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) = 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) = 2x + 2.

The function is a polynomial and is differentiable everywhere.

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

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

The function is a polynomial and is differentiable everywhere, hence there are no points where it is not differentiable.

The function is a polynomial and is differentiable everywhere, so there is no such x.

f(x) is a polynomial and is differentiable everywhere. The x-coordinate can be any real number.

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, ∞).

Setting the left limit (0 + 1 = 1) equal to the right limit (3 - 0 = 3), we find b = 1.

By definition, f(1) = 3, as given in the piecewise function.

Setting f(1-) = f(1+) and f'(1-) = f'(1+) gives k = 2 for differentiability.

For continuity, f(3) must equal the limit as x approaches 3, which is 9.

The function is not differentiable at x = -3 and x = 2, but the first point of interest is -3.