Continuity and Differentiability — Practice Quiz

f(x) = x² + 1, f(x) = |x|, f(x) = sin(x)

For continuity at x = 0, we need lim(x→0) f(x) = f(0). 1) f(x) = x² + 1: lim(x→0) (x² + 1) = 1 = f(0) ✓ 2) f(x) = |x|: lim(x→0) |x| = 0 = f(0) ✓ 3) f(x) = 1/x: lim(x→0) 1/x doesn't exist (approaches ±∞) ✗ 4) f(x) = sin(x): lim(x→0) sin(x) = 0 = f(0) ✓ 5) f(x) = [x]: lim(x→0⁻) [x] = -1, lim(x→0⁺) [x] = 0, f(0) = 0. Since left and right limits differ ✗

x = 1, x = 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: Solve: x = 1 or x = 3 Verification: f'(1) = 3(1)² - 12(1) + 9 = 3 - 12 + 9 = 0 ✓ f'(3) = 3(9) - 12(3) + 9 = 27 - 36 + 9 = 0 ✓

2x/(x² + 1)

Using chain rule for logarithmic functions: Step 1: d/dx[ln(u)] = (1/u) × du/dx, where u = x² + 1 Step 2: du/dx = d/dx(x² + 1) = 2x Step 3: f'(x) = (1/(x² + 1)) × 2x = 2x/(x² + 1) Remember: The derivative of ln(g(x)) is g'(x)/g(x).

Every differentiable function is continuous, f(x) = |x| is not differentiable at x = 0, f(x) = x² is differentiable everywhere

1) True: If f is differentiable at a point, then f is continuous at that point. 2) False: f(x) = |x| is continuous everywhere but not differentiable at x = 0. 3) True: f(x) = |x| has no unique tangent at x = 0 (left derivative = -1, right derivative = 1). 4) True: Polynomial functions are differentiable everywhere in their domain. 5) False: Differentiability implies continuity, so this is impossible.