If f(x) = x^2 + 2x + 1 for x < 0 and f(x) = kx + 1 for x >= 0, find k such that f is differentiable at x = 0.

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Q: If f(x) = x^2 + 2x + 1 for x < 0 and f(x) = kx + 1 for x >= 0, find k such that f is differentiable at x = 0.

Solution: Setting the left-hand derivative equal to the right-hand derivative at x = 0 gives k = 2.

Steps: 10

Step 1: Identify the function f(x) for x < 0, which is f(x) = x^2 + 2x + 1.
Step 2: Identify the function f(x) for x >= 0, which is f(x) = kx + 1.
Step 3: Find the left-hand derivative of f(x) at x = 0. This means we need to differentiate f(x) = x^2 + 2x + 1 and evaluate it at x = 0.
Step 4: Differentiate f(x) = x^2 + 2x + 1. The derivative is f'(x) = 2x + 2.
Step 5: Evaluate the left-hand derivative at x = 0: f'(0) = 2(0) + 2 = 2.
Step 6: Find the right-hand derivative of f(x) at x = 0. This means we need to differentiate f(x) = kx + 1 and evaluate it at x = 0.
Step 7: Differentiate f(x) = kx + 1. The derivative is f'(x) = k.
Step 8: Evaluate the right-hand derivative at x = 0: f'(0) = k.
Step 9: Set the left-hand derivative equal to the right-hand derivative: 2 = k.
Step 10: Solve for k. Therefore, k = 2.