5. [-/4 Points]
DETAILS SCALC9 3.3.012.
MY NOTES
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PRACTICE ANOTHER
Consider the following. (If an answer does not exist, enter DNE.)
\[
f(x)=x^{2 / 3}(x-8)
\]
(a) Find the interval(s) on which $f$ is increasing. (Enter your answer using interval notation.)
(b) Find the interval(s) on which $f$ is decreasing. (Enter your answer using interval notation.)
(c) Find the local minimum and maximum value of $f$. (Round your answer to two decimal places.)
local minimum value
local maximum value
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Step 1 ：First, we need to find the derivative of the function \(f(x) = x^{2/3}(x - 8)\). The derivative will tell us where the function is increasing or decreasing.

Step 2 ：Using the power rule and the product rule, we find that the derivative of \(f(x)\) is \(f'(x) = 0.666666666666667*(x - 8)/x^{0.333333333333333} + x^{0.666666666666667}\).

Step 3 ：The critical points of the function are where the derivative is zero or undefined. Solving for \(f'(x) = 0\), we find that the critical point is \(x = 3.2\).

Step 4 ：To determine whether this critical point is a local minimum or maximum, we can use the second derivative test. The second derivative of \(f(x)\) is \(f''(x) = -0.222222222222222*(x - 8)/x^{1.33333333333333} + 1.33333333333333/x^{0.333333333333333}\).

Step 5 ：Substituting \(x = 3.2\) into \(f''(x)\), we find that the second derivative is positive, indicating that \(x = 3.2\) is a local minimum.

Step 6 ：The function is increasing where the derivative is positive and decreasing where the derivative is negative. From our calculations, we find that the derivative is positive for \(x > 3.2\) and negative for \(x < 3.2\).

Step 7 ：Thus, the function is increasing on the interval \((3.2, \infty)\), decreasing on the interval \((-\infty, 3.2)\), and has a local minimum at \(x = 3.2\).

Step 8 ：\(\boxed{\text{The function is increasing on the interval }(3.2, \infty), \text{ decreasing on the interval }(-\infty, 3.2), \text{ and has a local minimum at }x = 3.2}\)