f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = 1. f(1) = 0.

f'(x) = 4x^3 - 8x. Setting f'(x) = 0 gives x = 0, ±2. f(0) = 0.

Area = 1/2 * base * height = 5h. Max area occurs when h is maximized, thus Area = 50 when h = 10.

The vertex occurs at x = 3. f(3) = -3^2 + 6*3 + 5 = 8.

The maximum height occurs at t = -b/(2a) = -64/(2*-16) = 2. h(2) = -16(2^2) + 64(2) + 80 = 80.

The maximum height occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 80 = 80.

The vertex is at x = -4/(2*(-2)) = 1. The maximum value is f(1) = -2(1)^2 + 4(1) + 1 = 3.

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

f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f(2) = -2^2 + 4(2) = 8.

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

The vertex is at x = -6/(2*(-1)) = 3. The maximum value is f(3) = -3^2 + 6*3 - 8 = 1.

Arrange the numbers: 1, 3, 5, 7, 9. The median is the middle value, which is 5.

Arrange the numbers: 1, 2, 3, 4, 5, 6, 7, 8. The median is the average of the 4th and 5th numbers: (4 + 5) / 2 = 4.5.

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 5 = 1.

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 7 = 3.

The minimum occurs at x = -b/(2a) = 6/(2*1) = 3. f(3) = 3^2 - 6(3) + 10 = 3.

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

The function is a upward-opening parabola. The minimum occurs at x = -b/(2a) = 4/(2*1) = 2. f(2) = 2^2 - 4(2) + 6 = 2.

The mode is 4, as it appears 3 times, more than any other number.

The moment of inertia of a solid sphere about an axis through its center is I = 2/5 MR^2.

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.

The function f(x) = |x - 3| is not differentiable at x = 3 because it has a sharp corner.

f(x) is not differentiable at x = -2 and x = 2, but is differentiable everywhere else.

f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x(12x - 24) = 0, so x = 0 or x = 2. Check f(1) = 3.

f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x = 0 and x = 2. The point of inflection is at (1, 3).

Find f''(x) = 12x^2 - 24x + 12. Setting f''(x) = 0 gives x = 1 and x = 2. Testing intervals shows a change in concavity at x = 1.

Setting 2x + 1 = -x + 4 gives 3x = 3, hence x = 1. Substituting back gives y = 3.

Set f'(x) = 0. f'(x) = 6x^2 - 18x + 12 = 0 gives x = 2. f(2) = 0.