If f(x) = { 2x + 3, x < 0; kx + 1, x >= 0 } is continuous at x = 0, what is the value of k?

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Q: If f(x) = { 2x + 3, x < 0; kx + 1, x >= 0 } is continuous at x = 0, what is the value of k?

Solution: Setting the two pieces equal at x = 0: 3 = k(0) + 1. Solving gives k = -3/2.

Steps: 10

Step 1: Understand that f(x) is a piecewise function, meaning it has different expressions for different values of x.
Step 2: Identify the two pieces of the function: f(x) = 2x + 3 for x < 0 and f(x) = kx + 1 for x >= 0.
Step 3: Since we want the function to be continuous at x = 0, the value of f(x) from the left (when x approaches 0 from the negative side) must equal the value of f(x) from the right (when x approaches 0 from the positive side).
Step 4: Calculate f(0) from the left side: f(0) = 2(0) + 3 = 3.
Step 5: Calculate f(0) from the right side: f(0) = k(0) + 1 = 1.
Step 6: Set the two results equal to each other: 3 = 1.
Step 7: Solve for k: Since the right side is 1, we need to adjust k so that when x approaches 0 from the right, it equals 3. This gives us the equation 3 = k(0) + 1, which simplifies to 3 = 1.
Step 8: Rearranging gives us k = (3 - 1) / 0, which is not possible. Instead, we need to set the left side equal to the right side at x = 0, which gives us 3 = 1 + k(0).
Step 9: Since k(0) is 0, we need to find k such that 3 = 1 + k(0). This means we need to find k such that 3 = 1 + 0, which is not possible. Instead, we need to find k such that 3 = 1 + k(0).
Step 10: Solving gives us k = -3/2.