Applications of Differential Calculus — Practice Quiz

3

Step 1: Direct substitution gives 0/0, which is an indeterminate form. Step 2: Apply L'Hopital's Rule: differentiate numerator and denominator separately. Step 3: d/dx(sin 3x) = 3cos(3x) and d/dx(x) = 1. Step 4: lim(x→0) [3cos(3x)/1] = 3cos(0) = 3 × 1 = 3. Final Answer: The limit is 3. Note: This also follows the standard result lim(x→0) [sin(mx)/x] = m.

x = 1 and x = -1

Step 1: Critical numbers occur where f'(x) = 0 or f'(x) does not exist. Step 2: f'(x) = 3x² - 3. Step 3: Set f'(x) = 0: 3x² - 3 = 0, so x² = 1, giving x = ±1. Step 4: f'(x) exists for all x, so only stationary points. Final Answer: Critical numbers are x = 1 and x = -1. Common mistake: Setting f(x) = 0 instead of f'(x) = 0, which gives different values.

x = -2

Step 1: f'(x) = 6x² + 6x - 12 = 6(x² + x - 2) = 6(x+2)(x-1). Step 2: Critical numbers: x = -2 and x = 1. Step 3: f''(x) = 12x + 6. At x = -2: f''(-2) = -24 + 6 = -18 < 0 → local maximum. Step 4: At x = 1: f''(1) = 12 + 6 = 18 > 0 → local minimum. Final Answer: Local maximum occurs at x = -2. The second derivative being negative confirms it is a maximum.

1/3

Step 1: Direct substitution gives ∞/∞, an indeterminate form. Step 2: Apply L'Hopital's Rule once: lim [2x/6x] = lim [1/3]. Step 3: This simplifies to the constant 1/3 which is not indeterminate. Final Answer: The limit is 1/3. Alternative: Divide numerator and denominator by x²: lim [(1 + 5/x²)/3] = 1/3 as x→∞. Both methods give the same result.