f'(x) = 4x^3 - 12x^2. f'(2) = 4(2^3) - 12(2^2) = 32 - 48 = -16.

Finding the derivative f'(x) = 4x^3 - 16x. Setting it to zero gives x = 0, ±2. The minimum value occurs at x = 2, f(2) = 0.

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

The function f(x) = 1/(x-1) is not defined at x = 1, hence it is discontinuous at that point.

The function f(x) = sqrt(x) is continuous at x = 0 as it is defined and the limit exists.

The function f(x) = |x| is continuous at x = 0 because the left limit, right limit, and f(0) all equal 0.

The function f(x) = 1/(x-1) is discontinuous at x = 1, hence it is continuous on (1, ∞).

f(x) = 2x + 1 is a linear function and is continuous over the entire real line (-∞, ∞).

f(x) = 2x + 3 is a linear function and is continuous over the entire real line (-∞, ∞).

f(x) = 2x + 3 is a linear function and is continuous for all x.

To find the maximum, set f'(x) = cos(x) - sin(x) = 0. This occurs at x = π/4.

f(x) = x^2 + 3 is a polynomial function and is continuous for all x in (-∞, ∞).

The function f(x) = x^2 is a polynomial function and is continuous at all points, including -1, 0, and 1.

The function f(x) = x^3 - 3x is a polynomial function and is continuous at all points, including -2, 0, and 2.

Left limit as x approaches 1 is 2, right limit is 1, but f(1) = 2. Hence, it is discontinuous at x = 1.

The left limit as x approaches 0 is 0, but the right limit is 1. Hence, it is not continuous at x = 0.

At x = 0, lim x→0- f(x) = 0 and lim x→0+ f(x) = 1, hence it is discontinuous at x = 0.

The vertex of the parabola occurs at x = -b/(2a) = 4/2 = 2. The minimum value is f(2) = 2^2 - 4*2 + 6 = 2.

The derivative y' = 2x. At x = 2, y' = 2(2) = 4.

The function f(x) = |x| is continuous at x = 0 since both left and right limits equal f(0) = 0.

The function f(x) = sqrt(x) is continuous at x = 0 as it is defined and the limit exists.

Set f'(x) = 2x - 4 = 0; solving gives x = 2.

Find f'(x) = 2x - 4. Set f'(x) = 0, giving 2x - 4 = 0, hence x = 2.

First derivative f'(x) = 4x^3 - 12x^2. Setting f'(x) = 0 gives x = 0, 1, 3.

Using the power rule, f'(x) = 12x^3 - 10x.

Using the power rule, f'(x) = 12x^3 - 15x^2 + 2.

Using the power rule, f'(x) = -4/x^2.

Using the power rule, f'(x) = 10x - 4.

The derivative f'(x) = 15x^2 - 2.

Using the power rule, f'(x) = 25x^4 - 3.