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Value of $c$ satisfying the conditions and conclusions of Rolle's theorem for the function $\mathrm{f}(x)=x \sqrt{x+6}, x \in[-6,0]$ is
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Verified Answer
The correct answer is:
$-4$
$$
\begin{aligned}
\mathrm{f}(x) & =x \sqrt{x+6} \\
\therefore \quad \mathrm{f}^{\prime}(x) & =x\left(\frac{1}{2 \sqrt{x+6}}\right)+\sqrt{x+6}(1) \\
& =\frac{x}{2 \sqrt{x+6}}+\sqrt{x+6}
\end{aligned}
$$
Since $\mathrm{f}(x)$ satisfies all the conditions of Rolle's Theorem,
There exists $c \in(-6,0)$ such that
$$
\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{c})=0 \\
& \Rightarrow \frac{\mathrm{c}}{2 \sqrt{\mathrm{c}+6}}+\sqrt{\mathrm{c}+6}=0 \\
& \Rightarrow \mathrm{c}+2 \mathrm{c}+12=0 \\
& \Rightarrow \mathrm{c}=-4
\end{aligned}
$$
\begin{aligned}
\mathrm{f}(x) & =x \sqrt{x+6} \\
\therefore \quad \mathrm{f}^{\prime}(x) & =x\left(\frac{1}{2 \sqrt{x+6}}\right)+\sqrt{x+6}(1) \\
& =\frac{x}{2 \sqrt{x+6}}+\sqrt{x+6}
\end{aligned}
$$
Since $\mathrm{f}(x)$ satisfies all the conditions of Rolle's Theorem,
There exists $c \in(-6,0)$ such that
$$
\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{c})=0 \\
& \Rightarrow \frac{\mathrm{c}}{2 \sqrt{\mathrm{c}+6}}+\sqrt{\mathrm{c}+6}=0 \\
& \Rightarrow \mathrm{c}+2 \mathrm{c}+12=0 \\
& \Rightarrow \mathrm{c}=-4
\end{aligned}
$$
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