Problem 7 - The Calculus with Analytic Geometry

Question

In Exercises 7 through 12, the functions \(f\) and \(g\) are defined. In each problem define the following functions and determine the domain of the resulting function: (a) \(f+g ;\) (b) \(f-g ;\) (c) \(f \cdot g ;\) (d) \(f / g ;\) (e) \(g / f\); (f) \(f \circ g ;(\mathrm{g}) g \circ f\). $$ f(x)=x-5 ; g(x)=x^{2}-1 $$

Step-by-Step Solution

Answer

Defined functions and their domains are: (a) \( f+g = x^2 + x - 6 \), domain is \( \mathbb{R} \) (b) \( f-g = -x^2 + x - 4 \), domain is \( \mathbb{R} \) (c) \( f \cdot g = x^3 - x - 5x^2 + 5 \), domain is \( \mathbb{R} \) (d) \( \frac{f}{g} = \frac{x-5}{x^2-1} \), domain is \( \mathbb{R} \setminus \{ \pm 1 \} \) (e) \( \frac{g}{f} = \frac{x^2-1}{x-5} \), domain is \( \mathbb{R} \setminus \{ 5 \} \) (f) \( f \circ g = x^2 - 6 \), domain is \( \mathbb{R} \) (g) \( g \circ f = x^2 - 10x + 24 \), domain is \( \mathbb{R} \)

Step 1 - Define functions

Given functions are: \( f(x) = x - 5 \)\( g(x) = x^2 - 1 \)

Step 2 - Define and find the domain of \( f+g \)

The function \( f+g \) is defined as \( (f+g)(x) = f(x) + g(x) \). Substituting the given functions:\( (f+g)(x) = (x - 5) + (x^2 - 1) = x^2 + x - 6 \).Since both functions are polynomials, their domains are all real numbers. Therefore, the domain of \( f+g \) is \( \mathbb{R} \).

Step 3 - Define and find the domain of \( f-g \)

The function \( f-g \) is defined as \( (f-g)(x) = f(x) - g(x) \). Substituting the given functions:\( (f-g)(x) = (x - 5) - (x^2 - 1) = -x^2 + x - 4 \).The domain of \( f-g \) is all real numbers since both functions are polynomials. Therefore, the domain is \( \mathbb{R} \).

Step 4 - Define and find the domain of \( f \cdot g \)

The function \( f \cdot g \) is defined as \( (f \cdot g)(x) = f(x) \cdot g(x) \). Substituting the given functions:\( (f \cdot g)(x) = (x - 5) \cdot (x^2 - 1) = x^3 - x - 5x^2 + 5 \).The domain of \( f \cdot g \) is all real numbers, so the domain is \( \mathbb{R} \).

Step 5 - Define and find the domain of \( \frac{f}{g} \)

The function \( \frac{f}{g} \) is defined as \( \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} \). Substituting the given functions:\( \left( \frac{f}{g} \right)(x) = \frac{x-5}{x^2-1} \).For the function to be defined, the denominator cannot be zero. Therefore, we solve:\( x^2 - 1 = 0 \Rightarrow x = \pm 1 \).So, the domain is all real numbers except \( x = 1 \) and \( x = -1 \), which is \( \mathbb{R} \setminus \{ \pm 1 \} \).

Step 6 - Define and find the domain of \( \frac{g}{f} \)

The function \( \frac{g}{f} \) is defined as \( \left( \frac{g}{f} \right)(x) = \frac{g(x)}{f(x)} \). Substituting the given functions:\( \left( \frac{g}{f} \right)(x) = \frac{x^2-1}{x-5} \).For the function to be defined, the denominator cannot be zero. Therefore, we solve:\( x - 5 = 0 \Rightarrow x = 5 \).So, the domain is all real numbers except \( x = 5 \), which is \( \mathbb{R} \setminus \{ 5 \} \).

Step 7 - Define and find the domain of \( f \circ g \)

The function \( f \circ g \) is defined as \( (f \circ g)(x) = f(g(x)) \). Substituting the given functions:\( (f \circ g)(x) = f(x^2-1) = (x^2-1) - 5 = x^2 - 6 \).Since \( g(x) \) maps all real numbers to all real numbers, and \( f(x) \) is defined for all real inputs, the domain of \( f \circ g \) is \( \mathbb{R} \).

Step 8 - Define and find the domain of \( g \circ f \)

The function \( g \circ f \) is defined as \( (g \circ f)(x) = g(f(x)) \). Substituting the given functions:\( (g \circ f)(x) = g(x-5) = (x-5)^2 - 1 = x^2 - 10x + 25 - 1 = x^2 - 10x + 24 \).Since \( f(x) \) maps all real numbers to all real numbers, and \( g(x) \) is defined for all real inputs, the domain of \( g \circ f \) is \( \mathbb{R} \).

Key Concepts

composite functions

Understanding composite functions is crucial in calculus and many other areas of math. A composite function is formed when one function is applied to the result of another function. We denote this as \( (f \circ g)(x) \), which means you first apply function \ g \ to \ x \, and then apply function \ f \ to the result. For example, if \ f(x) = x - 5 \ and \ g(x) = x^2 - 1 \, the composite function \ (f \circ g)(x) \ is \ f(g(x)) \, which simplifies to \ f(x^2 - 1) = x^2 - 6 \.
Another example would be \ g \circ f \, which means finding \ g(f(x)) \. Given the same \ f(x) \ and \ g(x) \, we have \ g(f(x)) = g(x - 5) = (x - 5)^2 - 1 = x^2 - 10x + 24 \.
When dealing with composite functions, it’s essential to remember that the order of composition matters. \ (f \circ g)(x) \ is usually different from \ (g \circ f)(x) \.

domain of functions

A function's domain is the set of all possible input values (x-values) for which the function is defined. To determine the domain of a function composed of two functions, we need to understand each function's domain and how the functions interact.
For the simple polynomial functions like \ f(x) = x - 5 \ and \ g(x) = x^2 - 1 \, the domains are all real numbers, denoted as \ \( \mathbb{R} \) \, because you can plug any real number into these functions.
However, when functions involve division, the domain must exclude values that make the denominator zero. For instance, the domain of \ \( \frac{f}{g} \) \ is all real numbers except \ x = 1 \ and \ x = -1 \, since \ g(x) = 0 \ when \ x \ takes these values. Similarly, for \ \( \frac{g}{f} \) \, the domain excludes \ x = 5 \, as this makes \ f(x) = 0 \

function operations

Function operations involve combining functions in various ways. The primary operations with functions include addition, subtraction, multiplication, and division.
Given functions \ f(x) = x - 5 \ and \ g(x) = x^2 - 1 \, here’s how these operations work:
• Addition: \ (f+g)(x) = (x - 5) + (x^2 - 1) = x^2 + x - 6 \.
Because both functions are polynomials, their sum is also a polynomial. The same domain of all real numbers applies.
• Subtraction: \ (f-g)(x) = (x - 5) - (x^2 - 1) = -x^2 + x - 4 \. Again, as a polynomial, the domain is all real numbers.
• Multiplication: \ (f \cdot g)(x) = (x - 5) \cdot (x^2 - 1) = x^3 - x - 5x^2 + 5 \. The domain remains all real numbers.
• Division: \ \( \left( \frac{f}{g} \right)(x) = \frac{x-5}{x^2-1} \) \. To avoid division by zero, exclude \ x = 1 \ and \ x = -1 \ from the domain.
Understanding these basics helps in tackling more complex problems involving functions.