Using the second derivative test, how do you find the local maximum and local minimum for #f(x) = x^(3) - 6x^(2) + 5#?

Minimum at x=4 and Maximum at x=0

To begin determining the critical points, solve f'(x)=0.

Find the second derivative, f"(x) = 6x-12, in order to use the second derivative test.

It would therefore be a minima for x=4 and a maxima for x=0, where f"(x) = 24-12 = 12 (>0).

To find local maxima and minima using the second derivative test for f(x) = x^3 - 6x^2 + 5:

1. Find the critical points by setting the first derivative equal to zero and solving for x.
2. Find the second derivative of f(x).
3. Evaluate the second derivative at each critical point:
   • If f''(x) > 0, the function has a local minimum at that point.
   • If f''(x) < 0, the function has a local maximum at that point.
   • If f''(x) = 0, the test is inconclusive.

In this case:

1. The first derivative is f'(x) = 3x^2 - 12x. Setting f'(x) = 0, we get: 3x^2 - 12x = 0 x(3x - 12) = 0 x = 0 or x = 4 So, the critical points are x = 0 and x = 4.

2. The second derivative is f''(x) = 6x - 12.

3. Evaluating the second derivative at the critical points:

   • At x = 0: f''(0) = -12 < 0, so there is a local maximum at x = 0.
   • At x = 4: f''(4) = 6(4) - 12 = 12 - 12 = 0, so the test is inconclusive at x = 4.