Problem 11 - 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) \(f \circ g ;(\mathrm{g}) g \circ f\). $$ f(x)=\sqrt{x^{2}-1} ; g(x)=\sqrt{x-1} $$

Step-by-Step Solution

Answer

Function definitions with domains: a) \( f+g \) and b) \( f-g \): \( [1, +\infty) \); c) \( f \cdot g \): \( [1, +\infty) \); d) \( f \/ g \): \( [1, +\infty) \); e) \( g \/ f \): \( [1, +\infty) \);\( f \circ g \): \( [2, +\infty) \);\( g \circ f \): \( (-\infty, -\sqrt{2}] \cup [\sqrt{2}, \infty) \)}

Step 1: Define the Functions

Given functions are: \( f(x) = \sqrt{x^2 - 1} \) \( g(x) = \sqrt{x - 1} \)

Step 2: Find Domain of \( f(x) \)

The domain of \( f(x) \) is where the expression inside the square root is non-negative. Solving \( x^2 - 1 \geq 0 \), we get: \( (x-1)(x+1) \geq 0 \) This inequality holds for: \( x \leq -1 \) or \( x \geq 1 \). Thus, the domain of \( f(x) \) is: \( (-\infty, -1] \cup [1, +\infty) \)

Step 3: Find Domain of \( g(x) \)

The domain of \( g(x) \) is where the expression inside the square root is non-negative. Solving \( x - 1 \geq 0 \), we get: \( x \geq 1 \). Thus, the domain of \( g(x) \) is: \( [1, +\infty) \)

Step 4: Define Function f+g

The function \( f+g \) is: \( (f+g)(x) = \sqrt{x^2 - 1} + \sqrt{x - 1} \) The domain of \( f+g \) is the intersection of domains of \( f(x) \) and \( g(x) \) which is: \( [1, +\infty) \)

Step 5: Define Function f-g

The function \( f-g \) is: \( (f-g)(x) = \sqrt{x^2 - 1} - \sqrt{x - 1} \) The domain of \( f-g \) is the same as \( f+g \): \( [1, +\infty) \)

Step 6: Define Function f \cdot g

The function \( f \cdot g \) is: \( (f \cdot g)(x) = \sqrt{x^2 - 1} \cdot \sqrt{x - 1} \) Simplifying further: \( (f \cdot g)(x) = \sqrt{(x^2 - 1)(x - 1)} \) The domain of \( f \cdot g \) is also \( [1, +\infty) \)

Step 7: Define Function f \/ g

The function \( f \/ g \) is: \( (f \/ g)(x) = \frac{\sqrt{x^2 - 1}}{\sqrt{x - 1}} \) The domain of \( f \/ g \) is the same as \( [1, +\infty) \) (since \( g(x) eq 0 \) in this interval).

Step 8: Define Function g \/ f

The function \( g \/ f \) is: \( (g \/ f)(x) = \frac{\sqrt{x - 1}}{\sqrt{x^2 - 1}} \) The domain of \( g \/ f \) is the same as \( [1, +\infty) \) (since \( f(x) eq 0 \) in this interval).

Step 9: Define Composition f \circ g

The function \( f \circ g \) is: \( (f \circ g)(x) = \sqrt{(\sqrt{x-1})^2 - 1} \) Simplifying further, we get: \( (f \circ g)(x) = \sqrt{x-1 - 1} = \sqrt{x-2} \) The domain of \( f \circ g \) requires: \( x - 2 \geq 0 \), thus: \( x \geq 2 \). So, the domain is: \( [2, +\infty) \)

Step 10: Define Composition g \circ f

The function \( g \circ f \) is: \( (g \circ f)(x) = \sqrt{(\sqrt{x^2 - 1}) - 1} \) The domain of \( g \circ f \) requires: \( \sqrt{x^2 - 1} - 1 \geq 0 \). Solving \( \sqrt{x^2 - 1} \geq 1 \), we get: \( f(x) \geq 1\) which simplifies to: \( x^2 - 1 \geq 1 \ <=> x^2 \geq 2 \ <=> x \geq \sqrt{2} \ or \ x \leq -\sqrt{2}\). Thus, the domain is: \( (-\infty, -\sqrt{2}] \cup [\sqrt{2}, \infty) \)

Key Concepts

Function Composition

Function composition involves plugging one function into another. This is like a math nesting doll! For two functions, say \(f(x)\) and \(g(x)\), the composition is written as \((f \circ g)(x)\), which means \(f(g(x))\). Essentially, you take the output of \(g(x)\) and use it as the input in \(f(x)\).

The domain of \(f \circ g\) is the set of all \(x\) values for which \(g(x)\) is within the domain of \(f\). For the given functions, \(f(x) = \sqrt{x^2 - 1}\) and \(g(x) = \sqrt{x - 1}\):

• First, solve \(f(g(x))\) and \(g(f(x))\).
• For \(f \circ g\), \(f(g(x)) = \sqrt{(\sqrt{x-1})^2 - 1} = \sqrt{x - 2}\). The domain is \([2, +\infty)\), where \(x-2 \geq 0\).
• For \(g \circ f\), \(g(f(x)) = \sqrt{\sqrt{x^2 - 1} - 1}\). The domain is \((-\infty, -\sqrt{2}] \cup [\sqrt{2}, +\infty)\), solving \(\sqrt{x^2 - 1} - 1 \geq 0\).

Remember, you must first ensure \(g(x)\) lies in the domain of \(f\) for \(f \circ g\), and similarly for \(g \circ f\).

Domain of a Function

Understanding the domain is crucial in determining where a function is defined. A domain is simply the set of all possible input values (\(x\)) for which the function produces a valid output.

For the given functions, let’s break this down:

• For \(f(x) = \sqrt{x^2 - 1}\), the expression inside the square root must be non-negative: \(x^2 - 1 \geq 0\). Solving, we get: \(x \leq -1\) or \(x \geq 1\), meaning the domain is \((-\infty, -1] \cup [1, +\infty)\).
• For \(g(x) = \sqrt{x - 1}\), the expression inside the square root must also be non-negative: \(x - 1 \geq 0\). Solving this, we get \(x \geq 1\), so the domain is \([1, +\infty)\).

Hence, determining the domain involves ensuring the function outputs are real and valid based on the mathematical operations inside it.

Function Operations

Function operations include addition, subtraction, multiplication, and division. Applying these operations often impacts the domain of the resulting function. Let’s explore the operations for given functions \(f(x) = \sqrt{x^2 - 1}\) and \(g(x) = \sqrt{x - 1}\):

• **Addition**: \((f + g)(x) = \sqrt{x^2 - 1} + \sqrt{x - 1}\) The domain is the common values where both \(f(x)\) and \(g(x)\) are defined: \([1, +\infty)\).
• **Subtraction**: \((f - g)(x) = \sqrt{x^2 - 1} - \sqrt{x - 1}\) The domain remains \([1, +\infty)\).
• **Multiplication**: \((f \cdot g)(x) = \sqrt{x^2 - 1} \cdot \sqrt{x - 1} = \sqrt{(x^2 - 1)(x - 1)}\) The domain is still \([1, +\infty)\).
• **Division**: \((f \/ g)(x) = \frac{\sqrt{x^2 - 1}}{\sqrt{x - 1}}\) The domain is \([1, +\infty)\), ensuring \(g(x) \eq 0\).
• **Reverse Division**: \((g \/ f)(x) = \frac{\sqrt{x - 1}}{\sqrt{x^2 - 1}}\) The domain remains \([1, +\infty)\).

Function operations define how you combine functions and influence the resulting domain.