Mathematics - Application of Derivative Question with Solution | TestHub

$f\left(x\right)=4\alpha x^2+\frac{1}{x};x>0$
$\Rightarrow f^{\text{'}}\left(x\right)=8\alpha x-\frac{1}{x^2}=\frac{8\alpha x^3-1}{x^2}$
$\Rightarrow f\left(x\right)$attains its minimum at$x={\left(\frac{1}{8\alpha }\right)}^{\frac{1}{3}}$
$f\left\{{\left(\frac{1}{8\alpha }\right)}^{\frac{1}{3}}\right\}=1$
$\Rightarrow$$4\alpha {\left(\frac{1}{8\alpha }\right)}^{\frac{2}{3}}+{\left(8\alpha \right)}^{\frac{1}{3}}=1$
$\Rightarrow$$3{\alpha }^{\frac{1}{3}}=1 \Rightarrow \alpha =\frac{1}{27}$