https://r2.erweima.ai/mathbot/image/17260339845777tfsau5e.jpg

Let's solve the problems step-by-step to find the domains and ranges of \( f, g, f + g, \) and \( f \cdot g \) for each exercise.

### Exercise 1
Given:
- \( f(x) = x \)
- \( g(x) = \sqrt{x - 1} \)

**1. Finding the Domain and Range of \( f(x) = x \):**

- **Domain of \( f(x) = x \):** All real numbers, \( (-\infty, \infty) \).
- **Range of \( f(x) = x \):** All real numbers, \( (-\infty, \infty) \).

**2. Finding the Domain and Range of \( g(x) = \sqrt{x - 1} \):**

- **Domain of \( g(x) = \sqrt{x - 1} \):** The expression under the square root must be non-negative, so:
  \[
  x - 1 \geq 0 \implies x \geq 1.
  \]
  Therefore, the domain of \( g(x) \) is \( [1, \infty) \).
  
- **Range of \( g(x) = \sqrt{x - 1} \):** Since a square root function outputs only non-negative values, the range is \( [0, \infty) \).

**3. Finding the Domain and Range of \( f(x) + g(x) = x + \sqrt{x - 1} \):**

- **Domain of \( f(x) + g(x) \):** The domain of \( f + g \) is the intersection of the domains of \( f(x) \) and \( g(x) \), which is \( [1, \infty) \).

- **Range of \( f(x) + g(x) \):** Since for \( x \geq 1 \), \( x \geq \sqrt{x - 1} \), the expression \( x + \sqrt{x - 1} \) will start from 1 (when \( x = 1 \)) and increase without bound as \( x \to \infty \). Therefore, the range is \( [1, \infty) \).

**4. Finding the Domain and Range of \( f(x) \cdot g(x) = x \cdot \sqrt{x - 1} \):**

- **Domain of \( f(x) \cdot g(x) \):** The domain is the intersection of the domains of \( f(x) \) and \( g(x) \), which is \( [1, \infty) \).

- **Range of \( f(x) \cdot g(x) \):** For \( x \geq 1 \), the product \( x \cdot \sqrt{x - 1} \) starts at 0 (when \( x = 1 \)) and increases without bound as \( x \to \infty \). Therefore, the range is \( [0, \infty) \).

### Exercise 2
Given:
- \( f(x) = \sqrt{x + 1} \)
- \( g(x) = \sqrt{x - 1} \)

**1. Finding the Domain and Range of \( f(x) = \sqrt{x + 1} \):**

- **Domain of \( f(x) = \sqrt{x + 1} \):** The expression under the square root must be non-negative:
  \[
  x + 1 \geq 0 \implies x \geq -1.
  \]
  Therefore, the domain of \( f(x) \) is \( [-1, \infty) \).

- **Range of \( f(x) = \sqrt{x + 1} \):** Since a square root function outputs only non-negative values, the range is \( [0, \infty) \).

**2. Finding the Domain and Range of \( g(x) = \sqrt{x - 1} \):**

- **Domain of \( g(x) = \sqrt{x - 1} \):** The expression under the square root must be non-negative:
  \[
  x - 1 \geq 0 \implies x \geq 1.
  \]
  Therefore, the domain of \( g(x) \) is \( [1, \infty) \).

- **Range of \( g(x) = \sqrt{x - 1} \):** The range is \( [0, \infty) \).

**3. Finding the Domain and Range of \( f(x) + g(x) = \sqrt{x + 1} + \sqrt{x - 1} \):**

- **Domain of \( f(x) + g(x) \):** The domain is the intersection of the domains of \( f(x) \) and \( g(x) \), which is \( [1, \infty) \).

- **Range of \( f(x) + g(x) \):** Since both \( \sqrt{x + 1} \geq 0 \) and \( \sqrt{x - 1} \geq 0 \), their sum starts at \( \sqrt{2} \) (when \( x = 1 \)) and increases without bound as \( x \to \infty \). Therefore, the range is \( [\sqrt{2}, \infty) \).

**4. Finding the Domain and Range of \( f(x) \cdot g(x) = \sqrt{x + 1} \cdot \sqrt{x - 1} \):**

- **Domain of \( f(x) \cdot g(x) \):** The domain is the intersection of the domains of \( f(x) \) and \( g(x) \), which is \( [1, \infty) \).

- **Range of \( f(x) \cdot g(x) \):** The product can be expressed as \( \sqrt{(x+1)(x-1)} = \sqrt{x^2 - 1} \). Since \( x \geq 1 \), \( x^2 - 1 \geq 0 \). As \( x \) goes from 1 to infinity, \( x^2 - 1 \) goes from 0 to infinity. Thus, the range is \( [0, \infty) \).

Would you like further details or explanations? 

### Related Questions
1. What happens to the domain and range when combining functions in different ways (like subtraction or division)?
2. How do we determine the domain and range for composite functions?
3. Can we always find a function's domain and range using algebraic methods?
4. How do domain restrictions affect the graph of a function?
5. How do we handle functions involving more complex expressions like absolute values or logarithms?

### Tip
When finding the domain of a combined function (like \( f + g \) or \( f \cdot g \)), always start by finding the intersection of the individual domains of \( f \) and \( g \).

Find the domains and ranges of functions f, g, f + g, and f * g for the given exercises.

Explore how to find the domains and ranges of functions f(x) and g(x), along with their combinations f + g and f * g, including square root functions. Step-by-step analysis for high school students covering algebraic principles and domain intersections.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation