Question

The range of the real-valued function \(f(x)=-\sqrt{-x^2-6x-5}\) is

• A. \( [-\infty,0] \)
• B. \( [-5,-1] \)
• C. \( [-2,0] \)
• D. \( [-3,-2] \) \bigskip

Hint:

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.

Answer 1

The Correct Option is C

Concept: Rewrite the quadratic in vertex form so that the maximum value of the radicand can be identified directly.

Step 1: Given \(f(x)=-\sqrt{-x^2-6x-5}\), write the radicand as \(-x^2-6x-5=4-(x+3)^2\). Since \((x+3)^2\ge0\), we have \(0\le4-(x+3)^2\le4\).

Step 2: Hence \(0\le\sqrt{4-(x+3)^2}\le2\). Multiplying by \(-1\) gives \(-2\le-\sqrt{4-(x+3)^2}\le0\).

Step 3: Therefore, the range of the function is \(\boxed{[-2,0]}\).