The function is a polynomial and is differentiable everywhere, hence there are no points where it is not differentiable.

The function is a polynomial and is differentiable everywhere, so there is no such x.

f(x) is a polynomial and is differentiable everywhere. The x-coordinate can be any real number.

First, find f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 3)(x - 1) = 0, so critical points are x = 1 and x = 3.

f'(x) = 3x^2 - 12x + 9. The critical points are x = 1 and x = 3. The function is increasing on (1, 3) and (3, ∞).

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

Find f''(x) = 12x^2 - 16. Setting f''(x) = 0 gives x^2 = 4, so x = ±2. f(2) = 0, thus the inflection point is (2, 0).

f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x(x^2 - 4) = 0, so x = -2, 0, 2. Test intervals: f' is positive in (-2, ∞).

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.

Setting the left limit (0 + 1 = 1) equal to the right limit (3 - 0 = 3), we find b = 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.

By definition, f(1) = 3, as given in the piecewise function.

Setting f(1-) = f(1+) and f'(1-) = f'(1+) gives k = 2 for differentiability.

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

For continuity, f(3) must equal the limit as x approaches 3, which is 9.

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.

The function is not differentiable at x = -3 and x = 2, but the first point of interest is -3.

The distance between the foci of the hyperbola is 2c, where c = √(a^2 + b^2) = √(25 + 16) = √41, so the distance is 2√41.

The slopes can be found by solving the quadratic equation derived from the given equation.

The product of the slopes of the lines can be found from the equation, which gives -1.

The sum of the slopes can be found using the relationship between the coefficients of the quadratic.

The condition for parallel lines is that the determinant of the coefficients must equal zero.

Factoring the equation gives the slopes as m1 = 1 and m2 = 3.

The lines are at an angle of 45 degrees as the determinant of the coefficients gives a non-zero value.

The sum of the slopes is given by - (coefficient of xy)/(coefficient of x^2) = -6/5.

The sum of the slopes of the lines is given by -b/a, which is -5/6.

The slopes can be found by solving the quadratic equation formed by the coefficients of x^2, xy, and y^2.

The angle can be calculated using the slopes derived from the equation.

The inverse of D is given by (1/det(D)) * adj(D). Here, det(D) = (4*3) - (2*1) = 10. The adjugate is [[3, -2], [-1, 4]]. Thus, D^(-1) = (1/10) * [[3, -2], [-1, 4]].

Using the determinant formula, det(E) = 1*(1*0 - 4*6) - 2*(0*0 - 4*5) + 3*(0*6 - 1*5) = 1*(-24) - 2*(-20) - 15 = -24 + 40 - 15 = 1.