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(A) Given function: $f(x) = x \sqrt{x+6}$ on $[-6, 0]$.First,check the conditions of Rolle's Theorem:$1$. $f(x)$ is continuous on $[-6, 0]$.$2$. $f(x)

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$-4$

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(A) Given function: $f(x) = x \sqrt{x+6}$ on $[-6, 0]$.First,check the conditions of Rolle's Theorem:$1$. $f(x)$ is continuous on $[-6, 0]$.$2$. $f(x)$ is differentiable on $(-6, 0)$.$3$. $f(-6) = -6 \sqrt{-6+6} = 0$ and $f(0) = 0 \sqrt{0+6} = 0$. Since $f(-6) = f(0)$,all conditions are satisfied.Now,find $f'(x)$:$f'(x) = x \cdot \frac{1}{2\sqrt{x+6}} + \sqrt{x+6} \cdot 1 = \frac{x + 2(x+6)}{2\sqrt{x+6}} = \frac{3x + 12}{2\sqrt{x+6}}$.According to Rolle's Theorem,there exists $c \in (-6, 0)$ such that $f'(c) = 0$.$\frac{3c + 12}{2\sqrt{c+6}} = 0$$3c + 12 = 0$$3c = -12$$c = -4$.Since $-4 \in (-6, 0)$,the value is $-4$.

The value of $c$ satisfying the conditions and conclusions of Rolle's theorem for the function $f(x) = x \sqrt{x+6}$ on the interval $x \in [-6, 0]$ is:

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