The least value of alpha in R for which 4 alp | vedclass

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(C) Let $f(x) = 4 \alpha x^2 + \frac{1}{x}$ for $x > 0$.To find the minimum value of $f(x)$,we differentiate with respect to $x$:$f'(x) = 8 \alpha x -

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$\frac{1}{27}$

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(C) Let $f(x) = 4 \alpha x^2 + \frac{1}{x}$ for $x > 0$.To find the minimum value of $f(x)$,we differentiate with respect to $x$:$f'(x) = 8 \alpha x - \frac{1}{x^2}$.Setting $f'(x) = 0$ for critical points:$8 \alpha x = \frac{1}{x^2}$ $\Rightarrow x^3 = \frac{1}{8 \alpha}$ $\Rightarrow x = \frac{1}{2 \alpha^{1/3}}$.Substituting this value into $f(x)$ to find the minimum:$f\left(\frac{1}{2 \alpha^{1/3}}\right) = 4 \alpha \left(\frac{1}{4 \alpha^{2/3}}\right) + \frac{1}{1/(2 \alpha^{1/3})} = \alpha^{1/3} + 2 \alpha^{1/3} = 3 \alpha^{1/3}$.Given $f(x) \geq 1$ for all $x > 0$,the minimum value must be at least $1$:$3 \alpha^{1/3} \geq 1$ $\Rightarrow \alpha^{1/3} \geq \frac{1}{3}$ $\Rightarrow \alpha \geq \frac{1}{27}$.Thus,the least value of $\alpha$ is $\frac{1}{27}$.

The least value of $\alpha \in R$ for which $4 \alpha x^2 + \frac{1}{x} \geq 1$,for all $x > 0$,is

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