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.

Setting 1^3 - 3(1) + b = 2(1) + 1 gives b = 2.

To ensure differentiability at x = -1, we find f'(-1) exists. Setting a = 0 ensures the derivative is defined.

The function is a polynomial and is differentiable for all real numbers, hence any value of a works.

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

f(x) is a polynomial and is differentiable for all a, hence any value of a works.

Setting f'(1) = 0 gives a = 1, ensuring differentiability at that point.

Setting f'(1) = 0 gives a = 1 for differentiability at x = 1.

Setting a = 1 gives continuity at x = 0.

Setting 3(2) + a = 4(2) - 1 gives a = 1.

Setting ax + 1 = 2 at x = 0 gives a = 2.

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

Setting ax + 2 = 3 at x = 1 gives a = 1.

To ensure continuity at x = 1, we set limit as x approaches 1 from left (1 + a) equal to f(1) = 3, thus a = 2.

Setting the two pieces equal at x = 0 gives -a = 1, so a = -1.

Setting the two pieces equal at x = 1 gives 1 - a = 1. Thus, a = 0.