Question:
The range of the real-valued function \(f(x)=-\sqrt{-x^2-6x-5}\) is
The range of the real-valued function \(f(x)=-\sqrt{-x^2-6x-5}\) is
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For functions involving square roots:
- First ensure that the expression inside the square root is non-negative.
- Find the maximum and minimum values of the radicand.
- Use the fact that \(\sqrt{x}\ge 0\).
- A negative sign outside the square root reflects the range about the \(x\)-axis.
Updated On: Jul 9, 2026
- \( [-\infty,0] \)
- \( [-5,-1] \)
- \( [-2,0] \)
- \( [-3,-2] \) \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
For a square root function to be real-valued, the expression inside the square root must be non-negative.
\[
g(x)\ge 0
\]
Also,
\[
\sqrt{g(x)}\ge 0
\]
Therefore,
\[
-\sqrt{g(x)}\le 0.
\]
To find the range, we first determine the possible values of the expression inside the square root.
Step 1: Find the domain of the function. Given, \[ f(x)=-\sqrt{-x^2-6x-5}. \] For real values of \(f(x)\), \[ -x^2-6x-5\ge 0. \] Multiplying throughout by \(-1\), \[ x^2+6x+5\le 0. \] Factorizing, \[ (x+1)(x+5)\le 0. \] Hence, \[ -5\le x\le -1. \] Thus, the domain is \[ [-5,-1]. \]
Step 2: Find the maximum value of the expression inside the square root. Completing the square, \[ -x^2-6x-5 =-(x^2+6x+9)+4 \] \[ =-(x+3)^2+4. \] Since \[ (x+3)^2\ge 0, \] we have \[ -(x+3)^2+4\le 4. \] Thus, the maximum value of the radicand is \[ 4, \] which occurs at \[ x=-3. \]
Step 3: Find the range of \(f(x)\). Since \[ 0\le -x^2-6x-5\le 4, \] taking square roots, \[ 0\le \sqrt{-x^2-6x-5}\le 2. \] Multiplying by \(-1\), \[ -2\le -\sqrt{-x^2-6x-5}\le 0. \] Therefore, \[ -2\le f(x)\le 0. \] Hence, the range of the function is \[ [-2,0]. \]
Step 4: Write the final answer. \[ \boxed{[-2,0]} \]
Step 1: Find the domain of the function. Given, \[ f(x)=-\sqrt{-x^2-6x-5}. \] For real values of \(f(x)\), \[ -x^2-6x-5\ge 0. \] Multiplying throughout by \(-1\), \[ x^2+6x+5\le 0. \] Factorizing, \[ (x+1)(x+5)\le 0. \] Hence, \[ -5\le x\le -1. \] Thus, the domain is \[ [-5,-1]. \]
Step 2: Find the maximum value of the expression inside the square root. Completing the square, \[ -x^2-6x-5 =-(x^2+6x+9)+4 \] \[ =-(x+3)^2+4. \] Since \[ (x+3)^2\ge 0, \] we have \[ -(x+3)^2+4\le 4. \] Thus, the maximum value of the radicand is \[ 4, \] which occurs at \[ x=-3. \]
Step 3: Find the range of \(f(x)\). Since \[ 0\le -x^2-6x-5\le 4, \] taking square roots, \[ 0\le \sqrt{-x^2-6x-5}\le 2. \] Multiplying by \(-1\), \[ -2\le -\sqrt{-x^2-6x-5}\le 0. \] Therefore, \[ -2\le f(x)\le 0. \] Hence, the range of the function is \[ [-2,0]. \]
Step 4: Write the final answer. \[ \boxed{[-2,0]} \]
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