Application of Derivatives

f'(x) = 6x² - 12x + 6, f has a critical point at x = 1, f'(1) = 0, f''(x) = 12x - 12

f'(x) = 6x² - 12x + 6 = 6(x² - 2x + 1) = 6(x - 1)². So f'(1) = 0, making x = 1 a critical point. f''(x) = 12x - 12. Since f'(x) = 6(x - 1)² ≥ 0 for all x, f is always increasing except at x = 1 where it has zero slope.

1.5 m/s

Let x = distance from wall to bottom, y = height up wall. Given: x² + y² = 100, dx/dt = 2 m/s, x = 6 m. Find dy/dt. When x = 6: y = √(100 - 36) = 8 m. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0. So 2(6)(2) + 2(8)(dy/dt) = 0, giving 24 + 16(dy/dt) = 0, so dy/dt = -1.5 m/s. The top slides down at 1.5 m/s.

At local maxima, f'(x) = 0, If f'(c) = 0 and f''(c) > 0, then c is a local minimum, If f'(c) = 0 and f''(c) < 0, then c is a local maximum

At local extrema of differentiable functions, f'(x) = 0. The second derivative test: f''(c) > 0 indicates local minimum, f''(c) < 0 indicates local maximum. However, some critical points may be inflection points, not extrema. Also, f''(x) need not be zero at extrema.

2

f'(x) = 3x² - 3 = 3(x² - 1). Critical points: x = ±1. Evaluate f at critical points and endpoints: f(-2) = -8 + 6 = -2, f(-1) = -1 + 3 = 2, f(1) = 1 - 3 = -2, f(2) = 8 - 6 = 2. The absolute maximum value is 2, occurring at x = -1 and x = 2.