Find the value of k such that the function f(x) = { kx + 1, x < 1; 2x - 1, x

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

Options:

Correct Answer: 1

Solution:

Setting k(1) + 1 = 2(1) - 1 gives k + 1 = 1, so k = 0.

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

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

Correct Answer: 0

Step 1: Understand that we want the function f(x) to be continuous at x = 1. This means the value of f(x) when approaching from the left (x < 1) should equal the value when approaching from the right (x >= 1).
Step 2: Write down the two parts of the function: f(x) = kx + 1 for x < 1 and f(x) = 2x - 1 for x >= 1.
Step 3: Find the value of f(x) as x approaches 1 from the left. This means we use the first part of the function: f(1) = k(1) + 1 = k + 1.
Step 4: Find the value of f(x) as x approaches 1 from the right. This means we use the second part of the function: f(1) = 2(1) - 1 = 2 - 1 = 1.
Step 5: Set the two values equal to each other to ensure continuity: k + 1 = 1.
Step 6: Solve for k by subtracting 1 from both sides: k = 1 - 1 = 0.

Continuity of Piecewise Functions – The question tests the understanding of how to ensure continuity at a point for piecewise functions by equating the limits from both sides.