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

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

Solution: Setting 1 + b = 2 gives b = 1.

Steps: 7

Step 1: Understand that the function f(x) has two parts: one for x < 1 and another for x >= 1.
Step 2: Identify the two parts of the function: f(x) = x^2 + b when x < 1 and f(x) = 3x - 1 when x >= 1.
Step 3: To find the value of b that makes the function continuous at x = 1, we need to ensure that the two parts of the function equal each other at x = 1.
Step 4: Calculate f(1) using the second part of the function (since 1 is included in x >= 1): f(1) = 3(1) - 1 = 2.
Step 5: Now, calculate the limit of f(x) as x approaches 1 from the left (using the first part of the function): f(1) = 1^2 + b = 1 + b.
Step 6: Set the two results equal to each other to ensure continuity: 1 + b = 2.
Step 7: Solve for b: Subtract 1 from both sides to get b = 2 - 1, which simplifies to b = 1.