What is the value of p for which the function f(x) = { 3x + p, x < 2; x^2 - 4, x >= 2 } is continuous at x = 2?

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Q: What is the value of p for which the function f(x) = { 3x + p, x < 2; x^2 - 4, x >= 2 } is continuous at x = 2?

Solution: Setting the two pieces equal at x = 2: 3(2) + p = 2^2 - 4. Solving gives p = -2.

Steps: 10

Step 1: Identify the two pieces of the function f(x). The first piece is 3x + p for x < 2, and the second piece is x^2 - 4 for x >= 2.
Step 2: To find the value of p that makes the function continuous at x = 2, we need to set the two pieces equal to each other at x = 2.
Step 3: Substitute x = 2 into the first piece: 3(2) + p.
Step 4: Substitute x = 2 into the second piece: 2^2 - 4.
Step 5: Set the two results equal: 3(2) + p = 2^2 - 4.
Step 6: Calculate the left side: 3(2) = 6, so we have 6 + p.
Step 7: Calculate the right side: 2^2 = 4, and 4 - 4 = 0, so we have 0.
Step 8: Now we have the equation: 6 + p = 0.
Step 9: Solve for p by subtracting 6 from both sides: p = 0 - 6.
Step 10: Therefore, p = -6.