R is symmetric because if (x,y) is in R, then (y,x) is also in R. It is not reflexive or transitive.

f(x) is a polynomial function, which is differentiable everywhere, including at x = 1.

The left limit is 0 and the right limit is undefined. Thus, f(x) is not continuous at x = 0.

At x = 1, f(1) = 3, but lim x->1- f(x) = 1 and lim x->1+ f(x) = 2. Thus, it is not continuous.

Both sides equal 2 at x = 1, hence it is continuous.

The left-hand derivative is -1 and the right-hand derivative is 1. Since they are not equal, f(x) is not differentiable at x = 1.

The area is given by the integral from 0 to 1 of (x - x^3) dx. This evaluates to [x^2/2 - x^4/4] from 0 to 1 = (1/2 - 1/4) = 1/4.

The area enclosed is found by integrating from -2 to 2: ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 16/3.

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

The area under the curve y = 1/x from x = 1 to x = 2 is given by ∫(from 1 to 2) (1/x) dx = [ln(x)] from 1 to 2 = ln(2) - ln(1) = ln(2).

The coefficient of x^2 is C(4,2) * (3)^2 * (-4)^2 = 6 * 9 * 16 = 864.

The coefficient of x^2 is C(6,2) * (3)^2 * (-4)^4 = 15 * 9 * 256 = 34560.

The coefficient of x^2 is C(6,2)(-2)^4 = 15 * 16 = 240.

The lines are perpendicular if 2h = a + b, which leads to h^2 = -ab.

The lines are parallel if the discriminant of the quadratic equation is zero, which leads to the condition h^2 = ab.

For the lines to be coincident, the discriminant must be zero, i.e., b^2 - 4ac = 0.

The lines are perpendicular if the condition a + b = 0 holds true.

The limit as x approaches 0 does not exist, hence f(x) is not continuous at x = 0.

The left limit is 0, the right limit is 2, and f(1) = 3. Thus, it is discontinuous.

The left limit as x approaches 1 is 1, and the right limit is also 1. Thus, f(1) = 1, making it continuous.

Setting f'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so critical points are x = -1 and x = 1.

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x = 0 and x = 2. Critical points are (0, 4) and (2, 1).

Setting f'(x) = 0 gives critical points at x = 1 and x = 2.

Setting f'(x) = 0 gives critical points at x = 0, ±2.

f'(x) = 4x^3 - 16x = 4x(x^2 - 4). Critical points are x = 0, ±2.

f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 1)(x - 3) = 0, so critical points are x = 1 and x = 3.

Using the power rule, f'(x) = -1/x^2.

Using the chain rule, f'(x) = (1/(x^2 + 1)) * (2x) = 2x/(x^2 + 1).

Equation of circle: (x - h)² + (y - k)² = r² => (x - 2)² + (y + 3)² = 5² = 25.

The slope m = (9 - 0) / (3 - 0) = 3. The equation is y = 3x.