Insurance and Annuity

[1, 3]

Step 1: For Rolle's theorem, f(a) = f(b). Step 2: f(x) = x² - 4x + 3 = (x-1)(x-3). Step 3: f(1) = 0 and f(3) = 0. Step 4: Since f(1) = f(3), interval [1,3] satisfies Rolle's theorem conditions.

f is continuous on [a,b], f is differentiable on (a,b), f'(c) = [f(b)-f(a)]/(b-a) for some c in (a,b)

LMVT requires: (1) f continuous on [a,b], (2) f differentiable on (a,b), (3) conclusion: ∃c∈(a,b) such that f'(c) = [f(b)-f(a)]/(b-a). Note: f(a) = f(b) is for Rolle's theorem, not LMVT.

x < -1, x > 3

Step 1: f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x+1)(x-3). Step 2: f'(x) > 0 when (x+1)(x-3) > 0. Step 3: This occurs when both factors have same sign: x < -1 or x > 3.

x = 1, 3

Step 1: Find f'(x) = 3x² - 12x + 9. Step 2: Set f'(x) = 0: 3x² - 12x + 9 = 0. Step 3: Divide by 3: x² - 4x + 3 = 0. Step 4: Factor: (x-1)(x-3) = 0. Step 5: Critical points are x = 1 and x = 3.