Question

Find the coordinates of the stationary points of the curve and use the secondderivative to determine the type of each.

Answer 1

Calculate the derivative of the function, as shown below

`\begin{gathered} y=3x+(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}`

Set y'=0 and solve for x, as shown below

`\begin{gathered} y^(\prime)=0 \\ \Rightarrow3-108x^(-2)=0,x\\e0 \\ \Rightarrow3=(108)/(x^2) \\ \Rightarrow x^2=(108)/(3) \\ \Rightarrow x^2=36 \\ \Rightarrow x=\pm\sqrt[]{36} \\ \Rightarrow x=\pm6 \end{gathered}`

Their corresponding y-coordinates are

`\begin{gathered} x=\pm6 \\ \Rightarrow y=3(6)+(108)/(6)=18+18=36 \\ \Rightarrow(6,36) \\ \text{and} \\ 3(-6)+(108)/(-6)=-18-18=-36 \\ \Rightarrow(-6,36) \end{gathered}`

Therefore, the two stationary points are (6,36) and (-6,-36).

Using the second derivative test,

`\begin{gathered} y^(\prime)=3-108x^(-2) \\ \Rightarrow y^(\doubleprime)=-108(-2x^(-2-1))=216x^(-3) \end{gathered}`

Then,

`\begin{gathered} y^(\doubleprime)(6)=(216)/((6)^3)=1>0\to\text{ local minimum at x=6} \\ \text{and} \\ y^(\doubleprime)(-6)=(216)/((-6)^3)=-1<0\to\text{ local maximum at x=-6} \end{gathered}`

(6,36) is a local minimum and (-6,-36) is a local maximum.