(1 point) Suppose the derivative of the function $y=f(x)$ is given by $y^{\prime}=(x-6)^{2}(x+4)(x+8)$. Where does $f$ have local extrema or points of inflection?
(Enter a comma-separated list of $\mathrm{x}$-coordinates. If there are no such points, enter "none".)

Local maxima occur at

Local minima occur at

Inflection points occur at

Step 1 ：To find the local extrema, we set the derivative equal to zero: \(y^{\prime} = (x-6)^{2}(x+4)(x+8) = 0\)

Step 2 ：Solving for \(x\), we find the critical points: \(x = -8, -4, 6\)

Step 3 ：To determine if these are local maxima or minima, we examine the sign of the second derivative at these points

Step 4 ：The second derivative is \(y^{\prime\prime} = (x - 6)^{2}(x + 4) + (x - 6)^{2}(x + 8) + (x + 4)(x + 8)(2x - 12)\)

Step 5 ：At \(x = -8\), the second derivative is positive, indicating a local minimum

Step 6 ：At \(x = -4\), the second derivative is negative, indicating a local maximum

Step 7 ：To find inflection points, we look for where the second derivative changes sign

Step 8 ：Potential inflection points are at \(x = 6, -3 + \sqrt{11}, -\sqrt{11} - 3\)

Step 9 ：Examining the sign of the second derivative around these points confirms they are inflection points

Step 10 ：Local maxima occur at \(\boxed{-8}\)

Step 11 ：Local minima occur at \(\boxed{-4}\)

Step 12 ：Inflection points occur at \(\boxed{-3 + \sqrt{11}}, \boxed{-\sqrt{11} - 3}\)