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

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

Solution: Setting x^2 - c = 2x + 1 at x = 1 gives c = 2.

Steps: 6

Step 1: Identify the function f(x) which has two parts: f(x) = x^2 - c for x < 1 and f(x) = 2x + 1 for x >= 1.
Step 2: To find the value of c 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 3: Calculate the value of f(x) when x = 1 using the second part of the function: f(1) = 2(1) + 1 = 3.
Step 4: Now, calculate the value of f(x) using the first part of the function when x is just less than 1. We set x = 1 in the first part: f(1) = 1^2 - c = 1 - c.
Step 5: Set the two results equal to each other to ensure continuity: 1 - c = 3.
Step 6: Solve for c: 1 - c = 3 leads to -c = 3 - 1, which simplifies to -c = 2, and thus c = -2.