The function f(x) = frac{x}{2} + frac{2}{x} h | vedclass

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(D) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.First,find the derivative $f'(x)$:$f'(x) = \frac{1}{2} - \frac{2}{x^2}$.For local maxima or m

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(D) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.First,find the derivative $f'(x)$:$f'(x) = \frac{1}{2} - \frac{2}{x^2}$.For local maxima or minima,set $f'(x) = 0$:$\frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{1}{2} = \frac{2}{x^2} \implies x^2 = 4 \implies x = \pm 2$.Now,find the second derivative $f''(x)$:$f''(x) = \frac{d}{dx}(\frac{1}{2} - 2x^{-2}) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$.Check the sign of $f''(x)$ at critical points:At $x = 2$,$f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0$.Since $f''(2) > 0$,the function has a local minimum at $x = 2$.At $x = -2$,$f''(-2) = \frac{4}{(-2)^3} = \frac{4}{-8} = -\frac{1}{2} Since $f''(-2) < 0$,the function has a local maximum at $x = -2$.

The function $f(x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at $x = $ ..........

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