The value of c in the Lagrange's mean value t | vedclass

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(C) Given,$f(x)=\sqrt{x-2}$ for $x \in [2,6]$.According to Lagrange's Mean Value Theorem,there exists at least one $c \in (2,6)$ such that $f'(c) = \f

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$3$

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(C) Given,$f(x)=\sqrt{x-2}$ for $x \in [2,6]$.According to Lagrange's Mean Value Theorem,there exists at least one $c \in (2,6)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.Here,$a=2$ and $b=6$.First,find the derivative: $f'(x) = \frac{1}{2\sqrt{x-2}}$.So,$f'(c) = \frac{1}{2\sqrt{c-2}}$.Now,calculate $f(b)$ and $f(a)$: $f(6) = \sqrt{6-2} = \sqrt{4} = 2$ and $f(2) = \sqrt{2-2} = 0$.Substitute these into the formula: $\frac{1}{2\sqrt{c-2}} = \frac{2-0}{6-2}$.$\frac{1}{2\sqrt{c-2}} = \frac{2}{4} = \frac{1}{2}$.This implies $\sqrt{c-2} = 1$.Squaring both sides,$c-2 = 1$,which gives $c = 3$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$k$	$2k$	$3k$	$2k$	$k$

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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