Question

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

Answer 1

Answer:

Find the first and second derivatives:

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

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

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

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

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

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

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

$f(3)=-3-\dfrac{9}{3}=-6 \implies (3,-6)$

$f(-3)=-(-3)-\dfrac{9}{-3}=6 \implies (-3,6)$

Therefore:

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