For a maximum, f'(x) = 2x + k = 0 at x = -2. Thus, k = 4.

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f''(1) < 0 indicates a local maximum at x = 1.

The vertex occurs at x = -b/(2a) = 4/2 = 2, which is the x-coordinate of the minimum point.

The vertex occurs at x = -b/(2a) = 4/2 = 2. This is where the local minimum occurs.

f''(x) = 12x - 18. Setting f''(x) = 0 gives x = 1.5. The inflection point is (1.5, f(1.5)).

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 3. Testing intervals shows f is increasing on (1, 3).

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f(1) = 5 is a local maximum.

The vertex is at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 7 = -1. So, the vertex is (2, -1).

f''(x) = 18x - 24. Setting f''(x) = 0 gives x = 4/3. This is the inflection point.

Finding f'(x) = 9x^2 - 24x + 9 and solving gives two critical points. The second derivative test confirms one maximum and one minimum.

To find the point of inflection, we compute f''(x) = e^x - 2. Setting f''(x) = 0 gives e^x = 2, leading to x = ln(2). The closest integer is x = 1.

f'(x) = cos(x) - sin(x). Setting f'(x) = 0 gives tan(x) = 1, so x = π/4 + nπ. In [0, 2π], the maximum occurs at x = 3π/4.

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, which is the minimum value.

The vertex is at x = -b/(2a) = 4/2 = 2. f(2) = 2^2 - 4(2) + 5 = 1, so the vertex is (2, 1).

The x-coordinate of the vertex is given by x = -b/(2a) = 6/(2*1) = 3.

First, find 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.

f'(x) = 3x^2 - 12x + 9. The critical points are x = 1 and x = 3. The function is increasing on (1, 3) and (3, ∞).

Find f''(x) = 12x^2 - 16. Setting f''(x) = 0 gives x^2 = 4, so x = ±2. f(2) = 0, thus the inflection point is (2, 0).

f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x(x^2 - 4) = 0, so x = -2, 0, 2. Test intervals: f' is positive in (-2, ∞).

Find f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. Testing intervals, f(x) is increasing on (1, 3).

Find f'(x) = e^x - 2x. Setting f'(x) = 0 gives a local maximum at x = 1.

The derivative f'(x) = 1/x + 2x. For f'(x) > 0, we need 1/x + 2x > 0, which holds for x > 0.

To find critical points, we set f'(x) = cos(x) - sin(x) = 0. This gives tan(x) = 1, leading to x = π/4 and x = 5π/4 in the interval [0, 2π].

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives 3x(x - 2) = 0, so x = 0 and x = 2 are critical points.

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f''(2) = 6 > 0, so (2, f(2)) = (2, 1) is a local minimum.

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.

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.

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.