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The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is
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The correct answer is:
$3$
Given, $f(x)=\sqrt{x-2}, x \in[2,6]$
We know that, by Lagrange's mean value theorem, their exist $c \in(2,6)$ such that
$$
\begin{aligned}
& f(c)=\frac{f(b)-f(a)}{b-a} \\
& =\frac{1}{2 \sqrt{c-2}}=\frac{f(6)-f(2)}{6-2} \\
\Rightarrow \quad & \frac{1}{2 \sqrt{c-2}}=\frac{\sqrt{6-2}-\sqrt{2-2}}{4}
\end{aligned}
$$
$\begin{array}{ll}\Rightarrow & \frac{1}{\sqrt{c-2}}=\frac{\sqrt{4}-0}{2} \\ \Rightarrow & \frac{1}{\sqrt{c-2}}=1 \Rightarrow \sqrt{c-2}=1 \\ \Rightarrow & c-2=1 \Rightarrow c=3\end{array}$
We know that, by Lagrange's mean value theorem, their exist $c \in(2,6)$ such that
$$
\begin{aligned}
& f(c)=\frac{f(b)-f(a)}{b-a} \\
& =\frac{1}{2 \sqrt{c-2}}=\frac{f(6)-f(2)}{6-2} \\
\Rightarrow \quad & \frac{1}{2 \sqrt{c-2}}=\frac{\sqrt{6-2}-\sqrt{2-2}}{4}
\end{aligned}
$$
$\begin{array}{ll}\Rightarrow & \frac{1}{\sqrt{c-2}}=\frac{\sqrt{4}-0}{2} \\ \Rightarrow & \frac{1}{\sqrt{c-2}}=1 \Rightarrow \sqrt{c-2}=1 \\ \Rightarrow & c-2=1 \Rightarrow c=3\end{array}$
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