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.

Find f'(x) = 6x^2 - 18x + 12, set to 0. Local maxima at x = 2 gives f(2) = 6.

The vertex is at x = -(-12)/(2*3) = 2. The minimum value is f(2) = 3(2^2) - 12(2) + 7 = -5. Thus, the coordinates are (2, -5).

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).

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 7 = -5.

The x-coordinate of the vertex is given by x = -b/(2a) = 12/(2*3) = 2.

The vertex is at x = -b/(2a) = 2. f(2) = 3.

The vertex occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

The vertex x-coordinate is found using -b/(2a) = 12/(2*3) = 2.

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.

The second derivative f''(x) = d/dx(e^x) = e^x.

f(x) = ln(x) is not defined for x ≤ 0, hence not differentiable at x = 0.

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.

f'(x) = cos(x) - sin(x). At x = π/4, f'(π/4) = cos(π/4) - sin(π/4) = √2/2 - √2/2 = 0.

f'(x) = cos(x); f'(π/2) = cos(π/2) = 0.

f'(x) = 2x + 2.

The function is a polynomial and is differentiable everywhere.

f'(x) = 2x + 2. Setting f'(x) = 0 gives x = -1. f(-1) = 1.

Setting the derivative f'(-1) = 0 gives k = 1 for differentiability.

f'(x) = 2x - 2. Thus, f'(1) = 2(1) - 2 = 0.

As a polynomial, f(x) is differentiable everywhere, including at x = 2.

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 vertex is at x = 6/2 = 3. The minimum value is f(3) = 3^2 - 6*3 + 10 = 1.

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

f(x) is a polynomial function and is continuous everywhere, hence no points of discontinuity.

As a polynomial, f(x) is differentiable everywhere, hence no points of non-differentiability.