Using the derivatives of sine and cosine, g'(x) = cos(x) - sin(x).

f'(x) = -2x + 6; setting to 0 gives x = 3; f(3) = -3^2 + 6(3) - 8 = 1.

The maximum occurs at x = -b/(2a) = -12/(-6) = 2. f(2) = -3(2^2) + 12(2) - 5 = 7.

The minimum occurs at x = -b/(2a) = 16/(2*4) = 2. f(2) = 4(2^2) - 16(2) + 15 = 1.

The minimum occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

First derivative f'(x) = 16x^3 - 6x^2 + 1. Second derivative f''(x) = 48x^2 - 12x.

First derivative f'(x) = 3x^2 - 6x; second derivative f''(x) = 6x - 6.

First derivative f'(x) = 3x^2 - 6x; Second derivative f''(x) = 6x - 6.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12. Setting 6x - 12 = 0 gives x = 2.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12. Setting 6x - 12 = 0 gives x = 2.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12. Setting it to zero gives x = 2.

The maximum occurs at x = -b/(2a) = -4/(-2) = 2. f(2) = -2^2 + 4(2) + 1 = 5.

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

f'(x) = cos(x); f'(π/2) = cos(π/2) = 0.

f(x) is a polynomial function and is continuous everywhere, hence no points of discontinuity.

The left limit as x approaches 0 is 0, the right limit is 1, and f(0) = 0. Since the limits do not match, f(x) is discontinuous at x = 0.

To make f(x) continuous at x = 1, we need c = 1^2 = 1. Thus, c must be 1.

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

To ensure continuity at x = 2, k(2) + 1 must equal 3. Thus, k = 1.

f(1) = 3(1) + 2 = 5. Since f(x) is a linear function, it is continuous everywhere.

f(2) = 3(2) + 2 = 8. Since f(x) is a polynomial, it is continuous everywhere.

First, find f'(x) = 6x + 2. Then, f'(2) = 6(2) + 2 = 12 + 2 = 14.

First, find f'(x) = 12x^2 - 4x + 1, then differentiate again to get f''(x) = 24x - 4.

The second derivative f''(x) = d^2/dx^2(e^x) = e^x.

f'(x) = e^x and f''(x) = e^x. Therefore, f''(0) = e^0 = 1.

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).

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