Question

Let \(f(x)=\sqrt{\frac{x+1}{x+3}}\) and \(g(x)=\sqrt{\frac{2-x}{x+3}}\) be two real valued functions. Then the domain of \(\frac{f}{g}\) is:

• A. \((-\infty,-3)\cup[-1,\infty)\)
• B. \([-1,2)\)
• C. \((-3,2)\)
• D. \((-\infty,-3)\cup[2,\infty)\)

Hint:

When finding the domain of a ratio of functions, find the intersection of the individual domains and exclude points where the denominator is zero.

Answer 1

The Correct Option is B

Step 1: Domain of \(f(x)\).
For \(f(x)=\sqrt{\frac{x+1}{x+3}}\) to be real, the argument of the square root must be non-negative:
\[ \frac{x+1}{x+3} \geq 0 \]
This inequality holds when either:
1. \(x+1\ge 0\) and \(x+3\gt 0 \Rightarrow x\ge -1\) (since denominator cannot be zero, \(x\neq -3\))
2. \(x+1\le 0\) and \(x+3\lt 0 \Rightarrow x\le -3\)
Hence, the domain of \(f\) is: \((-\infty,-3)\cup[-1,\infty)\)

Step 2: Domain of \(g(x)\).
For \(g(x)=\sqrt{\frac{2-x}{x+3}}\) to be real:
\[ \frac{2-x}{x+3} \ge 0 \]
1. \(2-x\ge 0\) and \(x+3\gt 0 \Rightarrow x\le 2 \text{ and } x\gt -3 \Rightarrow -3\lt x\le 2\)
2. \(2-x\le 0\) and \(x+3\lt 0 \Rightarrow x\ge 2 \text{ and } x\lt -3\), impossible
Hence, domain of \(g\) is: \((-3,2]\)

Step 3: Domain of \(\frac{f}{g}\).
The domain of \(\frac{f}{g}\) is the intersection of domains of \(f\) and \(g\), excluding points where \(g(x)=0\).
- \(g(x)=0 \Rightarrow \frac{2-x}{x+3}=0 \Rightarrow x=2\)
Intersection: \((-\infty,-3)\cup[-1,\infty)\) \(\cap\) \((-3,2]\)
\((-\infty,-3)\cap(-3,2]=\emptyset\)
\([-1,\infty)\cap(-3,2] = [-1,2)\) (exclude 2 since \(g(2)=0\))
Hence, the domain of \(\frac{f}{g}\) is \([-1,2)\)

Step 4: Final conclusion.
\[ \boxed{[-1,2)} \]