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.

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

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.

Scalar product = 0*3 + 1*4 + 2*5 = 0 + 4 + 10 = 14.

A · B = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

A · B = 2*3 + 2*3 + 2*3 = 6 + 6 + 6 = 18.

-2x + 6 > 0 => -2x > -6 => x < 3.

-2x + 6 ≥ 0 => -2x ≥ -6 => x ≤ 3.

-3x + 1 ≤ 4 => -3x ≤ 3 => x ≥ -1.

-3x + 4 ≤ 1 => -3x ≤ -3 => x ≥ 1.

2x + 3 ≤ 7 => 2x ≤ 4 => x ≤ 2.

6 - x > 2 => -x > -4 => x < 4.

6 - x ≤ 3 => -x ≤ -3 => x ≥ 3.

7 - 3x < 1 => -3x < -6 => x > 2.

7x - 2 ≤ 5x + 6 => 2x ≤ 8 => x ≤ 4.

2(x - 1) ≥ 3 => x - 1 ≥ 1.5 => x ≥ 2.

2x + 3 > 5 => 2x > 2 => x > 1.

2x + 3 > 7 => 2x > 4 => x > 2.

2x + 3 ≤ 7 => 2x ≤ 4 => x ≤ 2.

3x - 4 < 2x + 5 => x < 9.

4x - 1 > 3 => 4x > 4 => x > 1.

4x - 1 > 3x + 2 => x > 3.

4x - 1 < 3 => 4x < 4 => x < 1.

4x - 8 < 0 => 4x < 8 => x < 2.

5 - 2x ≤ 3 => -2x ≤ -2 => x ≥ 1.

5 - x ≥ 2 => -x ≥ -3 => x ≤ 3.