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

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(D) Given the function $f(x) = \frac{x}{2} + \frac{2}{x}$.To find the local extrema,we first find the derivative $f'(x)$:$f'(x) = \frac{1}{2} - \frac{

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(D) Given the function $f(x) = \frac{x}{2} + \frac{2}{x}$.To find the local extrema,we first find the derivative $f'(x)$:$f'(x) = \frac{1}{2} - \frac{2}{x^2}$.Setting $f'(x) = 0$ to find critical points:$\frac{1}{2} - \frac{2}{x^2} = 0 \Rightarrow \frac{2}{x^2} = \frac{1}{2} \Rightarrow x^2 = 4 \Rightarrow x = 2, -2$.Now,we find the second derivative $f''(x)$ to test for local minima:$f''(x) = \frac{d}{dx}(\frac{1}{2} - 2x^{-2}) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$.Evaluating 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 $x = 2$.

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

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Statement-$I$: Among the numbers $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6}, 7^{1/7}$,the maximum is $3^{1/3}$.
Statement-$II$: The function $f(x) = x^{1/x}$ increases for $0 < x < e$ and decreases for $x > e$.

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