To find local maxima and minima, we first find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. Evaluating f(1) = 2 shows it is a local minimum.

To find local maxima, we first find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. Checking the second derivative f''(x) = 6x - 6, we find f''(2) < 0, indicating a local maxima at x = 2.

To find local maxima, we first find f'(x) = 3x^2 - 6. Setting f'(x) = 0 gives x^2 - 2 = 0, so x = ±√2. Evaluating f''(x) at x = 1 gives f''(1) = 0, indicating a point of inflection. Thus, local maxima occurs at x = 1.

First, find f'(x) = 3x^2 - 6x. Then, f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.

First derivative f'(x) = 3x^2 - 4, then f''(x) = 6x.

f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 3)(x - 1) = 0, so critical points are x = 1 and x = 3.

Find f''(x) = 6x - 12. Set f''(x) = 0 gives x = 2. The inflection point is (2, f(2)) = (2, 0).

First, find f'(x) = 4x^3 - 6x^2 + 1. Then, f'(1) = 4(1)^3 - 6(1)^2 + 1 = 4 - 6 + 1 = -1.

Calculating f'(x) = 4x^3 - 12x^2 + 12x - 4. Thus, f'(1) = 4 - 12 + 12 - 4 = 0.

f'(x) = 4x^3 - 12x^2 + 12x - 4; f'(2) = 0.

f'(x) = 4x^3 - 12x^2 + 12x - 4; f'(1) = 0.

f'(x) = 4x^3 - 12x^2 + 12x; f'(2) = 4(2^3) - 12(2^2) + 12(2) = 32 - 48 + 24 = 8.

f'(x) = 4x^3 - 12x^2 + 12x; f'(2) = 4(2^3) - 12(2^2) + 12(2) = 32 - 48 + 24 = 8.

f'(x) = 4x^3 - 12x^2; thus, f'(2) = 4(2^3) - 12(2^2) = 32 - 48 = -16.

f'(x) = 4x^3 - 12x^2. f'(2) = 4(2^3) - 12(2^2) = 32 - 48 = -16.

To find points of inflection, we find f''(x) = 12x^2 - 16. Setting f''(x) = 0 gives x^2 = 4, so x = ±2 are points of inflection.

Finding the derivative f'(x) = 4x^3 - 16x. Setting it to zero gives x = 0, ±2. The minimum value occurs at x = 2, f(2) = 0.

Setting the two pieces equal at x = 0: 3 = k(0) + 1. Solving gives k = -3/2.

For continuity at x = 0, we need the left limit (1) to equal k. Thus, k = 1.

To be continuous at x = 0, k must equal the limit from the left, which is 1.

For continuity at x = 0, k must equal the limit as x approaches 0, which is 1.

For continuity at x = 0, we need 1 = 2, thus k must be 1.

For continuity at x = 0, we need 1 = 3 and 1 = -1 + 3k, solving gives k = 1.

At x = 0, we use the second piece: f(0) = 2(0) + 3 = 3.

Setting the derivatives equal at x = 0 gives k = 0.

To ensure continuity at x = 0, we set k(0) + 1 = 0^2, leading to k = 2.

Setting the two pieces equal at x = 1 gives 1 = k + 1, hence k = 0.

For continuity at x = 2, f(2) must equal the limit from both sides, which is 4.

For continuity at x = 3, we need limit as x approaches 3 from left (9) to equal f(3) = k, thus k = 9.

For continuity at x = 3, we need k to equal the limit from both sides, which is 9.