Setting the two pieces equal at x = 1 gives 1 = a(1) + 1, leading to a = 0.

Setting the left limit (2(1) + 1 = 3) equal to the right limit (1^2 + 1 = 2), we find b = 3.

Setting limit as x approaches 1 gives b = 2 for continuity.

Setting the two pieces equal at x = 2 gives us 0 = 2b + 2. Solving for b gives b = -1.

Setting 1 - 3 + b = 2 gives b = 4 for continuity.

Setting 3(1) + c = 2(1)^2 gives c = -1 for continuity.

Setting the two pieces equal at x = c: c^2 - 4 = 3c - 5. Solving gives c = 3.

Setting x^2 - c = 2x + 1 at x = 1 gives c = 2.

Setting x^2 = 2x + 1 at x = c gives c = 2.

f'(x) = 2kx + 2; f'(-1) = -2k + 2 must exist for any k.

Setting 3 + m = 2 and 2 = m + 1 gives m = 1 for continuity.

Setting the two pieces equal at x = 0 gives us p = -1.

The function is discontinuous at x = 1 because it leads to division by zero.

Find f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. Testing intervals, f(x) is increasing on (1, 3).

f'(x) = 10x + 3; f'(1) = 10*1 + 3 = 13.

Using the chain rule, f'(x) = 2e^(2x).

Find f'(x) = e^x - 2x. Setting f'(x) = 0 gives a local maximum at x = 1.

f'(x) = e^x; f'(0) = e^0 = 1.

f''(x) = e^x, thus f''(0) = e^0 = 1.

The derivative f'(x) = 1/x + 2x. For f'(x) > 0, we need 1/x + 2x > 0, which holds for x > 0.

f'(x) = 1/x; f'(1) = 1, hence f is differentiable at x = 1.

f'(x) = 1/x.

To find critical points, we set f'(x) = cos(x) - sin(x) = 0. This gives tan(x) = 1, leading to x = π/4 and x = 5π/4 in the interval [0, 2π].

Setting the left-hand derivative equal to the right-hand derivative at x = 0 gives k = 2.

f'(x) = 2x + 2, thus f'(1) = 2(1) + 2 = 4.

Calculating the derivative f'(x) = 2x + 2, we find f'(1) = 4.

f'(x) = 2x + 2. Therefore, f'(1) = 2(1) + 2 = 4.

The limit as x approaches 2 is f(2) = 2^2 - 4 = 0.

f'(x) = 2x - 4. Thus, f'(2) = 2(2) - 4 = 0.

f'(1) from left = 2 and from right = 2; hence f is differentiable at x = 1.