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

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

Solution: Setting ax + 1 = 2 and x^2 + a = 2 at x = 1 gives a = 0.

Steps: 10

Step 1: Understand that we need to find the value of 'a' so that the function f(x) is continuous at x = 1.
Step 2: Recall that for a function to be continuous at a point, the left-hand limit and right-hand limit at that point must equal the function's value at that point.
Step 3: Identify the function's value at x = 1, which is f(1) = 2.
Step 4: For x < 1, the function is f(x) = ax + 1. We need to find the limit as x approaches 1 from the left: lim (x -> 1-) f(x) = a(1) + 1 = a + 1.
Step 5: For x > 1, the function is f(x) = x^2 + a. We need to find the limit as x approaches 1 from the right: lim (x -> 1+) f(x) = 1^2 + a = 1 + a.
Step 6: Set the left-hand limit equal to the function value: a + 1 = 2.
Step 7: Solve for 'a': a + 1 = 2 gives a = 2 - 1, so a = 1.
Step 8: Set the right-hand limit equal to the function value: 1 + a = 2.
Step 9: Solve for 'a' again: 1 + a = 2 gives a = 2 - 1, so a = 1.
Step 10: Since both limits give the same value for 'a', we conclude that a = 1 is the value that makes the function continuous at x = 1.