Question:
The domain of the real valued function
\[
f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}
\]
is
The domain of the real valued function
\[
f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}
\]
is
Show Hint
For square root functions, the expression inside the root must be non-negative. If the square root is in the denominator, it must be strictly positive.
Updated On: Jun 26, 2026
- \([-1,2]\)
- \((-1,2)\)
- \([-1,\infty)\)
- \([2,\infty)\)
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The Correct Option is A
Solution and Explanation
Step 1: Condition for \(\sqrt{2-x}\).
For square roots to be real, \[ 2-x\geq 0 \] So, \[ x\leq 2 \]
Step 2: Condition for \(\sqrt{1+x}\).
For square root to be real, \[ 1+x\geq 0 \] So, \[ x\geq -1 \]
Step 3: Condition for denominator \(\sqrt{x+3}\).
Since \(\sqrt{x+3}\) is in the denominator, we need: \[ x+3\gt 0 \] So, \[ x\gt -3 \]
Step 4: Find common domain.
Combining all conditions: \[ x\leq 2,\quad x\geq -1,\quad x\gt -3 \] The common interval is: \[ [-1,2] \]
Step 5: Final conclusion.
Hence, the domain is \[ \boxed{[-1,2]} \]
For square roots to be real, \[ 2-x\geq 0 \] So, \[ x\leq 2 \]
Step 2: Condition for \(\sqrt{1+x}\).
For square root to be real, \[ 1+x\geq 0 \] So, \[ x\geq -1 \]
Step 3: Condition for denominator \(\sqrt{x+3}\).
Since \(\sqrt{x+3}\) is in the denominator, we need: \[ x+3\gt 0 \] So, \[ x\gt -3 \]
Step 4: Find common domain.
Combining all conditions: \[ x\leq 2,\quad x\geq -1,\quad x\gt -3 \] The common interval is: \[ [-1,2] \]
Step 5: Final conclusion.
Hence, the domain is \[ \boxed{[-1,2]} \]
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