Problem 15 - 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) \((f / g)(x)\) $$ f(x)=x-\sqrt{x-1}, \quad g(x)=x+\sqrt{x-1} $$

Step-by-Step Solution

Answer

(f+g)(x) = 2x; (f-g)(x) = -2\sqrt{x-1}; (fg)(x) = x^2 - x + 1; (f/g)(x) = \frac{x - \sqrt{x-1}}{x + \sqrt{x-1}}. Domain for all is \( x \geq 1 \).

Step 1: Understand the Functions

We are given \( f(x) = x - \sqrt{x-1} \) and \( g(x) = x + \sqrt{x-1} \). Both functions involve the square root \( \sqrt{x-1} \), which is only defined when \( x \geq 1 \). Therefore, the domain for both \( f(x) \) and \( g(x) \) is \( x \geq 1 \).

Step 2: Calculate \((f+g)(x)\)

To find \((f+g)(x)\), simply add \( f(x) \) and \( g(x) \):\[(f+g)(x) = (x - \sqrt{x-1}) + (x + \sqrt{x-1}) = 2x\]Thus, \((f+g)(x) = 2x\), and the domain is \( x \geq 1 \) since it's derived from the original functions.

Step 3: Calculate \((f-g)(x)\)

To find \((f-g)(x)\), subtract \( g(x) \) from \( f(x) \):\[(f-g)(x) = (x - \sqrt{x-1}) - (x + \sqrt{x-1}) = -2\sqrt{x-1}\]Thus, \((f-g)(x) = -2\sqrt{x-1}\), and the domain remains \( x \geq 1 \).

Step 4: Calculate \((f \cdot g)(x)\)

To find \((f \cdot g)(x)\), multiply \( f(x) \) and \( g(x) \):\[(f \cdot g)(x) = (x - \sqrt{x-1})(x + \sqrt{x-1}) = x^2 - (\sqrt{x-1})^2 = x^2 - (x-1) = x^2 - x + 1\]Thus, \((f \cdot g)(x) = x^2 - x + 1\), with the domain \( x \geq 1 \).

Step 5: Calculate \((f / g)(x)\)

To find \((f / g)(x)\), divide \( f(x) \) by \( g(x) \):\[(f / g)(x) = \frac{x - \sqrt{x-1}}{x + \sqrt{x-1}}\]This function is undetermined when the denominator is zero, i.e., \( x + \sqrt{x-1} = 0 \). This happens when \( x = -\sqrt{x-1} \), which is not possible because \( \sqrt{x-1} \geq 0 \) and \( x \geq 1 \). Thus, there are no additional domain restrictions other than \( x \geq 1 \).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial in mathematics, particularly when dealing with algebraic expressions that include operations such as addition, subtraction, multiplication, or division. The domain of a function refers to the set of all possible input values, often represented as 'x', for which the function is defined.
In the exercise above, both functions given involve the expression \( \sqrt{x-1} \). A square root function is only defined for non-negative values; thus, \( x-1 \) must be greater than or equal to zero. To find this, solve the inequality:

• \( x - 1 \geq 0 \)
• \( x \geq 1 \)

Therefore, the domain for each original function, \( f(x) \) and \( g(x) \), is \( x \geq 1 \). This condition ensures no negative values are under the square root, which is why it's an essential part of identifying the domain.

Algebraic Expressions

Algebraic expressions use numbers, variables, and operations to represent a value or concept. They often involve addition, subtraction, multiplication, and division. In solving the original exercise, we perform these operations on the given functions \( f(x) \) and \( g(x) \).
For example, the expressions like \((f+g)(x)\) and \((f-g)(x)\) involve adding and subtracting terms of the functions. When expressing \((f+g)(x)\), sum each corresponding term:

• The term without the square root \(2x\).

Similarly, for subtraction, the square root terms cancel each other leading to an expression \(-2\sqrt{x-1}\). Understanding how to manipulate these algebraic expressions helps consolidate the concept of function operations, which requires combining and simplifying terms for such functions.

Square Root Function

A square root function, \( \sqrt{x-1} \), is a special type of radical function. Its primary characteristic is that it only accepts non-negative inputs, reflecting on the restricted domain conditions.
In our exercise, the square root functions \( \sqrt{x-1} \) play a crucial role in determining the overall expression. Calculating values of expressions involving the square root requires cautious handling, especially in multiplication or subtraction. For instance, with \((f-g)(x) = -2\sqrt{x-1}\), the square root term only provides real numbers if \( x \geq 1 \).

• The graph of a square root function is typically a curve that starts at the defined threshold (here, \( x = 1 \)) and stretches out to the right, indicating that it takes a defined range of positive x-values.

Overall, comprehending the behavior of the square root function is crucial for grasping function operations, particularly when deriving or simplifying expressions involving such terms.