Find the value of a for which the function f(x) = { ax + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.

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Q: Find the value of a for which the function f(x) = { ax + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.

Solution: Setting the two pieces equal at x = 2: 2a + 1 = 2^2 - 3. Solving gives a = 2.

Steps: 8

Step 1: Identify the two pieces of the function f(x). The first piece is ax + 1 for x < 2, and the second piece is x^2 - 3 for x >= 2.
Step 2: To find the value of a 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: f(2) = a(2) + 1 = 2a + 1.
Step 4: Substitute x = 2 into the second piece: f(2) = 2^2 - 3 = 4 - 3 = 1.
Step 5: Set the two expressions equal to each other: 2a + 1 = 1.
Step 6: Solve for a by subtracting 1 from both sides: 2a = 1 - 1 = 0.
Step 7: Divide both sides by 2: a = 0 / 2 = 0.
Step 8: Check the solution by substituting a back into the function to ensure continuity at x = 2.