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 the minimum, set f'(x) = e^x + 2x = 0. The minimum occurs at x = 0.

Set f'(x) = e^x - 2x = 0. The solution is approximately x = 1.

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.

Setting f'(x) = 3x^2 - 12x + 9 = 0 gives x = 1, 3. Testing shows x = 2 is a local minimum.

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.

The vertex is at x = -(-12)/(2*3) = 2. The minimum value is f(2) = 3(2^2) - 12(2) + 7 = -5. Thus, the coordinates are (2, -5).

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

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

The vertex is at x = 6/2 = 3. The minimum value is f(3) = 3^2 - 6*3 + 10 = 1.

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

The left limit as x approaches 1 is 3, the right limit is 1, and f(1) = 3. Since the limits do not match, f(x) is discontinuous at x = 1.

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.

At x = 2, left limit is 4 and right limit is 4, but f(2) = 4. Hence, f(x) is continuous at x = 2.

The left limit as x approaches 3 is 9, the right limit is also 9, and f(3) = 9. Therefore, f(x) is continuous at x = 3.

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 be continuous at x = 2, k(2) + 1 must equal 3. Thus, 2k + 1 = 3, leading to k = 1.

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

To be continuous at x = 2, k(2) + 1 must equal 3. Thus, 2k + 1 = 3, giving 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 first derivative f'(x) = 10x + 3, and the second derivative f''(x) = 10.

First, find f'(x) = 10x + 3. Then, f'(2) = 10(2) + 3 = 20 + 3 = 23.