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

f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 0, which is the minimum value.

Integrating gives y = x^2 + C. Using the initial condition y(0) = 1, we find C = 1.

The general solution is y = e^x + C. Using the initial condition y(0) = 1, we find C = 1.

f''(x) = 6x - 12. Setting f''(x) = 0 gives x = 2. The point of inflection is (2, f(2)) = (2, 3).

f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x(x - 2) = 0, so x = 0 or x = 2. The point of inflection is at (2, f(2)) = (2, 2).

f''(x) = e^x, thus f''(0) = e^0 = 1.

First derivative f'(x) = (2x)/(x^2 + 1). Second derivative f''(x) = (2(x^2 + 1) - 4x^2)/(x^2 + 1)^2 = -2/(x^2 + 1)^2.

f'(x) = 3x^2 - 12x + 9; f''(x) = 6x - 12. At x = 2, f''(2) = 6(2) - 12 = 0.

First derivative f'(x) = 4x^3 - 12x^2 + 12. Second derivative f''(x) = 12x^2 - 24.

The derivative f'(x) = cos(x). At x = π/4, f'(π/4) = cos(π/4) = √2/2.

This is a linear first-order equation. The general solution is y = 3/2 + Ce^(-2x).

This is a second-order linear homogeneous differential equation. The characteristic equation has roots ±2i.

Using an integrating factor e^x, we solve to get y = e^x + Ce^(-x).

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

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

Setting the two pieces equal at x = 2 gives us 2a + 1 = 1. Solving for a gives a = 0.

Setting the two pieces equal at x = 2: 2a + 1 = 2^2 - 3. Solving gives a = 2.

Set the left-hand limit equal to the right-hand limit and their derivatives at x = 2.

Setting the left limit (1 + a) equal to the right limit (3), we find a = 2.

Setting 1 + b = 2 + 3 gives b = 4.

Setting 1 + b = 2 gives b = 1.

Setting the left-hand limit equal to the right-hand limit gives c = 1.

Setting the two pieces equal at x = 2 gives 4 = 4 + c, hence c = 0.

To ensure continuity at x = 1, we set the left limit (1 - 3 + 2 = 0) equal to the right limit (1 + 1 = 2), leading to c = 2.

Setting limit as x approaches c from left equal to 4 and from right gives c = 1.

f'(x) = 2kx + 2. At x = 0, f'(0) = 2. The function is differentiable for any k, but k = 0 gives a constant function.

The function is a polynomial and is differentiable for all k, hence k can be any real number.

For f(x) to be differentiable at x = k, f'(k) must exist. Setting k = 1 makes f'(k) continuous.

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