lim x→-1 f(x) = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0.

f(-1) = (-1)^2 + 3(-1) + 2 = 0. Since f(x) is a polynomial, it is continuous everywhere.

The first derivative f'(x) = 2x + 3, and the second derivative f''(x) = 2.

f(x) = x^2 - 4 is a polynomial function, which is continuous everywhere, including at x = 2.

At x = 1, f(1) = 1^2 = 1 and the limit from the left is also 1, hence f(x) is continuous at x = 1.

At x = 1, f(1) = 3 and limit from left is 1^2 = 1. Since they are not equal, f(x) is discontinuous at x = 1.

f(1) = 1^3 - 3*1 + 2 = 0. The function is a polynomial and hence continuous everywhere, including at x = 1.

f(1) = 1^3 - 3(1) + 2 = 0. Since f(x) is a polynomial, it is continuous everywhere.

First, find f'(x) = 3x^2 - 6x. Then, f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.

First derivative f'(x) = 3x^2 - 4, then f''(x) = 6x.

Find f''(x) = 6x - 12. Set f''(x) = 0 gives x = 2. The inflection point is (2, f(2)) = (2, 0).

First, find f'(x) = 4x^3 - 6x^2 + 1. Then, f'(1) = 4(1)^3 - 6(1)^2 + 1 = 4 - 6 + 1 = -1.

f'(x) = 4x^3 - 12x^2 + 12x; f'(2) = 4(2^3) - 12(2^2) + 12(2) = 32 - 48 + 24 = 8.

f'(x) = 4x^3 - 12x^2; thus, f'(2) = 4(2^3) - 12(2^2) = 32 - 48 = -16.

f'(x) = 4x^3 - 12x^2. f'(2) = 4(2^3) - 12(2^2) = 32 - 48 = -16.

Finding the derivative f'(x) = 4x^3 - 16x. Setting it to zero gives x = 0, ±2. The minimum value occurs at x = 2, f(2) = 0.

The eigenvalues are found by solving the characteristic equation det(G - λI) = 0. The eigenvalues are λ = 1, 8.

The determinant of H is calculated as (0*-0) - (1*-1) = 0 + 1 = 1.

Using the determinant formula for 3x3 matrices, det(H) = 1(1*0 - 4*6) - 2(0 - 4*5) + 3(0 - 1*5) = 0 - 40 - 15 = -55.

The rank of H is 1 because the second row is a multiple of the first row.

The eigenvalues are found by solving the characteristic equation: det(H - λI) = 0, which gives λ^2 - 9λ + 1 = 0.

Using the chain rule, h'(x) = 2e^(2x).

The rank of I is 1 because all rows are linearly dependent.

Using the determinant formula for 3x3 matrices, det(I) = 1(1*0 - 4*6) - 2(0 - 4*5) + 3(0 - 1*5) = 0 - 40 - 15 = -55.

The trace of a matrix is the sum of its diagonal elements. Thus, trace(I) = 2 + 2 = 4.

The rank of J is 1 because both rows are linearly dependent (they are identical).

The determinant of J is 0, as the rows are linearly dependent.

log_10(0.01) = log_10(10^-2) = -2, so x = -2.

log_10(10^x) = x. Therefore, x = 2.

log_10(20) = log_10(2) + log_10(10) = 0.301 + 1 = 1.301.