https://r2.erweima.ai/mathbot/image/1726033605786noh38le1.jpg

Let's solve the problem of finding the domain and range for each function in the image. 

### 1. \( f(x) = 1 + x^2 \)
- **Domain:** There are no restrictions for \( x \), as any real number can be squared. Hence, the domain is all real numbers: \( (-\infty, \infty) \).
- **Range:** Since \( x^2 \) is always non-negative and the minimum value of \( x^2 \) is 0, the minimum value of \( f(x) = 1 + x^2 \) is 1. Hence, the range is \( [1, \infty) \).

### 2. \( f(x) = 1 - \sqrt{x} \)
- **Domain:** The square root function is defined only for \( x \geq 0 \), so the domain is \( [0, \infty) \).
- **Range:** The square root of \( x \) is non-negative, so \( \sqrt{x} \geq 0 \). The function \( f(x) = 1 - \sqrt{x} \) decreases from 1 when \( x = 0 \) and approaches negative infinity as \( x \to \infty \). Hence, the range is \( (-\infty, 1] \).

### 3. \( F(x) = \sqrt{5x + 10} \)
- **Domain:** The expression inside the square root, \( 5x + 10 \), must be non-negative: 
  \[
  5x + 10 \geq 0 \implies x \geq -2
  \]
  Thus, the domain is \( [-2, \infty) \).
- **Range:** Since the square root function produces non-negative results, the minimum value is 0, which happens when \( x = -2 \). The function increases without bound as \( x \to \infty \). Therefore, the range is \( [0, \infty) \).

### 4. \( g(x) = \sqrt{x^2 - 3x} \)
- **Domain:** The expression inside the square root, \( x^2 - 3x \), must be non-negative. Factor the quadratic:
  \[
  x(x - 3) \geq 0
  \]
  The solutions occur when \( x \leq 0 \) or \( x \geq 3 \). Therefore, the domain is \( (-\infty, 0] \cup [3, \infty) \).
- **Range:** Since the square root function outputs non-negative values, the range is \( [0, \infty) \).

### 5. \( f(t) = \frac{4}{3 - t} \)
- **Domain:** The denominator cannot be zero, so \( 3 - t \neq 0 \), which implies \( t \neq 3 \). Thus, the domain is \( (-\infty, 3) \cup (3, \infty) \).
- **Range:** The function is a rational function and will take all real values except zero. Thus, the range is \( (-\infty, 0) \cup (0, \infty) \).

### 6. \( G(t) = \frac{2}{t^2 - 16} \)
- **Domain:** The denominator cannot be zero, so \( t^2 - 16 \neq 0 \), which implies \( t \neq \pm 4 \). Therefore, the domain is \( (-\infty, -4) \cup (-4, 4) \cup (4, \infty) \).
- **Range:** Since \( t^2 - 16 \) can be both positive and negative (except at \( t = \pm 4 \)), and the function approaches zero as \( |t| \to \infty \), the range is \( (-\infty, 0) \cup (0, \infty) \).

Would you like further details on any of these functions?

Here are 5 related questions you might consider:
1. How do you find the inverse of each of these functions?
2. What are the graphs of these functions and how do their domains and ranges affect the graphs?
3. How would you apply these functions to real-world scenarios?
4. How do you compute the limits of these functions as \( x \to \infty \) or \( t \to 3 \)?
5. How would the range change if you alter each function slightly (e.g., shifting, scaling)?

**Tip:** Always analyze the domain first, as it dictates the valid input values, and then consider the range by evaluating the output behavior based on the function's structure.

Find the domain and range of each function provided in the exercises.

This problem explores how to find the domain and range of functions involving square roots, quadratics, and rational expressions. The exercises focus on key mathematical concepts related to domains, ranges, and solving inequalities.

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