Question

Find the local maximum and minimum values using the Second Derivative Test.​

Answer 1

Find the first and second derivatives:

`\begin{aligned}f(x) & =-x-(9)/(x)\\& =-x-9x^(-1)\\\\\implies f'(x) & =-1-(-1)9x^((-1-1))\\ & =-1+9x^(-2)\\ & = -1+(9)/(x^2)\\\\\implies f''(x) & = 0+(-2)9x^((-2-1))\\& = -18x^(-3)\\& = -(18)/(x^3)\end{aligned}`

To find the stationary points (local minimum and maximum) set the first derivative to zero and solve for x:

`\begin{aligned}f'(x) & = 0 \\\\\implies -1+(9)/(x^2) & = 0 \\\\(9)/(x^2) & = 1 \\\\9 & =x^2\\\\\implies x & = \pm 3\end{aligned}`

To determine the type of stationary points, input the found values of x into the second derivative.

`f''(3)=-(18)/(3^3)=-(2)/(3) < 0 \implies \textsf{maximum}`

`f''(-3)=-(18)/((-3)^3)=(2)/(3) > 0 \implies \textsf{minimum}`

Finally, to find the y-values of the stationary points, input the found values of x into the original function:

`f(3)=-3-(9)/(3)=-6 \implies (3,-6)`

`f(-3)=-(-3)-(9)/(-3)=6 \implies (-3,6)`

Therefore:

`\large \begin{array} c \textsf{At} \: x= & \textsf{and} \: y= & \textsf{sign of} \: f''(x) & \textsf{conclusion}\\ \cline{1-4} -3 & 6 & + & \textsf{minimum} \\ \cline{1-4} 3 & -6 & - & \textsf{maximum}\end{array}`