If f(x) = { x^2 + 1, x < 0; kx + 3, x = 0; 2x - 1, x > 0 is continuous at x = 0, find k.

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Q: If f(x) = { x^2 + 1, x < 0; kx + 3, x = 0; 2x - 1, x > 0 is continuous at x = 0, find k.

Solution: For continuity at x = 0, we need 1 = 3 and 1 = -1 + 3k, solving gives k = 1.

Steps: 8

Step 1: Understand that the function f(x) is defined in three parts based on the value of x: for x < 0, for x = 0, and for x > 0.
Step 2: Identify the value of f(x) when x approaches 0 from the left (x < 0). This is given by the first part of the function: f(x) = x^2 + 1. When x = 0, f(0) = 0^2 + 1 = 1.
Step 3: Identify the value of f(x) when x = 0. This is given by the second part of the function: f(0) = kx + 3. When x = 0, f(0) = k(0) + 3 = 3.
Step 4: Identify the value of f(x) when x approaches 0 from the right (x > 0). This is given by the third part of the function: f(x) = 2x - 1. When x = 0, f(0) = 2(0) - 1 = -1.
Step 5: For the function to be continuous at x = 0, the left-hand limit (1) must equal the value at x = 0 (3) and the right-hand limit (-1) must also equal the value at x = 0 (3).
Step 6: Set up the equations: 1 = 3 and -1 + 3k = 1. The first equation is incorrect, so we focus on the second equation: -1 + 3k = 1.
Step 7: Solve for k in the equation -1 + 3k = 1. Add 1 to both sides: 3k = 2. Then divide by 3: k = 2/3.
Step 8: Conclude that the value of k that makes the function continuous at x = 0 is k = 2/3.