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

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Question

(A) Given the function $f(x)=-\sqrt{-x^2-6x-5}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$-x^2-6x-5 \geq 0$

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$[-2,0]$

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(A) Given the function $f(x)=-\sqrt{-x^2-6x-5}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$-x^2-6x-5 \geq 0$$x^2+6x+5 \leq 0$$(x+1)(x+5) \leq 0$This inequality holds for $x \in [-5, -1]$.Now,let $g(x) = -x^2-6x-5$. The vertex of this parabola occurs at $x = -\frac{b}{2a} = -\frac{-6}{2(-1)} = -3$.The value of $g(x)$ at $x = -3$ is $-(-3)^2-6(-3)-5 = -9+18-5 = 4$.At the endpoints $x = -5$ and $x = -1$,$g(x) = 0$.Thus,the range of $g(x)$ is $[0, 4]$.Since $f(x) = -\sqrt{g(x)}$,the range of $f(x)$ is $-\sqrt{[0, 4]} = [-\sqrt{4}, -\sqrt{0}] = [-2, 0]$.

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

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