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

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

Solution: Setting limit as x approaches 1 gives b = 2 for continuity.

Steps: 10

Step 1: Understand that for a function to be continuous at a point, the limit of the function as it approaches that point from both sides must equal the value of the function at that point.
Step 2: Identify the function f(x) and the point of interest, which is x = 1.
Step 3: Calculate the limit of f(x) as x approaches 1 from the left (x < 1). This means using the part of the function x^2 - 1.
Step 4: Substitute x = 1 into the left-side function: f(1) = 1^2 - 1 = 0.
Step 5: Calculate the limit of f(x) as x approaches 1 from the right (x > 1). This means using the part of the function 3x - 2.
Step 6: Substitute x = 1 into the right-side function: f(1) = 3(1) - 2 = 1.
Step 7: For the function to be continuous at x = 1, the left limit (0) must equal the right limit (1) and also equal f(1) (which is b).
Step 8: Set the left limit equal to b: 0 = b. This means b must be 0 for continuity.
Step 9: However, we need to ensure that the right limit also matches. Set the right limit equal to b: 1 = b. This means b must be 1 for continuity.
Step 10: Since both limits must equal b, we find that b must equal 2 for the function to be continuous at x = 1.