Find the value of k such that the function f(x) = { kx + 2, x < 1; 3, x = 1; 2x + 1, x > 1 } is continuous at x = 1.

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Q: Find the value of k such that the function f(x) = { kx + 2, x < 1; 3, x = 1; 2x + 1, x > 1 } is continuous at x = 1.

Solution: Setting k(1) + 2 = 3 gives k = 1.

Steps: 7

Step 1: Understand that the function f(x) has different expressions depending on the value of x. We need to check the value of f(x) at x = 1.
Step 2: Identify the value of f(x) when x = 1. According to the function, when x = 1, f(x) = 3.
Step 3: For the function to be continuous at x = 1, the limit of f(x) as x approaches 1 from the left (x < 1) must equal f(1).
Step 4: Write the expression for f(x) when x < 1, which is kx + 2. We need to find the limit as x approaches 1 from the left: f(1) = k(1) + 2.
Step 5: Set the limit equal to the value of f(1): k(1) + 2 = 3.
Step 6: Solve the equation k + 2 = 3 by subtracting 2 from both sides: k = 3 - 2.
Step 7: Simplify the equation to find k: k = 1.