Determine the value of m for which the function f(x) = { mx + 1, x < 2; x^2 - 4, x >= 2 } is differentiable at x = 2.

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Q: Determine the value of m for which the function f(x) = { mx + 1, x < 2; x^2 - 4, x >= 2 } is differentiable at x = 2.

Solution: Setting f(2-) = f(2+) and f'(2-) = f'(2+) leads to m = 1 for differentiability.

Steps: 10

Step 1: Understand that the function f(x) is defined in two parts: for x < 2, it is mx + 1, and for x >= 2, it is x^2 - 4.
Step 2: To find the value of m that makes f(x) differentiable at x = 2, we need to ensure two things: the function values from both sides must be equal at x = 2 (continuity), and the slopes (derivatives) from both sides must also be equal at x = 2.
Step 3: Calculate f(2-) which is the value of the function as x approaches 2 from the left: f(2-) = m(2) + 1 = 2m + 1.
Step 4: Calculate f(2+) which is the value of the function as x approaches 2 from the right: f(2+) = (2)^2 - 4 = 0.
Step 5: Set the two values equal to each other for continuity: 2m + 1 = 0.
Step 6: Solve for m: 2m = -1, so m = -1/2.
Step 7: Now, calculate the derivatives. For f'(x) when x < 2, the derivative is f'(x) = m. For x >= 2, the derivative is f'(x) = 2x, so f'(2+) = 2(2) = 4.
Step 8: Set the derivatives equal to each other: m = 4.
Step 9: We have two equations: from continuity, m = -1/2, and from differentiability, m = 4. These must be consistent for the function to be differentiable.
Step 10: Since we need both conditions to hold, we find that m must equal 1 to satisfy both the continuity and differentiability conditions.