Problem 8 - Precalculus Mathematics for Calculus

Question

Find \(f+g, f-g, f g,\) and \(f / g\) and their domains. $$f(x)=\sqrt{9-x^{2}}, \quad g(x)=\sqrt{x^{2}-4}$$

Step-by-Step Solution

Answer

Find expressions by adding, subtracting, multiplying, and dividing functions with domain \([-3,-2] \cup [2,3]\) for fg, f+g, f-g; and \([-3,-2) \cup (2,3]\) for f/g.

Step 1: Identify Operations

To solve this problem, we need to find the expressions for \(f+g, f-g, fg,\) and \(f/g\) by using the functions \(f(x) = \sqrt{9-x^2}\) and \(g(x) = \sqrt{x^2-4}\). We will also determine their domains.

Step 2: Addition of Functions

The expression for \(f+g\) is given by adding the two functions:\[(f+g)(x) = \sqrt{9-x^2} + \sqrt{x^2-4}.\] The domain of \(f+g\) is determined by the intersection of the individual domains, \(f(x)\) and \(g(x)\). For \(f(x) = \sqrt{9-x^2}\), \(9-x^2\geq 0\) implies \(-3 \leq x \leq 3\). For \(g(x) = \sqrt{x^2-4}\), \(x^2-4 \geq 0\) implies \(|x|\geq 2\), so \(x \leq -2\) or \(x \geq 2\). The intersection is \([-3,-2] \cup [2,3]\).

Step 3: Subtraction of Functions

The expression for \(f-g\) is given by subtracting the functions:\[(f-g)(x) = \sqrt{9-x^2} - \sqrt{x^2-4}.\] The domain is the same as in Step 2, \([-3,-2] \cup [2,3]\), because the same conditions apply to both functions.

Step 4: Multiplication of Functions

The expression for \(fg\) is given by multiplying the functions:\[(fg)(x) = \sqrt{9-x^2} \times \sqrt{x^2-4}.\] The domain remains \([-3,-2] \cup [2,3]\) as it is derived from the intersection of individual domains.

Step 5: Division of Functions

The expression for \(f/g\) is given by dividing the functions:\[(f/g)(x) = \frac{\sqrt{9-x^2}}{\sqrt{x^2-4}}.\] We need \(g(x) eq 0\) to avoid division by zero. Since \(g(x) = \sqrt{x^2-4}\), \(g(x) = 0\) when \(x=\pm 2\). Hence, the domain is \([-3,-2) \cup (2,3]\).

Key Concepts

Domain of Function

Understanding the domain of a function is crucial as it defines the set of possible input values the function can accept. For square root functions like we have here, it's important to ensure that what's under the square root is non-negative, since taking the square root of a negative number results in a non-real outcome.

• For the function \( f(x) = \sqrt{9 - x^2} \), the expression under the square root, \( 9-x^2 \), must be greater than or equal to zero. Solving \( 9 - x^2 \geq 0 \), we get \(-3 \leq x \leq 3\).
• For the function \( g(x) = \sqrt{x^2 - 4} \), we require \( x^2-4 \geq 0 \). Solving this inequality, we find that \( |x| \geq 2 \), or more clearly, \( x \leq -2 \) or \( x \geq 2 \).

When considering both functions at once, the domain of the combined function operations \( f+g, f-g, fg \), and \( f/g \) is the intersection of their individual domains: \([-3, -2] \cup [2, 3]\). This means only values of \( x \) in these intervals satisfy both functions simultaneously.

Addition of Functions

Adding functions involves simply taking the sum of their expressions and determining where this sum is valid. For the functions \( f(x) \) and \( g(x) \), the addition is represented by \((f+g)(x) = \sqrt{9-x^2} + \sqrt{x^2-4}\).The key to finding the domain for this operation is ensuring that both functions are defined together, which means sticking to their intersected domain, \([-3,-2] \cup [2,3]\). Inside these intervals, both square roots return real numbers, enabling the addition itself to result in real outputs.

Subtraction of Functions

Similar to addition, subtracting two functions involves taking their difference. Here, we express it as \((f-g)(x) = \sqrt{9-x^2} - \sqrt{x^2-4}\).The domain remains \([-3,-2] \cup [2,3]\) because we are still constrained by the domains of \( f(x) \) and \( g(x) \). Within these intervals, each function provides valid, real outputs, allowing the subtraction operation to yield real results without encountering issues with undefined elements.

Multiplication of Functions

Multiplying functions involves creating a new function by taking the product of the two expressions. In our case, \((fg)(x) = \sqrt{9-x^2} \times \sqrt{x^2-4}\). Again, the critical part is ensuring that both original functions are defined over the same set of \( x \)-values.

• The domain here is still \([-3,-2] \cup [2,3]\). Within these ranges, the square roots fetch real numbers, meaning their product also results in real number outputs.

The domain ensures you are only working with input values where both \( f(x) \) and \( g(x) \) are real and valid.

Division of Functions

When dividing functions, the situation becomes slightly more complex due to the potential for division by zero. The division here is expressed as \((f/g)(x) = \frac{\sqrt{9-x^2}}{\sqrt{x^2-4}}\).To safely divide, \( g(x) \) must not be zero. Since \( g(x) = \sqrt{x^2 - 4} \), this function becomes zero when \( x = \pm 2 \). Hence, we must exclude these points to avoid undefined behavior.Thus, the domain for \( f/g \) excludes these critical points, resulting in a domain of \([-3,-2) \cup (2,3]\). This range encompasses all available values except where \( g(x) \) leads to division by zero.