Mathematics - Application of Derivative Question with Solution | TestHub

$\begin{array}{rl}& f(x)=\frac{x}{3}+\frac{3}{x}+3,x\ne 0\\ & f^{\prime }(x)=\frac{1}{3}-\frac{3}{x^2}=0\Rightarrow x=\pm 3\\ & f^{\prime }(x)=\frac{x^2-3}{3x^2}\end{array}$
$\begin{array}{rl}& {\mathrm{f}}^{\prime }(\mathrm{x})>0\forall (-\infty ,-3)\cup (3,\infty )\to \text{ increasing }\\ & {\mathrm{f}}^{\prime }(\mathrm{x})<0\forall (-3,0)\cup (0,3)\to \text{ decreasing }\end{array}$
${\sum }_{i=1}^5{\alpha }_i^2=(-3{)}^2+(3{)}^2+(-3{)}^2+(0{)}^2+(3{)}^2$
$=36$