Let f(x) = alpha x^2 - 2 + frac{1}{x},where a | vedclass

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(D) Given $f(x) = \alpha x^2 - 2 + \frac{1}{x} = \frac{\alpha x^3 - 2x + 1}{x}$.For $f(x) \geq 0$ and $x > 0$,we must have $g(x) = \alpha x^3 - 2x + 1

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$\frac{2^5}{3^3}$

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(D) Given $f(x) = \alpha x^2 - 2 + \frac{1}{x} = \frac{\alpha x^3 - 2x + 1}{x}$.For $f(x) \geq 0$ and $x > 0$,we must have $g(x) = \alpha x^3 - 2x + 1 \geq 0$ for all $x > 0$.To find the minimum of $g(x)$,we calculate $g'(x) = 3\alpha x^2 - 2$.Setting $g'(x) = 0$,we get $x^2 = \frac{2}{3\alpha}$,so $x = \sqrt{\frac{2}{3\alpha}}$ (since $x > 0$).The second derivative $g''(x) = 6\alpha x > 0$ at this point,confirming a local minimum.Substituting $x = \sqrt{\frac{2}{3\alpha}}$ into $g(x)$:$g\left(\sqrt{\frac{2}{3\alpha}}\right) = \alpha \left(\frac{2}{3\alpha}\right) \sqrt{\frac{2}{3\alpha}} - 2\sqrt{\frac{2}{3\alpha}} + 1 \geq 0$.$\sqrt{\frac{2}{3\alpha}} \left( \frac{2}{3} - 2 \right) + 1 \geq 0$.$1 - \frac{4}{3} \sqrt{\frac{2}{3\alpha}} \geq 0 \Rightarrow 1 \geq \frac{4}{3} \sqrt{\frac{2}{3\alpha}}$.$\frac{3}{4} \geq \sqrt{\frac{2}{3\alpha}} \Rightarrow \frac{9}{16} \geq \frac{2}{3\alpha}$.$27\alpha \geq 32 \Rightarrow \alpha \geq \frac{32}{27} = \frac{2^5}{3^3}$.Thus,the smallest value is $\frac{2^5}{3^3}$.

Let $f(x) = \alpha x^2 - 2 + \frac{1}{x}$,where $\alpha$ is a real constant. The smallest $\alpha$ for which $f(x) \geq 0$ for all $x > 0$ is

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