The integral evaluates to [x^3 - 2x] from 1 to 2 = (8 - 4) - (1 - 2) = 4 + 1 = 5.

The integral evaluates to [x^3 - x^2 + x] from 1 to 2 = (8 - 4 + 2) - (1 - 1 + 1) = 5.

The integral evaluates to [x - x^3/3] from 0 to 1 = (1 - 1/3) = 2/3.

∫_0^1 (4x^3) dx = [x^4] from 0 to 1 = 1.

∫_0^1 (x^2 + 1) dx = [x^3/3 + x] from 0 to 1 = (1/3 + 1) = 4/3.

The integral evaluates to [x^5/5 + (1/2)x^4 + (1/3)x^3] from 0 to 1 = 1/5 + 1/2 + 1/3 = 31/30.

∫_0^1 (x^4 + 2x^3) dx = [x^5/5 + (1/2)x^4] from 0 to 1 = (1/5 + 1/2) = 7/10.

∫_0^1 x^4 dx = [x^5/5] from 0 to 1 = 1/5.

Using the identity sin(2x) = 2sin(x)cos(x), the integral becomes (1/2)∫_0^π sin(2x) dx = 0.

∫_0^π sin(x) dx = [-cos(x)] from 0 to π = -(-1 - 1) = 2.

The integral evaluates to [x/2 + sin(2x)/4] from 0 to π/2 = π/4.

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f''(1) < 0 indicates a local maximum at x = 1.

The vertex occurs at x = -b/(2a) = 4/2 = 2, which is the x-coordinate of the minimum point.

The vertex occurs at x = -b/(2a) = 4/2 = 2. This is where the local minimum occurs.

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

f''(x) = 12x - 18. Setting f''(x) = 0 gives x = 1.5. The inflection point is (1.5, f(1.5)).

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 3. Testing intervals shows f is increasing on (1, 3).

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f(1) = 5 is a local maximum.

The vertex is at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 7 = -1. So, the vertex is (2, -1).

f''(x) = 18x - 24. Setting f''(x) = 0 gives x = 4/3. This is the inflection point.

Finding f'(x) = 9x^2 - 24x + 9 and solving gives two critical points. The second derivative test confirms one maximum and one minimum.

To find the point of inflection, we compute f''(x) = e^x - 2. Setting f''(x) = 0 gives e^x = 2, leading to x = ln(2). The closest integer is x = 1.

f(x) = ln(x) is not defined for x ≤ 0, hence not differentiable at x = 0.

f'(x) = cos(x) - sin(x). Setting f'(x) = 0 gives tan(x) = 1, so x = π/4 + nπ. In [0, 2π], the maximum occurs at x = 3π/4.

f'(x) = 2x + 2.

The function is a polynomial and is differentiable everywhere.

Setting the derivative f'(-1) = 0 gives k = 1 for differentiability.

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

As a polynomial, f(x) is differentiable everywhere, including at x = 2.

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, which is the minimum value.