Problem 17 - 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) \((f / g)(x)\) $$ f(x)=\sqrt{x}-1, \quad g(x)=\sqrt{x}+1 $$

Step-by-Step Solution

Answer

The formulas are: (a) \(2\sqrt{x}\), (b) \(-2\), (c) \(x-1\), (d) \(\frac{\sqrt{x}-1}{\sqrt{x}+1}\); all have domain \([0, \infty)\).

Step 1: Understand the Problem

We are given two functions: \(f(x) = \sqrt{x} - 1\) and \(g(x) = \sqrt{x} + 1\). Our task is to find the formulas for the expressions \((f+g)(x)\), \((f-g)(x)\), \((f \cdot g)(x)\), and \((f/g)(x)\), and determine the domain for each resulting function.

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

To find \((f+g)(x)\), sum the expressions for \(f(x)\) and \(g(x)\).\[(f+g)(x) = f(x) + g(x) = (\sqrt{x} - 1) + (\sqrt{x} + 1) = 2\sqrt{x}\]**Domain:** Both \(f(x)\) and \(g(x)\) require \(x \geq 0\) to keep the values real, so the domain of \((f+g)(x)\) is \([0, \infty)\).

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

To find \((f-g)(x)\), subtract \(g(x)\) from \(f(x)\):\[(f-g)(x) = f(x) - g(x) = (\sqrt{x} - 1) - (\sqrt{x} + 1) = -2\]**Domain:** Since the expression is constant for all \(x\), the domain is still constrained by \(f(x)\) and \(g(x)\), thus \([0, \infty)\).

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

To find \((f \cdot g)(x)\), multiply \(f(x)\) and \(g(x)\):\[(f \cdot g)(x) = f(x) \cdot g(x) = (\sqrt{x} - 1)(\sqrt{x} + 1) = x - 1\]**Domain:** The multiplication doesn't add any new restrictions, so the domain remains \([0, \infty)\).

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

To find \((f / g)(x)\), divide \(f(x)\) by \(g(x)\):\[(f / g)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x} - 1}{\sqrt{x} + 1}\]**Domain:** For the division expression \(g(x) = \sqrt{x} + 1 eq 0\), ensuring no division by zero. However, \(\sqrt{x} + 1 > 0\) for \(x \geq 0\), so the domain remains \([0, \infty)\).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial to solving problems in calculus and algebra. The domain refers to the set of input values (or x-values) for which a function is defined and returns real, valid results. A function can sometimes have restrictions based on mathematical operations involved. For instance:

• Square roots require the value inside to be non-negative.
• Divisions cannot have zero in the denominator.

In our specific problem, both functions, \( f(x) = \sqrt{x} - 1 \) and \( g(x) = \sqrt{x} + 1 \), involve square roots. This means their domains are constrained to non-negative numbers, specifically \( x \geq 0 \). Consequently, when we combine these functions via addition, subtraction, multiplication, or division, we must ensure that these domain restrictions are observed. This ensures that all resulting operations are valid within the defined domain of \([0, \infty)\).

Square Root Function

The square root function is quite simple yet often misunderstood due to its inherent domain restriction. The function \( y = \sqrt{x} \) represents the positive square root of \( x \) and is only defined when \( x \geq 0 \). This is because square roots of negative numbers are not real numbers.
Consider the expressions given in the original exercise: \( f(x) = \sqrt{x} - 1 \) and \( g(x) = \sqrt{x} + 1 \). Each expression involves a square root, meaning their operations are constrained by the same domain restriction. When you perform operations like addition or subtraction on square root functions, it is key to remember:

• Keep the domain restricted to where \( x \geq 0 \) to ensure the function is valid.
• Addition or subtraction of square roots keeps the domain constraint unchanged.

These guidelines help maintain the integrity of the function by ensuring the values stay real and valid.

Function Addition and Subtraction

Function operations, such as addition and subtraction, are processes where you work with two functions to produce a new one. When adding or subtracting functions, their operations are straightforward:

• Addition: \((f+g)(x) = f(x) + g(x)\)
• Subtraction: \((f-g)(x) = f(x) - g(x)\)

For both operations, it’s important to consider the overlapping domains of each function to determine the domain of the resulting function.
In the given problem, \( f(x) = \sqrt{x} - 1 \) and \( g(x) = \sqrt{x} + 1 \):

• When you add them, you combine the like terms, resulting in \( 2\sqrt{x} \) with a domain \([0, \infty)\).
• When you subtract them, constant terms cancel out to give \( -2 \), but the domain remains \([0, \infty)\) since the individual functions’ restrictions dictate available \( x \)-values.

This understanding is essential for correctly identifying valid functions and evaluating them effectively.