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: Find where f is defined using a sign chart.
\(f(x)=\sqrt{\frac{x+1}{x+3}}\) needs \(\frac{x+1}{x+3}\ge 0\). The critical points are \(x=-3\) and \(x=-1\).
Testing the three intervals: for \(x<-3\) both terms are negative so the ratio is positive; for \(-3<x<-1\) the ratio is negative; for \(x>-1\) both terms are positive.
Including the zero at \(x=-1\) and excluding the undefined point \(x=-3\), the domain of \(f\) is \((-\infty,-3)\cup[-1,\infty)\).

Step 2: Find where g is defined the same way.
\(g(x)=\sqrt{\frac{2-x}{x+3}}\) needs \(\frac{2-x}{x+3}\ge 0\), with critical points \(x=-3\) and \(x=2\).
The ratio is positive only for \(-3<x<2\), so with \(x=2\) included and \(x=-3\) excluded, the domain of \(g\) is \((-3,2]\).

Step 3: Intersect the domains and remove where g=0.
The domain of \(f/g\) needs both functions defined and \(g\neq 0\).
\((-\infty,-3)\cup[-1,\infty)\) intersected with \((-3,2]\) leaves only \([-1,2]\), since the piece before \(-3\) has no overlap with \((-3,2]\).
Since \(g(2)=0\), we drop the endpoint \(x=2\), giving the final domain \([-1,2)\).
\[ \boxed{[-1,2)} \]