Problem 20 - College Algebra with Modeling and Visualization

Question

Exercises \(11-30:\) Use \(f(x)\) and \(g(x)\) to find a formula for each expression. Identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d) \((f / g)(x)\) $$ f(x)=x^{1 / 2}, \quad g(x)=3 $$

Step-by-Step Solution

Answer

(a) \((f+g)(x) = x^{1/2} + 3\), domain: \(x \geq 0\); (b) \((f-g)(x) = x^{1/2} - 3\), domain: \(x \geq 0\); (c) \((f\cdot g)(x) = 3x^{1/2}\), domain: \(x \geq 0\); (d) \((f/g)(x) = x^{1/2}/3\), domain: \(x \geq 0\).

Step 1: Understanding Functions

The functions given to us are \( f(x) = x^{1/2} \) and \( g(x) = 3 \). These are the functions we will use to create new expressions.

Step 2: Addition of Functions

To find \((f+g)(x)\), we use the formula \((f+g)(x) = f(x) + g(x)\). By substituting the given functions: \((f+g)(x) = x^{1/2} + 3\).

Step 3: Domain of (f+g)(x)

The domain of \( f(x) = x^{1/2} \) is \( x \geq 0 \), and \( g(x) = 3 \) is defined for all real numbers. The domain of \( (f+g)(x) \) is restricted by \( f(x) \), so it is \( x \geq 0 \).

Step 4: Subtraction of Functions

To find \((f-g)(x)\), we use \((f-g)(x) = f(x) - g(x)\). By substituting the functions: \((f-g)(x) = x^{1/2} - 3\).

Step 5: Domain of (f-g)(x)

The domain for \((f-g)(x) = x^{1/2} - 3\) is the same as for \(f(x)\) since \(g(x)\) is defined everywhere. So, the domain is \( x \geq 0 \).

Step 6: Multiplication of Functions

For \((f \cdot g)(x)\), use \((f \cdot g)(x) = f(x) \cdot g(x)\). With the functions given: \((f \cdot g)(x) = x^{1/2} \cdot 3 = 3x^{1/2}\).

Step 7: Domain of (f \\cdot g)(x)

The domain is again determined by the most restrictive function, which is \( f(x) \). Therefore, \((f \cdot g)(x)\) has domain \( x \geq 0 \).

Step 8: Division of Functions

For \((f/g)(x)\), use \((f/g)(x) = f(x) / g(x)\). By substituting: \((f/g)(x) = x^{1/2} / 3\).

Step 9: Domain of (f/g)(x)

Since \( g(x) = 3 \) is never zero, the restriction solely comes from \( f(x) \). Thus, the domain of \((f/g)(x)\) is \( x \geq 0 \).

Key Concepts

Domain of a Function

Functions have a domain, which is the set of all possible input values (usually represented as \( x \)) for which the function is defined. In other words, the domain includes every number you are allowed to plug into a function without causing any mathematical mishaps like division by zero or taking the square root of a negative number.

For example:

• The function \( f(x) = x^{1/2} \) only accepts non-negative numbers (\( x \geq 0 \)) because we cannot take the square root of a negative number in the set of real numbers.
• The function \( g(x) = 3 \) does not have these restrictions; it is defined for all real numbers because it is a constant function with no variables to limit it.

When combining functions, their domains need to coincide; this often means the domain of the new function will be the intersection of their individual domains. This ensures the new function is also correctly defined. Always take note of what restrictions each function may have before performing operations.

Addition of Functions

To add two functions \( f \) and \( g \), simply sum their outputs for the same input \( x \). This operation is stated mathematically as \((f+g)(x) = f(x) + g(x)\).

Here's an example:

• If \( f(x) = x^{1/2} \) and \( g(x) = 3 \), then \((f+g)(x) = x^{1/2} + 3\).

It's like combining the results of both functions into one expression.
The critical part about addition is its domain. The new function's domain must respect the domains of \( f \) and \( g \). So, if there's any restriction from either function, it will also apply to \((f+g)(x)\). For our example, since \( f(x) = x^{1/2} \) requires \( x \geq 0 \), the domain of \((f+g)(x)\) is also \( x \geq 0 \). Always ensure both functions are defined at any chosen \( x \).

Subtraction of Functions

Subtraction of functions works similarly to addition. You subtract the output of \( g \) from the output of \( f \) for the same input \( x \). This is written as \((f-g)(x) = f(x) - g(x)\).

For example:

• With \( f(x) = x^{1/2} \) and \( g(x) = 3 \), the expression would be \((f-g)(x) = x^{1/2} - 3\).

The domain of \((f-g)(x)\) follows the same rule as addition: respect the steepest restriction of \( f \) and \( g \). Since \( f(x) = x^{1/2} \) works only for \( x \geq 0 \), this restriction carries over to \((f-g)(x)\).
Make sure the functions overlap correctly in their domains, which ensures that \((f-g)(x)\) is valid wherever both initial functions are defined.

Multiplication of Functions

When multiplying functions, you simply multiply their outputs for the same input. This is written as \((f \cdot g)(x) = f(x) \cdot g(x)\).

Take this example:

• Let \( f(x) = x^{1/2} \) and \( g(x) = 3 \), then \((f \cdot g)(x) = 3x^{1/2}\).

When you multiply these two functions, the domain concerns remain paramount. The domain of \((f \cdot g)(x)\) comes from overlapping the domains of \( f \) and \( g \).

Since \( f(x) = x^{1/2} \) imposes an \( x \geq 0 \) condition, this restriction is inherited by the product function \((f \cdot g)(x)\). Always ensure you respect the limitations implied by specific function definitions.

Division of Functions

Division involves dividing the output of \( f \) by the output of \( g \) for the same input \( x \), and is written as \((f/g)(x) = f(x) / g(x)\).

Example:

• If \( f(x) = x^{1/2} \) and \( g(x) = 3 \), then \((f/g)(x) = x^{1/2} / 3\).

Here’s an important note: make sure the divisor, \( g(x) \), is never zero as division by zero is undefined. Since \( g(x) = 3 \) for our example, there is no risk of zero, but this is a crucial check for other divisions.

The domain of \((f/g)(x)\) will follow the domain of both functions, except for values making \( g(x) = 0 \). In our case, since \( f(x) = x^{1/2} \) requires \( x \geq 0 \), this also becomes the domain for the division. Domain overlaps and any zero conditions in a real setting will drive the limits of this newly formed function.