The Lagrange mean-value theorem is satisifed for $f (x) = x^{3} + 5 x$ , in the interval $[1, 4]$ at a value (round off to the second deciaml place) of $x$ equal to
2.645

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Solution

The correct option is A 2.645
$f (x) = x^{3} + 5 x, f^{'} (x) = 3 x^{2}, x ϵ [1, 4]$
By Lagrange's mean value thorem, thereexists $C ϵ (1, 4)$ such that
$f^{'} (C) = \frac{f (4) - f (1)}{4 - 1}$
$3 C^{2} = \frac{69 - 6}{3}$
$3 C^{2} = 21$
$C^{2} = 7$
$C = \pm \sqrt{7}$
$C = \sqrt{7} (∵ - \sqrt{7} \notin [1, 4])$
$C = 2.645$

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