Determine the value of n for which the function f(x) = { n^2 - 1, x < 0; 2x + 3, x >= 0 } is continuous at x = 0.

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Q: Determine the value of n for which the function f(x) = { n^2 - 1, x < 0; 2x + 3, x >= 0 } is continuous at x = 0.

Solution: Setting the two pieces equal at x = 0: n^2 - 1 = 3. Solving gives n = ±2.

Steps: 8

Step 1: Understand that the function f(x) has two parts: one for x < 0 and one for x >= 0.
Step 2: Identify the part of the function for x < 0, which is f(x) = n^2 - 1.
Step 3: Identify the part of the function for x >= 0, which is f(x) = 2x + 3.
Step 4: Find the value of f(0) using the part of the function for x >= 0. Substitute x = 0 into 2x + 3: f(0) = 2(0) + 3 = 3.
Step 5: For the function to be continuous at x = 0, the value from the left side (n^2 - 1) must equal the value from the right side (3).
Step 6: Set up the equation: n^2 - 1 = 3.
Step 7: Solve the equation: Add 1 to both sides to get n^2 = 4.
Step 8: Take the square root of both sides to find n: n = ±2.