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

The determinant is calculated as 1*(1*1 - 0*3) - 2*(0*1 - 0*2) + 1*(0*3 - 1*2) = 1 - 0 - 2 = -1.

The determinant is (2*3) - (1*1) = 6 - 1 = 5.

The determinant is (4*1) - (2*3) = 4 - 6 = -2.

The integral of 3x^2 is x^3. Evaluating from 1 to 3 gives 27 - 1 = 26.

The integral of x^3 is (1/4)x^4. Evaluating from 1 to 2 gives (4 - 1/4) = 3.75.

f(x) = |x| is not differentiable at x = 0 because the left-hand and right-hand derivatives are not equal.

First, find the derivative: y' = 6x^2 - 6. At x = 1, y' = 6(1)^2 - 6 = 0.

The integral of (2x + 3) is x^2 + 3x. Evaluating from 0 to 2 gives (4 + 6) - (0 + 0) = 10.

The determinant is 1 because it is an upper triangular matrix.

The determinant is 0 because the rows are linearly dependent.

The determinant is calculated using the rule of Sarrus or cofactor expansion, which results in 0.

The integral of e^x is e^x. Evaluating from 0 to 1 gives e^1 - e^0 = e - 1.

The vertex occurs at x = 3. f(3) = -3^2 + 6(3) - 8 = 6.

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

For continuity at x = 1, f(1) = k(1)^2 + 2(1) + 1 = k + 3 must be defined for all k.

sin(-π/2) = -1, therefore sin^(-1)(-1) = -π/2.

sin(π/3) = √3/2, therefore sin^(-1)(√3/2) = π/3.

The vertex occurs at x = -b/(2a) = 12/(2*3) = 2, which is a local maximum.

Find f'(x) = e^x - 2x. Setting f'(x) = 0 gives e^x = 2x. The critical point is approximately (1, e-1).

f(x) is differentiable for all values of a since it is a polynomial function.

The sum A + B is calculated as [[1+0, 2+1], [3+1, 4+0]] = [[1, 3], [4, 4]].

The product AB is calculated as [[(1*5 + 2*7), (1*6 + 2*8)], [(3*5 + 4*7), (3*6 + 4*8)]] = [[19, 22], [43, 50]].

cos(π) = -1, hence cos^(-1)(-1) = π.

f'(x) = 9x^2 - 12x; f''(x) = 18x - 12. Therefore, f''(1) = 18(1) - 12 = 6.

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

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

sin(π/3) = √3/2, hence x = √3/2.

tan(π/4) = 1, hence x = 1.

The identity matrix squared is still the identity matrix: A^2 = A.