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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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{f}(x) & =x \sqrt{x+6} \ 	herefore \quad \mathrm{f}^{\prime}(x) & =x\left(\frac{1}{2 \sqrt{x+6}}ight)+\sqrt{x+6}(1) \ & =\frac{x}{2 \sqrt{x+6}}+\sqrt{x+6} \end{aligned}$ Since $\mathrm{f}(x)$ satisfies all the conditions of Rolle's Theorem, There exists $c \in(-6,0)$ such that $\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{c})=0 \ & \Rightarrow \frac{\mathrm{c}}{2 \sqrt{\mathrm{c}+6}}+\sqrt{\mathrm{c}+6}=0 \ & \Rightarrow \mathrm{c}+2 \mathrm{c}+12=0 \ & \Rightarrow \mathrm{c}=-4 \end{aligned}$

Value of $c$ satisfying the conditions and conclusions of Rolle's theorem for the function $\mathrm{f}(x)=x…

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