Calculate the derivative of the function, as shown below
[tex]\begin{gathered} y=3x+\frac{108}{x}=3x+108x^{-1} \\ \Rightarrow y^{\prime}=3+108((-1)x^{-1-1})=3-108x^{-2} \\ \Rightarrow y^{\prime}=3-108x^{-2} \end{gathered}[/tex]Set y'=0 and solve for x, as shown below
[tex]\begin{gathered} y^{\prime}=0 \\ \Rightarrow3-108x^{-2}=0,x\ne0 \\ \Rightarrow3=\frac{108}{x^2} \\ \Rightarrow x^2=\frac{108}{3} \\ \Rightarrow x^2=36 \\ \Rightarrow x=\pm\sqrt[]{36} \\ \Rightarrow x=\pm6 \end{gathered}[/tex]Their corresponding y-coordinates are
[tex]\begin{gathered} x=\pm6 \\ \Rightarrow y=3(6)+\frac{108}{6}=18+18=36 \\ \Rightarrow(6,36) \\ \text{and} \\ 3(-6)+\frac{108}{-6}=-18-18=-36 \\ \Rightarrow(-6,36) \end{gathered}[/tex]Therefore, the two stationary points are (6,36) and (-6,-36).
Using the second derivative test,
[tex]\begin{gathered} y^{\prime}=3-108x^{-2} \\ \Rightarrow y^{\doubleprime}=-108(-2x^{-2-1})=216x^{-3} \end{gathered}[/tex]Then,
[tex]\begin{gathered} y^{\doubleprime}(6)=\frac{216}{(6)^3}=1>0\to\text{ local minimum at x=6} \\ \text{and} \\ y^{\doubleprime}(-6)=\frac{216}{(-6)^3}=-1<0\to\text{ local maximum at x=-6} \end{gathered}[/tex](6,36) is a local minimum and (-6,-36) is a local maximum.
