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Let's go through each function one by one to determine its domain.

### 8a. \( f(x) = \sqrt{x + 3} \)
The square root function \( \sqrt{y} \) is defined only for non-negative values of \( y \). Thus, for \( \sqrt{x + 3} \) to be defined, we need:

\[
x + 3 \geq 0
\]
\[
x \geq -3
\]

**Domain:** \( x \geq -3 \), or \( [-3, \infty) \).

---

### 8b. \( g(x) = \frac{3}{x^2 - 4} \)
The function \( \frac{a}{b} \) is undefined when the denominator \( b = 0 \). So, we need to find when \( x^2 - 4 = 0 \):

\[
x^2 - 4 = (x - 2)(x + 2) = 0
\]
\[
x = 2 \quad \text{or} \quad x = -2
\]

Thus, the function is undefined at \( x = 2 \) and \( x = -2 \).

**Domain:** \( x \in (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \).

---

### 8c. \( h(x) = x^2 + 7x - 3 \)
This is a quadratic function, which is defined for all real values of \( x \). Therefore, there are no restrictions on the domain.

**Domain:** \( (-\infty, \infty) \).

---

### 8d. \( p(x) = \sqrt{7 - 4x} \)
For the square root to be defined, we need the expression inside the square root to be non-negative:

\[
7 - 4x \geq 0
\]
\[
4x \leq 7
\]
\[
x \leq \frac{7}{4}
\]

**Domain:** \( (-\infty, \frac{7}{4}] \).

---

### 8e. \( k(x) = \frac{\sqrt{x - 2}}{x - 9} \)
1. The square root in the numerator \( \sqrt{x - 2} \) requires:
   \[
   x - 2 \geq 0 \quad \Rightarrow \quad x \geq 2
   \]

2. The denominator \( x - 9 \neq 0 \), so \( x \neq 9 \).

Thus, the domain is \( x \geq 2 \) but \( x \neq 9 \).

**Domain:** \( [2, 9) \cup (9, \infty) \).

---

### 8f. \( n(x) = \frac{x}{x + 2} \)
The function is undefined when the denominator is zero, so \( x + 2 \neq 0 \):

\[
x \neq -2
\]

**Domain:** \( x \in (-\infty, -2) \cup (-2, \infty) \).

---

### 8g. \( j(x) = \frac{x}{9} \)
This is a linear function, which is defined for all real values of \( x \).

**Domain:** \( (-\infty, \infty) \).

---

### 8h. \( m(x) = \frac{9}{x} \)
The function is undefined when the denominator is zero, so \( x \neq 0 \).

**Domain:** \( x \in (-\infty, 0) \cup (0, \infty) \).

---

Do you need more details for any of the solutions? 

Here are some relative questions to extend your understanding:

1. How do we determine the domain for a function involving a square root?
2. Why do rational functions have undefined values at certain points?
3. What makes quadratic functions have domains over all real numbers?
4. How can we apply graphing to visualize the domain of these functions?
5. How does solving inequalities help in determining the domain?

**Tip:** Always check for restrictions like square roots and division by zero when determining the domain of a function.

Determine the domain of the following functions: a) f(x) = sqrt(x + 3) b) g(x) = 3 / (x^2 - 4) c) h(x) = x^2 + 7x - 3 d) p(x) = sqrt(7 - 4x) e) k(x) = sqrt(x - 2) / (x - 9) f) n(x) = x / (x + 2) g) j(x) = x / 9 h) m(x) = 9 / x

Learn how to determine the domain of various functions, including those involving square roots, rational expressions, and quadratic functions. This step-by-step guide is suitable for high school students in grades 9-12.

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