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

f(1) = 1^2 + 3(1) + 2 = 6. Since f(x) is a polynomial function, it is continuous everywhere.

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.

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

f(x) = x^2 - 4 is a polynomial function, which is continuous everywhere, including at 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.

f'(x) = 2x - 4. Thus, f'(2) = 2(2) - 4 = 0.

(f ∘ g)(2) = f(g(2)) = f(2 + 1) = f(3) = 3^2 = 9.

f'(1) from left = 2 and from right = 2; hence f is differentiable at x = 1.

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.

Using the limit definition of the derivative, f'(0) exists, hence f is differentiable at x = 0.

f(-3) = (-3)^2 = 9.

f'(x) = 3x^2 - 3. Thus, f'(1) = 3(1)^2 - 3 = 0.

Set f'(x) = 3x^2 - 3 = 0 and solve for x.

The function is a polynomial and is differentiable everywhere, hence no points of non-differentiability.

f(x) is a polynomial function and is continuous for all x.

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

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.

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

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives 3x(x - 2) = 0, so x = 0 and x = 2 are critical points.

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2.

The function is a polynomial and is differentiable everywhere, but checking critical points shows f'(x) = 0 at x = 2.

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f''(2) = 6 > 0, so (2, f(2)) = (2, 1) is a local minimum.

f'(x) = 3x^2 - 6x; f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3.

f'(x) = 3x^2 - 6x; f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.