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

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.

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 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.

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.

Using the product rule, f'(x) = e^x * ln(x) + e^x * (1/x) = e^x * (ln(x) + 1/x).

Using the product rule, f'(x) = e^x * x^2 + e^x * 2x = e^x * (x^2 + 2x).

The derivative f'(x) = d/dx(ln(x)) = 1/x.

Using the product rule, f'(x) = 2x * sin(x) + x^2 * cos(x).

Using the product rule, f'(x) = 3x^2 * e^x + x^3 * e^x.

Using the product rule, f'(x) = 3x^2 * ln(x) + x^2.

Using the power rule, f'(x) = 4x^3 - 12x.

The derivative f'(x) = d/dx(3x^2 + 5x - 7) = 6x + 5.

The derivative f'(x) = 1/x.

Using the identity 1 - cos(x) = 2sin^2(x/2), we have lim (x -> 0) (1 - cos(x))/(x^2) = lim (x -> 0) (2sin^2(x/2))/(x^2) = 1/2.

Using the Taylor series expansion for cos(x), we find that lim (x -> 0) (cos(x) - 1)/x^2 = -1/2.

Using the derivative of e^x at x = 0, we find that lim (x -> 0) (e^x - 1)/x = 1.