Find the value of k such that the function f(x) = { kx + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.
Practice Questions
1 question
Q1
Find the value of k such that the function f(x) = { kx + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.
1
2
3
4
Setting k(2) + 1 = 2^2 - 3 gives 2k + 1 = 1, leading to k = 0.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the value of k such that the function f(x) = { kx + 1, x < 2; x^2 - 3, x >= 2 } is continuous at x = 2.
Solution: Setting k(2) + 1 = 2^2 - 3 gives 2k + 1 = 1, leading to k = 0.
Steps: 6
Step 1: Understand that we want the function f(x) to be continuous at x = 2. This means the value of f(x) when approaching from the left (x < 2) must equal the value of f(x) when approaching from the right (x >= 2).
Step 2: Identify the two parts of the function. For x < 2, the function is f(x) = kx + 1. For x >= 2, the function is f(x) = x^2 - 3.
Step 3: Calculate the value of f(x) as x approaches 2 from the left. Substitute x = 2 into the first part: f(2) = k(2) + 1 = 2k + 1.
Step 4: Calculate the value of f(x) as x approaches 2 from the right. Substitute x = 2 into the second part: f(2) = 2^2 - 3 = 4 - 3 = 1.
Step 5: Set the two results equal to each other to ensure continuity: 2k + 1 = 1.
Step 6: Solve for k. Subtract 1 from both sides: 2k = 1 - 1, which simplifies to 2k = 0. Then divide both sides by 2: k = 0.
