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).

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.

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, ∞).

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.

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 = 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.

The function is a polynomial and is differentiable everywhere, hence nowhere.

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

Setting f(1-) = f(1+) and f'(1-) = f'(1+) leads to a = 1 for differentiability.

Setting the two pieces equal at x = 1 gives us 3 + c = 1. Thus, c = -2.

To ensure continuity at x = 1, k(1) + 1 = 2(1) - 3, solving gives k = 2.

To ensure continuity at x = 1, we need to set the two pieces equal: k + 1^2 = 2(1) + 3. This gives k + 1 = 5, so k = 4.

To ensure continuity at x = 1, we need to set the two pieces equal: 1^2 + k = 2(1) + 1. This gives k = 2.

For f(x) to be continuous at x = 2, we need limit as x approaches 2 from left to equal limit as x approaches 2 from right and equal to f(2). Thus, k = 4.

For f(x) to be continuous at x = 2, we need limit as x approaches 2 from left to equal limit as x approaches 2 from right. Thus, k must equal 0.

Setting f(2-) = f(2+) and f'(2-) = f'(2+) leads to m = 1 for differentiability.