1. Local min at x = -2.

1. We calculate the derivative: f ‘(x) = 2x + 4. This is zero at x = -2, so this is our critical point. To classify it, note that f”(x) = 2 > 0, so we have a local minimum.

1. No local min or max.

1. The derivative is f'(x) = 3x² + 9, which is never zero. This function is increasing everywhere, with no critical points.

1. Local min at x = -1.

1. The derivative is f'(x) = 4x³ + 4. This is zero at x = -1. The second derivative is f”(x) = 12x². This is positive everywhere, so our critical point is a local minimum.

1. Global min at x = 0. No global max.

1. The derivative is f'(x) = 2x, which is zero at x = 0. This is our critical point candidate.For x far away from zero, in either direction, the function goes very very large, so there is no global maximum. The point x = 0 is a global minimum.

1. Global max is at x = 5, and global min is at x = -2.

1. The derivative is f'(x) = 3x², which is zero at x = 0. The point x = 0 is inside our interval, so it is our critical point candidate for global max/min. There is only one critical point, so it is obvious that max, or min, or both, occur on the boundary. The boundary points are x = -2 and x = 5, and these are our other candidates. We compare the values of all of them:

a. f(0) = 2

b. f(-2) = -6

c. f(5) = 127                                                                                                                                                                                                                                                                                                                         Global max is at x = 5, and global min is at x = -2. Note that the critical point x = 0 is neither in     this case.