2# uses the second derivative criterion to calculate the local maxima and minima of the following functions: af(x) = x3 -3x +2

The first derivative of the function f(x) = x^3 - 3x + 2 is found by differentiating with respect to x :
f'(x) = 3x^2 - 3

To find the critical points, we set the first derivative equal to 0 and solve for x :
3x^2 - 3 = 0
x^2 - 1 = 0
x = -1 \text{ or } x = 1

The second derivative of the function is found by differentiating the first derivative:
f''(x) = 6x

Now, we evaluate the second derivative at each critical point:
1. At x = -1 :
f''(-1) = 6(-1) = -6 < 0
Since the second derivative is negative at x = -1 , there is a local maximum at this point.

2. At x = 1 :
f''(1) = 6(1) = 6 > 0
Since the second derivative is positive at x = 1 , there is a local minimum at this point.

Therefore, the function f(x) = x^3 - 3x + 2 has a local maximum at x = -1 and a local minimum at x = 1 .

\boxed{\text{Answer: The function has a local maximum at } x = -1 \text{ and a local minimum at } x = 1.}