Problem 40 - 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. $$ f(x)=\sqrt{x}, g(x)=x-2 $$

Step-by-Step Solution

Answer

1. \( f+g = \sqrt{x} + x - 2 \), Domain: \( x \geq 0 \). 2. \( f-g = \sqrt{x} - x + 2 \), Domain: \( x \geq 0 \). 3. \( f \cdot g = \sqrt{x}(x - 2) \), Domain: \( x \geq 0 \). 4. \( \frac{f}{g} = \frac{\sqrt{x}}{x - 2} \), Domain: \([0, 2) \cup (2, \infty) \).

Step 1: Define the Functions

First, let's define the functions. \( f(x) = \sqrt{x} \) and \( g(x) = x - 2 \). We're tasked with finding different combinations of these functions: \( f+g \), \( f-g \), \( f \cdot g \), and \( \frac{f}{g} \). We'll also determine the domain for each combination.

Step 1: Find \( f+g \)

The sum of two functions is defined as \( (f+g)(x) = f(x) + g(x) \). Thus, \[(f+g)(x) = \sqrt{x} + (x - 2)\]. The domain of \( f(x) = \sqrt{x} \) is \( x \geq 0 \) because square roots are not defined for negative numbers. The domain of \( g(x) = x - 2 \) is all real numbers. Therefore, the common domain where both functions are defined is \( x \geq 0 \). So the domain of \( f+g \) is \( x \geq 0 \).

Step 2: Find \( f-g \)

The difference of two functions is defined as \( (f-g)(x) = f(x) - g(x) \). This gives us \[(f-g)(x) = \sqrt{x} - (x - 2) = \sqrt{x} - x + 2\]. Using the same logic as in Step 1, the domain of \( f-g \) is also where both functions are defined, which is \( x \geq 0 \).

Step 3: Find \( f \cdot g \)

The product of two functions is defined as \( (f \cdot g)(x) = f(x) \cdot g(x) \). Therefore, \[(f \cdot g)(x) = \sqrt{x} \cdot (x - 2)\]. The domain of \( f \cdot g \) is the same as the individual functions: \( x \geq 0 \).

Step 4: Find \( \frac{f}{g} \)

The quotient of two functions is defined as \( \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} \). Thus,\[\left( \frac{f}{g} \right)(x) = \frac{\sqrt{x}}{x-2}\]. The function \( g(x) \) cannot be zero, so we set \( x - 2 = 0 \) leading to \( x = 2 \). Hence the domain excludes this value. Thus, the domain of \( \frac{f}{g} \) is \( x \geq 0 \), except \( x eq 2 \). The domain is \([0, 2) \cup (2, \infty) \).

Key Concepts

Function Operations

Function operations involve combining two or more functions through addition, subtraction, multiplication, or division. When performing operations with functions, like the ones in our exercise, it is important to remember how these combinations impact the resulting function and their domain.
For example:

• For addition, \[(f+g)(x) = f(x) + g(x),\]we simply add the outputs of the functions.
• For subtraction, \[(f-g)(x) = f(x) - g(x),\]we subtract the outputs.
• Multiplication is done by multiplying the outputs: \[(f \cdot g)(x) = f(x) \cdot g(x).\]
• Division requires dividing the outputs, but with an additional condition: \[ \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)}\]and importantly, the denominator cannot be zero.

By understanding these operations, we can determine new functions, provided we are mindful of any pitfalls, especially regarding division.

Domain

The domain of a function is all the possible input values (x-values) that allow the function to be defined without resulting in any mathematical errors. When combining functions, finding a common domain is crucial to ensure all operations are valid.
For each operation:

• When adding or subtracting functions, consider the domain where both original functions are defined. This common domain forms the new function's domain.
• The domain for multiplication remains similar to that of addition and subtraction, focusing on where both functions are valid.
• For division, it's crucial not only to find the common domain but also to avoid any x-values that make the denominator zero. This introduces constraints within the domain.

By carefully evaluating these conditions, we ensure each of our new functions remains valid and useful.

Composite Functions

Composite functions involve plugging one function into another. For example, if we have two functions \(f(x)\) and \(g(x)\), a composite function is written as \(f(g(x))\). This means you first calculate \(g(x)\) and then use that result as the input for \(f\).
Composite functions require special attention to their domain. The domain must satisfy two conditions:

• The inner function \(g(x)\) must be defined for its inputs.
• The form created by \(g(x)\) must also be valid within the domain of \(f(x)\).

This ensures that the entire process of substitution doesn't lead to undefined operations. It ties closely to understanding the domains of both involved functions.

Square Root Function

The square root function \(f(x) = \sqrt{x}\) is a basic yet essential type of function. It is only defined for values of \(x\) that are greater than or equal to zero, because the square root of a negative number is not a real number.
Moving beyond its domain, the square root function has:

• A range of \([0, \infty)\), meaning its outputs will always be non-negative.
• A basic form that can be combined using operations to create more complex functions.

Its properties make it a wonderful tool in mathematical calculations, particularly when seeking non-negative solutions. When working with square roots, always check that your inputs respect its domain to avoid errors.