The range of the function f(x) = -sqrt{-x^2-6 | vedclass

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(B) Let $g(x) = -x^2-6x-5$. This is a downward-opening parabola. The maximum value of $g(x)$ is given by $-\frac{D}{4a}$,where $D = b^2-4ac = (-6)^2 -

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$[-2, 0]$

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(B) Let $g(x) = -x^2-6x-5$. This is a downward-opening parabola. The maximum value of $g(x)$ is given by $-\frac{D}{4a}$,where $D = b^2-4ac = (-6)^2 - 4(-1)(-5) = 36 - 20 = 16$. The maximum value is $-\frac{16}{4(-1)} = 4$. Thus,the range of $g(x)$ is $(-\infty, 4]$. Since the function is $f(x) = -\sqrt{g(x)}$,the expression inside the square root must be non-negative,so $g(x) \in [0, 4]$. Taking the square root,$\sqrt{g(x)} \in [0, 2]$. Multiplying by $-1$,we get $f(x) \in [-2, 0]$.

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

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