How do you use the second derivative test to find the local maximum and minimum for #f(x)=x^4-4x^3+4x^2+6#?

There is a local maxima at x=1 and a minima at x=0 and at x=2

f '(x)=0 gives x=0, 1,2

At x=0, f "(x)= 8 (>0), hence it is a minima there At x=1, f"(x)= -4 (<0), hence it is a maxima there At x=2, f"(x)= 8 (>0), hence it is a minima there.

To use the second derivative test to find the local maximum and minimum for ( f(x) = x^4 - 4x^3 + 4x^2 + 6 ), follow these steps:

1. Find the first derivative of ( f(x) ) and determine the critical points by setting it equal to zero and solving for ( x ).

2. Find the second derivative of ( f(x) ).

3. Evaluate the second derivative at each critical point.

4. If the second derivative is positive at a critical point, then the function has a local minimum at that point. If the second derivative is negative at a critical point, then the function has a local maximum at that point. If the second derivative is zero or undefined, the test is inconclusive.

5. Determine whether each critical point corresponds to a local maximum, local minimum, or if the test is inconclusive.

Following these steps, you can identify the local maximum and minimum points for the given function.