f(x) = frac{x}{2} + frac{2}{x}, x neq 0 is st | vedclass

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(A) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.To find the intervals where the function is strictly decreasing,we find the derivative $f'(x)

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$(-2, 0) \cup (0, 2)$

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(A) Given function is $f(x) = \frac{x}{2} + \frac{2}{x}$.To find the intervals where the function is strictly decreasing,we find the derivative $f'(x)$.$f'(x) = \frac{d}{dx} (\frac{x}{2} + 2x^{-1}) = \frac{1}{2} - 2x^{-2} = \frac{1}{2} - \frac{2}{x^2}$.For the function to be strictly decreasing,we require $f'(x) $\frac{1}{2} - \frac{2}{x^2} Since $2x^2 > 0$ for all $x 
eq 0$,the inequality holds when $x^2 - 4 $(x - 2)(x + 2) This inequality holds for $x \in (-2, 2)$.Since $x 
eq 0$,the function is strictly decreasing in the interval $(-2, 0) \cup (0, 2)$.

$f(x) = \frac{x}{2} + \frac{2}{x}, x \neq 0$ is strictly decreasing in

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