Setting f'(x) = 0 gives critical points at x = 0, ±2.

f'(x) = 4x^3 - 16x = 4x(x^2 - 4). Critical points are x = 0, ±2.

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

Using the power rule, f'(x) = -1/x^2.

Using the chain rule, f'(x) = (1/(x^2 + 1)) * (2x) = 2x/(x^2 + 1).

Using the product rule, f'(x) = d/dx(x^2 * e^x) = e^x * (x^2 + 2x).

Equation of circle: (x - h)² + (y - k)² = r² => (x - 2)² + (y + 3)² = 5² = 25.

The slope m = (9 - 0) / (3 - 0) = 3. The equation is y = 3x.

f'(x) = 2x + 2. At x = 1, f'(1) = 4. The point is (1, 4). The tangent line is y - 4 = 4(x - 1) => y = 4x - 4 + 4 => y = 4x - 2.

The equation x^2 - y^2 = c represents a family of hyperbolas with varying values of c.

The equation x^2/a^2 + y^2/b^2 = 1 represents a family of ellipses with varying semi-major and semi-minor axes.

The equation y = ax^2 + bx + c represents a family of parabolas with varying coefficients a, b, and c.

The equation y = ax^3 + bx represents a family of cubic functions where a and b are constants.

The equation y = ax^3 + bx^2 + cx + d represents a family of cubic functions.

The equation y = e^(kx) represents a family of exponential curves with varying growth rates determined by k.

The equation y = k/x represents a family of hyperbolas with varying values of 'k'.

The equation y = kx^2 represents a family of parabolas that open upwards or downwards depending on the sign of k.

The equation x^2 = 4py gives 4p = 12, hence p = 3. The focus is at (0, p) = (0, 3).

The standard form of the parabola is x^2 = 4py. Here, 4p = 8, so p = 2. The focus is at (0, p) = (0, 2).

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f'(x) > 0 for x > 1, so f(x) is increasing on (1, ∞).

f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). The function is increasing where f'(x) > 0, which is in the intervals (-∞, 0) and (3, ∞).

The length of the latus rectum for the parabola y^2 = 4px is given by 4p. Here, p = 4, so the length is 16.

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f''(1) = 6 > 0 (min), f''(-1) = 6 > 0 (min). Local maxima at (0, 0).

f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f''(1) > 0 (min), f''(3) < 0 (max).

f'(x) = 4x^3 - 12x^2 + 4. Setting f'(x) = 0 gives x = 0 and x = 2. f''(0) = 4 > 0 (min), f''(2) = -8 < 0 (max).

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

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

The discriminant indicates that the lines intersect at two distinct points.

Find f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(x - 2) = 0, so x = 0 or x = 2. f''(2) = 6 > 0, so (2, 1) is a local minimum.

The function |x - 1| is not differentiable at x = 1 due to a cusp.