Maximum value of the function f(x)=frac{x}{8} | vedclass

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(D) Given function is $f(x) = \frac{x}{8} + \frac{2}{x}$.First,find the derivative $f'(x) = \frac{1}{8} - \frac{2}{x^2} = \frac{x^2 - 16}{8x^2}$.To fi

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$\frac{17}{8}$

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(D) Given function is $f(x) = \frac{x}{8} + \frac{2}{x}$.First,find the derivative $f'(x) = \frac{1}{8} - \frac{2}{x^2} = \frac{x^2 - 16}{8x^2}$.To find critical points,set $f'(x) = 0$,which gives $x^2 - 16 = 0$,so $x = 4$ or $x = -4$.Since the interval is $[1, 6]$,we only consider $x = 4$.Now,evaluate $f(x)$ at the critical point and the endpoints of the interval $[1, 6]$:$f(1) = \frac{1}{8} + \frac{2}{1} = \frac{1}{8} + 2 = \frac{17}{8} = 2.125$.$f(4) = \frac{4}{8} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$.$f(6) = \frac{6}{8} + \frac{2}{6} = \frac{3}{4} + \frac{1}{3} = \frac{9+4}{12} = \frac{13}{12} \approx 1.083$.Comparing the values $f(1) = \frac{17}{8}$,$f(4) = 1$,and $f(6) = \frac{13}{12}$,the maximum value is $\frac{17}{8}$.

Maximum value of the function $f(x)=\frac{x}{8}+\frac{2}{x}$ on the interval $[1,6]$ is

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