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Let's analyze each of the functions provided to determine their **domain**, **range**, and whether they are **even**, **odd**, or **neither**.

### a. \( f(x) = x^2 + 2x - 3 \)

1. **Domain**: The function \( f(x) = x^2 + 2x - 3 \) is a polynomial, which means it is defined for all real numbers.  
   **Domain**: \( (-\infty, \infty) \).

2. **Range**: Since \( f(x) \) is a quadratic function opening upwards (the coefficient of \( x^2 \) is positive), the range is all real numbers greater than or equal to the minimum value of the function. The vertex form of the quadratic function is \( f(x) = a(x - h)^2 + k \).  
   Completing the square:  
   \[ f(x) = x^2 + 2x - 3 = (x+1)^2 - 4. \]  
   The minimum value is \(-4\), so the range is \([ -4, \infty) \).

3. **Parity (Even or Odd)**:  
   To determine if the function is even or odd, we need to check:
   - Even: \( f(-x) = f(x) \).
   - Odd: \( f(-x) = -f(x) \).  
   
   Calculating \( f(-x) \):  
   \[ f(-x) = (-x)^2 + 2(-x) - 3 = x^2 - 2x - 3. \]  
   Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), the function is **neither even nor odd**.

### b. \( f(x) = \frac{2}{\sqrt{x+1}} \)

1. **Domain**: The square root function inside the denominator imposes a restriction:
   \[ x + 1 > 0 \implies x > -1. \]  
   Also, the denominator must not be zero:
   \[ \sqrt{x+1} \neq 0 \implies x \neq -1. \]  
   **Domain**: \( (-1, \infty) \).

2. **Range**: Since the square root is in the denominator, the function values will be positive and will approach zero as \( x \to \infty \). As \( x \to -1^+ \), \( f(x) \to \infty \). Thus, the range is \( (0, \infty) \).

3. **Parity (Even or Odd)**:  
   Calculating \( f(-x) \):  
   \[ f(-x) = \frac{2}{\sqrt{-x+1}}. \]  
   Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), the function is **neither even nor odd**.

### c. \( f(x) = \frac{1}{1 - 3x^2} \)

1. **Domain**: The function has a restriction when the denominator equals zero:
   \[ 1 - 3x^2 \neq 0 \implies x^2 \neq \frac{1}{3} \implies x \neq \pm \frac{1}{\sqrt{3}}. \]  
   **Domain**: \( (-\infty, -\frac{1}{\sqrt{3}}) \cup (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty) \).

2. **Range**: The function \( f(x) = \frac{1}{1 - 3x^2} \) can take any value except 0 (since the numerator is constant and never zero). Thus, the range is all real numbers except 0:  
   **Range**: \( (-\infty, 0) \cup (0, \infty) \).

3. **Parity (Even or Odd)**:  
   Calculating \( f(-x) \):  
   \[ f(-x) = \frac{1}{1 - 3(-x)^2} = \frac{1}{1 - 3x^2} = f(x). \]  
   Since \( f(-x) = f(x) \), the function is **even**.

### Summary:

- **Function a**: Domain: \( (-\infty, \infty) \), Range: \([-4, \infty)\), Parity: Neither.
- **Function b**: Domain: \( (-1, \infty) \), Range: \( (0, \infty) \), Parity: Neither.
- **Function c**: Domain: \( (-\infty, -\frac{1}{\sqrt{3}}) \cup (-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty) \), Range: \( (-\infty, 0) \cup (0, \infty) \), Parity: Even.

Would you like further details or explanations on any part of this? Here are some related questions to explore:

1. How do you complete the square for a quadratic function?
2. What makes a function even, odd, or neither?
3. How do you determine the domain and range of functions with square roots or denominators?
4. What is the significance of a function's parity in calculus?
5. How do different transformations affect the domain and range of functions?

**Tip:** When determining if a function is even or odd, always evaluate \( f(-x) \) and compare it with \( f(x) \) and \(-f(x) \).

Determine the domain, range, and whether the following functions are even or odd in the reals:
a. f(x) = x^2 + 2x - 3
b. f(x) = 2 / sqrt(x + 1)
c. f(x) = 1 / (1 - 3x^2)

Analyze the domain, range, and parity of three different types of functions: quadratic, square root, and rational functions. Detailed solutions for f(x) = x^2 + 2x - 3, f(x) = 2 / sqrt(x + 1), and f(x) = 1 / (1 - 3x^2). Suitable for students in Grades 10-12.

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