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

Using the derivative rules, f'(x) = cos(x) - sin(x).

f'(x) = cos(x) - sin(x). At x = π/4, f'(π/4) = cos(π/4) - sin(π/4) = √2/2 - √2/2 = 0.

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

First derivative f'(x) = 2x + 2. Second derivative f''(x) = 2.

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

lim x→-1 f(x) = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0.

f(-1) = (-1)^2 + 3(-1) + 2 = 0. Since f(x) is a polynomial, it is continuous everywhere.

The first derivative f'(x) = 2x + 3, and the second derivative f''(x) = 2.

At x = 1, f(1) = 1^2 = 1 and the limit from the left is also 1, hence f(x) is continuous at x = 1.

At x = 1, f(1) = 3 and limit from left is 1^2 = 1. Since they are not equal, f(x) is discontinuous at x = 1.

f(1) = 1^3 - 3*1 + 2 = 0. The function is a polynomial and hence continuous everywhere, including at x = 1.

f(1) = 1^3 - 3(1) + 2 = 0. Since f(x) is a polynomial, it is continuous everywhere.

First, find f'(x) = 3x^2 - 6x. Then, f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.

Find f''(x) = 6x - 12. Set f''(x) = 0 gives x = 2. The inflection point is (2, f(2)) = (2, 0).

First, find f'(x) = 4x^3 - 6x^2 + 1. Then, f'(1) = 4(1)^3 - 6(1)^2 + 1 = 4 - 6 + 1 = -1.

f'(x) = 4x^3 - 12x^2 + 12x; f'(2) = 4(2^3) - 12(2^2) + 12(2) = 32 - 48 + 24 = 8.

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

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