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

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

Solution: Setting the left limit (2(1) + 1 = 3) equal to the right limit (1^2 + 1 = 2), we find b = 3.

Steps: 8

Step 1: Understand that we want the function f(x) to be continuous at x = 1.
Step 2: Recall that for a function to be continuous at a point, the left limit, right limit, and the function value at that point must all be equal.
Step 3: Identify the left limit as x approaches 1 from the left (x < 1). This is given by the function 2x + 1.
Step 4: Calculate the left limit: 2(1) + 1 = 2 + 1 = 3.
Step 5: Identify the right limit as x approaches 1 from the right (x > 1). This is given by the function x^2 + 1.
Step 6: Calculate the right limit: (1)^2 + 1 = 1 + 1 = 2.
Step 7: Set the left limit equal to the right limit to find the value of b: 3 = b.
Step 8: Conclude that for the function to be continuous at x = 1, b must equal 3.