Find the critical points of $f(x)=-6 x^{4}+2 x^{2}-8$ and apply the Second Derivative Test (if possible) to determine whether each of them corresponds to a local minimum or maximum. (Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. Enter DNE if there are no critical points.) the critical points with a local minimum at $x=$ the critical points with a local maximum at $x=$
Solution
Step 1 ：Given the function \(f(x) = -6x^4 + 2x^2 - 8\), we first need to find the derivative of the function.
Step 2 ：The derivative of the function \(f'(x) = -24x^3 + 4x\).
Step 3 ：To find the critical points, we set the derivative equal to zero and solve for \(x\). The solutions are the critical points: \(x = 0, -\sqrt{6}/6, \sqrt{6}/6\).
Step 4 ：We then find the second derivative of the function \(f''(x) = 4 - 72x^2\).
Step 5 ：Applying the Second Derivative Test, we find that the second derivative at \(x = 0\) is positive, so the function has a local minimum at that point.
Step 6 ：The second derivative at \(x = -\sqrt{6}/6, \sqrt{6}/6\) is negative, so the function has a local maximum at those points.
Step 7 ：\(\boxed{\text{The critical points with a local minimum at } x=0 \text{ and the critical points with a local maximum at } x=-\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{6}}\)
