Find all relative extrema of the function. Use the second derivative test where applicable. (If an answer does not exist, enter DNE. .......................................................................................................... $$ f(x)=x+\frac{64}{x} $$ relative maximum $$ \quad(x, y)=(\square) $$

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Analyze the function features with analysis type as extrema for the expression $$ f(x)=x+\frac{64}{x} $$.
Function by following steps:
- step0: Find the local extrema:
  $$f\left(xight)=x+\frac{64}{x}$$
- step1: Find the domain:
  $$f\left(xight)=x+\frac{64}{x},x
eq 0$$
- step2: Find the derivative:
  $$f^{\prime}\left(xight)=\frac{x^{2}-64}{x^{2}}$$
- step3: Find the domain:
  $$f^{\prime}\left(xight)=\frac{x^{2}-64}{x^{2}},x
eq 0$$
- step4: Substitute $$f^{\prime}\left(xight)=0:$$
  $$0=\frac{x^{2}-64}{x^{2}}$$
- step5: Swap the sides:
  $$\frac{x^{2}-64}{x^{2}}=0$$
- step6: Cross multiply:
  $$x^{2}-64=x^{2}	imes 0$$
- step7: Simplify the equation:
  $$x^{2}-64=0$$
- step8: Move the constant to the right side:
  $$x^{2}=64$$
- step9: Simplify the expression:
  $$x=\pm \sqrt{64}$$
- step10: Simplify:
  $$x=\pm 8$$
- step11: Separate into possible cases:
  $$\begin{align}&x=8\&x=-8\end{align}$$
- step12: Check if the solution is in the defined range:
  $$\begin{align}&x=8\&x=-8\end{align},x
eq 0$$
- step13: Find the intersection:
  $$\begin{align}&x=8\&x=-8\end{align}$$
- step14: Determine the intervals:
  $$\begin{align}&\left(0,8ight),\left(8,+\inftyight)\&\left(-\infty,-8ight),\left(-8,0ight)\end{align}$$
- step15: Choose the points:
  $$\begin{align}&x_{1}=4,x_{2}=9\&x_{3}=-9,x_{4}=-4\end{align}$$
- step16: Find the values of the derivatives:
  $$\begin{align}&f^{\prime}\left(4ight)=-3,f^{\prime}\left(9ight)\approx 0.209877\&f^{\prime}\left(-9ight)\approx 0.209877,f^{\prime}\left(-4ight)=-3\end{align}$$
- step17: The local minimum is at $$x=8:$$
  $$\begin{align}&f\left(xight)=x+\frac{64}{x},x=8\&f\left(xight)=x+\frac{64}{x},x=-8\end{align}$$
- step18: Calculate the function value:
  $$\begin{align}&f\left(8ight)=16\&f\left(-8ight)=-16\end{align}$$
- step19: The local minimum of the function is $$16$$ at $$x = 8:$$
  $$\begin{align}&	extrm{The local minimum is }16	extrm{ at }x = 8\&	extrm{The local maximum is }-16	extrm{ at }x = -8\end{align}$$
The relative extrema of the function $$ f(x) = x + \frac{64}{x} $$ are a local minimum at $$ x = 8 $$ with a value of 16 and a local maximum at $$ x = -8 $$ with a value of -16.

Therefore, the relative maximum is at $$ (x, y) = (-8, -16) $$.
The relative maximum is at $$ (-8, -16) $$.

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