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.

By substituting x = 1 into the equation, we find that it satisfies the equation, hence 1 is a root.

The roots can be found using the Rational Root Theorem and synthetic division, confirming that all roots are real.

The mode is 4, as it appears 3 times, more than any other number.

Mean = (1 + 2 + 3 + 4 + 5) / 5 = 3. Variance = [(1-3)² + (2-3)² + (3-3)² + (4-3)² + (5-3)²] / 5 = 2.

Mean = (10 + 12 + 14 + 16 + 18) / 5 = 14. Variance = [(10-14)² + (12-14)² + (14-14)² + (16-14)² + (18-14)²] / 5 = 8.

Mean = (10 + 12 + 14 + 16) / 4 = 13. Variance = [(10-13)² + (12-13)² + (14-13)² + (16-13)²] / 4 = 2.

Mean = (2 + 4 + 6 + 8 + 10) / 5 = 6. Variance = [(2-6)² + (4-6)² + (6-6)² + (8-6)² + (10-6)²] / 5 = 8.

All values are the same, so variance = 0.

Mean = 9. Variance = [(5-9)² + (7-9)² + (9-9)² + (11-9)² + (13-9)²] / 5 = 8.

Arranging the numbers: 5, 8, 12, 15, 20. The median is the middle number, which is 12.

The mode is 9, as it appears most frequently (three times) in the data set.

The discriminant is 2^2 - 4(1)(1) = 0, indicating that the roots are real and equal.

The discriminant must be non-negative: 4^2 - 4*1*k >= 0 leads to 16 - 4k >= 0, thus k <= 4.

The condition for no real roots is that the discriminant must be less than zero: 6^2 - 4*1*k < 0 => 36 < 4k => k > 9.

The equation can be factored as (x-1)^3 = 0, which has one real root (x = 1) with multiplicity 3.

By substituting x = 2 into the equation, we find that it satisfies the equation, thus x = 2 is a root.

The product of the roots of the cubic equation ax^3 + bx^2 + cx + d = 0 is given by -d/a. Here, d = -6 and a = 1, so the product is -(-6)/1 = 6.

By substituting x = 2 into the equation, we find that 2 is a root since 2^3 - 6(2^2) + 11(2) - 6 = 0.

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.