The size of the power set is 2^n where n is the number of elements in the set. Here, n = 3, so the size is 2^3 = 8.

The subsets of F that do not contain 'y' are {∅, {x}}, totaling 2 subsets.

The function is discontinuous at x = 1 because it leads to division by zero.

Substituting x = 4 into the function gives f(4) = 2(4) + 3 = 8 + 3 = 11.

f(5) = 2(5) + 3 = 10 + 3 = 13.

f(2) = 2(2^2) - 3(2) + 1 = 8 - 6 + 1 = 3.

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).

f(3) = 2^3 = 8.

f(-2) = 3(-2) - 4 = -6 - 4 = -10.

f(-2) = 3(-2) - 4 = -6 - 4 = -10.

To find the inverse, set y = 3x - 4, solve for x: x = (y + 4)/3, thus f^(-1)(x) = (x + 4)/3.

First, find f(1) = 3(1) - 5 = -2. Then, f(-2) = 3(-2) - 5 = -6 - 5 = -11.

f'(x) = 10x + 3; f'(1) = 10*1 + 3 = 13.

Using the chain rule, f'(x) = 2e^(2x).

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

f'(x) = e^x; f'(0) = e^0 = 1.

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

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

f'(x) = 1/x; f'(1) = 1, hence f is differentiable at x = 1.

f'(x) = 1/x.

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(π/2) = sin(π/2) = 1.

Setting the left-hand derivative equal to the right-hand derivative at x = 0 gives k = 2.

f'(x) = 2x + 2, thus f'(1) = 2(1) + 2 = 4.

Calculating the derivative f'(x) = 2x + 2, we find f'(1) = 4.

The vertex form is f(x) = (x + 1)^2, so the vertex is (-1, 0).

f'(x) = 2x + 2. Therefore, f'(1) = 2(1) + 2 = 4.

To find x-intercepts, set f(x) = 0: x^2 - 4 = 0, which gives x = ±2.

The limit as x approaches 2 is f(2) = 2^2 - 4 = 0.

f(2) = 2^2 - 4*2 + 3 = 4 - 8 + 3 = -1.