For the functionx + {1 over x},x in [1,,3], t | vedclass

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Question

(b) $f'(x) = 1 - \frac{1}{{{x^2}}} \Rightarrow f'(c) = 1 - \frac{1}{{{c^2}}}$

$	herefore 1 - \frac{1}{{{c^2}}} = \frac{{\frac{{10}}{3} - 2

Vedclass · Accepted Answer

$\sqrt 3 $

Vedclass · Answer

(b) $f'(x) = 1 - \frac{1}{{{x^2}}} \Rightarrow f'(c) = 1 - \frac{1}{{{c^2}}}$

$	herefore 1 - \frac{1}{{{c^2}}} = \frac{{\frac{{10}}{3} - 2}}{2} $

$\Rightarrow 1 - \frac{1}{{{c^2}}} = \frac{2}{3} \Rightarrow {c^2} = 3$.

For the function$x + {1 \over x},x \in [1,\,3]$, the value of $ c$ for the mean value theorem is

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