f(x) is a polynomial function, which is differentiable everywhere.

f(x) is a polynomial function, hence it is differentiable everywhere including at x = 2.

f(x) is a polynomial function, hence differentiable everywhere including at x = 2.

f(x) is a polynomial function, hence it is differentiable everywhere.

f'(1) from left = 2 and from right = 2; hence, f is continuous but not differentiable at x = 1.

True, as the limit of f'(x) as x approaches 0 exists and equals 0.

f(x) is a polynomial function, hence it is differentiable everywhere including at x = 1.

f'(x) = 3x^2 - 3 = 0 gives x = ±1, thus critical points are x = -1 and x = 1.

f'(x) = 3x^2 - 3, thus f'(1) = 0.

Finding f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1 and x = 3. Checking the second derivative shows one local maximum and one local minimum.

The limit as x approaches 0 does not equal f(0) = 0, hence it is not continuous at x = 0.

The function is not continuous at x = 0 since the limit does not equal f(0).

To check continuity at x = 1, we find the left limit (5) and the right limit (2). They are not equal, hence f(x) is not continuous at x = 1.

For continuity at x = 1, the left limit (3) must equal f(1) (2), which is not true.

For continuity at x = 1, both pieces must equal 4, hence the function is continuous.

To check continuity at x = 1, we find the left limit (3) and the right limit (3). Both equal 3, hence f(x) is continuous at x = 1.

Limit as x approaches 0 from left is 0, and f(0) = 1, hence it is not continuous at x = 0.

To check continuity at x = 0, we find f(0) = 1 and limit as x approaches 0 is also 1.

The function is not continuous at x = 1 because the left-hand limit does not equal the function value.

To check continuity at x = 1, we find f(1) = 1, limit as x approaches 1 from left is 1, and from right is also 1.

To check continuity at x = 1, we find the limit from both sides. Both limits equal 1, hence f(x) is continuous at x = 1.

For differentiability, the left and right derivatives must equal at x = 1, hence f'(1) = 1.

Both conditions must hold true for continuity at x = 2.

To be continuous at x = 2, k must equal f(2) = 2^2 = 4.

f(x) = |x| is not differentiable at x = 0 because the left-hand and right-hand derivatives do not match.

To find the maximum, we calculate f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, this is a maximum point.

The vertex form of a parabola gives the maximum value at x = -b/(2a) = 2. Evaluating f(2) = -2^2 + 4*2 + 1 = 9.

The vertex of the parabola given by f(x) = -x^2 + 4x + 1 occurs at x = -b/(2a) = -4/(-2) = 2, which gives the maximum value.

Finding the derivative f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. Evaluating f(0) = 16, f(2) = 0, and f(-2) = 0, the minimum value is 0.

The derivative f'(x) = cos(x). At x = π/4, f'(π/4) = cos(π/4) = √2/2.