine the domain of each of these new functions. $$ f(x)=9-x^{2}, g(x)=x^{2}-2 x-3 $$

Step-by-Step Solution

Answer

a. \(6-2x\), b. \(12-2x^2+2x\), c. \(-x^4+2x^3+12x^2-18x-27\), d. \(\frac{9-x^2}{x^2-2x-3}\), domains: \(\mathbb{R}\), \(\mathbb{R}\), \(\mathbb{R}\), \({x \in \mathbb{R} \setminus \{3, -1\}}\).

Step 1: Addition of Functions

To find \(f+g\), simply add the functions together:\[ f(x) + g(x) = (9-x^2) + (x^2-2x-3) \]Combine like terms:\[ 9 - x^2 + x^2 - 2x - 3 = 6 - 2x \]Therefore, \(f+g = 6 - 2x\). The domain of this function is all real numbers \(\mathbb{R}\), as there are no restrictions for polynomials.

Step 2: Subtraction of Functions

To find \(f-g\), subtract \(g(x)\) from \(f(x)\):\[ f(x) - g(x) = (9-x^2) - (x^2-2x-3) \]Distribute the negative:\[ 9 - x^2 - x^2 + 2x + 3 = 12 - 2x^2 + 2x \]Thus, \(f-g = 12 - 2x^2 + 2x\). The domain is all real numbers \(\mathbb{R}\), since it's a polynomial function.

Step 3: Multiplication of Functions

To find \(f \cdot g\), multiply the functions:\[ f(x) \cdot g(x) = (9-x^2)(x^2-2x-3) \]Expand this using the distributive property:\[ = 9x^2 - 18x - 27 - x^4 + 2x^3 + 3x^2 \]Combine like terms:\[ = -x^4 + 2x^3 + 12x^2 - 18x - 27 \]Therefore, \(f \cdot g = -x^4 + 2x^3 + 12x^2 - 18x - 27 \) with domain \(\mathbb{R}\), as it remains a polynomial.

Step 4: Division of Functions and Domain

To find \(f / g\), divide \(f(x)\) by \(g(x)\):\[ \frac{f(x)}{g(x)} = \frac{9-x^2}{x^2-2x-3} \]Ensure that the denominator is not zero:Solve \(x^2-2x-3 = 0\) by factoring:\[ (x-3)(x+1) = 0 \]Thus, \(x = 3\) or \(x = -1\). The domain of \(f / g\) is all real numbers except \(x = 3\) and \(x = -1\), given by \({x \in \mathbb{R} \setminus \{3, -1\}}\).

Key Concepts

Polynomial Addition

Adding polynomials is much like assembling blocks to make a new structure. When you add two polynomial functions, you combine their like terms to create a new polynomial. In the exercise, you have two functions:

• \( f(x) = 9 - x^2 \)
• \( g(x) = x^2 - 2x - 3 \)

To determine \( f+g \), combine the expressions:\[(9 - x^2) + (x^2 - 2x - 3)\]Notice how \( -x^2 \) and \( x^2 \) cancel each other out, simplifying your expression to:\[6 - 2x\]This result, \( f+g = 6 - 2x \), is a linear polynomial. The domain of any polynomial is all real numbers \( \mathbb{R} \), meaning that this function is valid for every real number. Think of it like a never-ending road with no stops or breaks.

Polynomial Multiplication

Polynomial multiplication can initially seem daunting, but it's simply a matter of distributing each term in one polynomial across each term in the other. When multiplying two polynomial functions like \( f(x) \) and \( g(x) \), use the distributive property:\[(9-x^2)(x^2-2x-3)\]First, multiply \( 9 \) by each term in \( g(x) \). Then multiply \( -x^2 \) by each term in \( g(x) \). Combine all these terms:\[9x^2 - 18x - 27 - x^4 + 2x^3 + 3x^2\]After grouping like terms, you get:\[-x^4 + 2x^3 + 12x^2 - 18x - 27\]This is the new polynomial, \( f \cdot g \). Its domain is all real numbers \( \mathbb{R} \) because it's a polynomial, and polynomials are defined everywhere on the real number line.

Function Division

Dividing functions involves placing one function over another as a fraction, and then considering the domain carefully. For \( f(x) \) divided by \( g(x) \), the expression is:\[\frac{f(x)}{g(x)} = \frac{9-x^2}{x^2-2x-3}\]The trick here is ensuring the denominator is never zero because division by zero is undefined. Solve \( x^2 - 2x - 3 = 0 \) using factoring:\[(x-3)(x+1) = 0\]Giving solutions \( x = 3 \) and \( x = -1 \). Therefore, the domain includes all real numbers except \( x = 3 \) and \( x = -1 \). This ensures that the function remains meaningful and operational at every point.

Function Domains

The domain of a function is like the terrain a car can drive on. It tells us where the function is operative and valid. For polynomial functions, the terrain is often wide open because they usually stretch across all real numbers \( \mathbb{R} \) without restrictions.
However, for functions involving division, such as \( \frac{f(x)}{g(x)} \), we must be cautious. These functions can't operate where their denominator equals zero. Finding these zero points involves solving the equation where the denominator is set to zero. The results, like in our problem's example, indicate the points that must be excluded from the domain.
Understanding the domain is essential for correctly applying these functions to real-world problems or further mathematical inquiries.