Question

The domain of the function $f(x)=\sqrt{\frac{x+1}{2-x}}$ is

• A. [1,2)
• B. [-1,2]
• C. [-1,2)
• D. [-1,3)
• E. (-1,3)

Hint:

Logic Tip: The critical point derived from the denominator of a rational function will ALWAYS be excluded with a parenthesis `)` or `(`, because division by zero is undefined. Knowing $x=2$ is the denominator's root instantly eliminates options B, which uses a bracket `]`.

Answer 1

The Correct Option is C

Concept:
For a real-valued function involving an even root (like a square root), the expression inside the radicand must be non-negative ($\ge 0$). Additionally, for rational expressions, the denominator cannot be zero.
Step 1: Establish the necessary conditions for the domain.
Condition 1: The expression inside the square root must be greater than or equal to zero. $$\frac{x+1}{2-x} \ge 0$$ Condition 2: The denominator cannot be zero. $$2 - x \neq 0 \implies x \neq 2$$
Step 2: Identify the critical points.
Critical points occur where the numerator is zero or the denominator is zero. Numerator: $x + 1 = 0 \implies x = -1$ Denominator: $2 - x = 0 \implies x = 2$
Step 3: Test the intervals created by the critical points.
The critical points $x = -1$ and $x = 2$ divide the real number line into three intervals: $(-\infty, -1)$, $(-1, 2)$, and $(2, \infty)$. We test a value in each interval to see if the inequality $\frac{x+1}{2-x} \ge 0$ holds.

• Interval $(-\infty, -1)$: Let $x = -2$. $$\frac{-2+1}{2-(-2)} = \frac{-1}{4}<0 \quad \text{(Invalid)}$$
• Interval $(-1, 2)$: Let $x = 0$. $$\frac{0+1}{2-0} = \frac{1}{2}>0 \quad \text{(Valid)}$$
• Interval $(2, \infty)$: Let $x = 3$. $$\frac{3+1}{2-3} = \frac{4}{-1} = -4<0 \quad \text{(Invalid)}$$

Step 4: Determine the boundary inclusions.
The valid interval is between $-1$ and $2$. We must now check the endpoints. At $x = -1$: The numerator is $0$, making the fraction $0$, which satisfies $\ge 0$. So, $-1$ is included (bracket $[-1$). At $x = 2$: The denominator is $0$, which makes the fraction undefined. So, $2$ is strictly excluded (parenthesis $2)$). Combining these gives the final domain: $$[-1, 2)$$