Question 8 Find the domain for each of the following functions. Present your answer in interval form. Use exact values. \( f(x)=\frac{1}{x+3} \) domain: \( \square \) \( f(x)=\sqrt{x+3} \) domain: \( \square \) \( f(x)=x+3 \) domain: \( \square \)

The function \( f(x)=\frac{1}{x+3} \) has a domain that excludes any value that makes the denominator zero. Since \( x+3=0 \) when \( x=-3 \), the domain is all real numbers except -3: \( (-\infty, -3) \cup (-3, \infty) \). For \( f(x)=\sqrt{x+3} \), the expression under the square root must be non-negative. Therefore, \( x+3 \geq 0 \) implies \( x \geq -3 \). The domain is \( [-3, \infty) \). Lastly, the function \( f(x)=x+3 \) is a linear function with no restrictions at all, so its domain is all real numbers: \( (-\infty, \infty) \). In summary: 1. \( f(x)=\frac{1}{x+3} \) domain: \( (-\infty, -3) \cup (-3, \infty) \) 2. \( f(x)=\sqrt{x+3} \) domain: \( [-3, \infty) \) 3. \( f(x)=x+3 \) domain: \( (-\infty, \infty) \) Ever wondered how these functions came to be? The use of rational functions like \( \frac{1}{x+3} \) dates back to the study of ratios and proportions in ancient civilizations. They played a crucial role in developing algebra, making it easier to understand relationships between quantities! When working with square roots, a common mistake is forgetting that the expression inside must be non-negative. Always check for values that might make your root undefined. It’s an easy slip, but rigor in checking domains will save you from headaches later on!