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

1 question

1 item

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

Solution: To ensure continuity at x = 1, k(1) + 1 = 2(1) - 3, solving gives k = 2.

Steps: 9

Step 1: Understand that the function f(x) has two parts: one for x < 1 and another for x >= 1.
Step 2: Identify the value of x where we want to check continuity, which is x = 1.
Step 3: For x < 1, the function is f(x) = kx + 1. We need to find the value of this function as x approaches 1 from the left.
Step 4: Substitute x = 1 into the first part of the function: f(1) = k(1) + 1 = k + 1.
Step 5: For x >= 1, the function is f(x) = 2x - 3. We need to find the value of this function at x = 1.
Step 6: Substitute x = 1 into the second part of the function: f(1) = 2(1) - 3 = 2 - 3 = -1.
Step 7: For the function to be continuous at x = 1, the two values we found must be equal: k + 1 = -1.
Step 8: Solve the equation k + 1 = -1 by subtracting 1 from both sides: k = -1 - 1 = -2.
Step 9: Therefore, the value of k that makes the function continuous at x = 1 is k = -2.