locate the critical points of the following function. then use the second derivative test to determine…

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=\frac{1}{4}x^{4}-\frac{8}{3}x^{3}-32x^{2}+512x$ using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. $f'(x)=x^{3}-8x^{2}-64x + 512$.

Step2: Set the first - derivative equal to zero

$x^{3}-8x^{2}-64x + 512 = 0$. Factor by grouping: $x^{2}(x - 8)-64(x - 8)=0$, then $(x - 8)(x^{2}-64)=0$, and further $(x - 8)(x - 8)(x + 8)=0$. The critical points are $x=-8$ and $x = 8$.

Step3: Find the second - derivative

Differentiate $f'(x)=x^{3}-8x^{2}-64x + 512$ to get $f''(x)=3x^{2}-16x-64$.

Step4: Evaluate the second - derivative at the critical points

For $x=-8$: $f''(-8)=3(-8)^{2}-16(-8)-64=3\times64 + 128-64=192 + 128-64=256>0$, so $x=-8$ is a local minimum. For $x = 8$: $f''(8)=3(8)^{2}-16(8)-64=3\times64-128 - 64=192-128-64=0$, the second - derivative test is inconclusive. Since the second - derivative test is inconclusive for $x = 8$ and $x=-8$ is a local minimum, there are no critical points that are local maxima according to the second - derivative test.

Answer:

B. There are no critical points that are local maxima according to the Second Derivative Test.