For which value of a is the function f(x) = { x^2 - a, x < 1; 3x - 2, x >= 1 } continuous at x = 1?
Practice Questions
1 question
Q1
For which value of a is the function f(x) = { x^2 - a, x < 1; 3x - 2, x >= 1 } continuous at x = 1?
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Setting the two pieces equal at x = 1 gives 1 - a = 1. Thus, a = 0.
Questions & Step-by-step Solutions
1 item
Q
Q: For which value of a is the function f(x) = { x^2 - a, x < 1; 3x - 2, x >= 1 } continuous at x = 1?
Solution: Setting the two pieces equal at x = 1 gives 1 - a = 1. Thus, a = 0.
Steps: 6
Step 1: Identify the two pieces of the function f(x). The function is defined as f(x) = x^2 - a for x < 1 and f(x) = 3x - 2 for x >= 1.
Step 2: To find the value of a that makes the function continuous at x = 1, we need to ensure that the two pieces of the function are equal at x = 1.
Step 3: Calculate the value of f(x) when x = 1 for both pieces. For the first piece (x < 1), we use f(1) = 1^2 - a = 1 - a. For the second piece (x >= 1), we use f(1) = 3(1) - 2 = 3 - 2 = 1.
Step 4: Set the two expressions equal to each other: 1 - a = 1.
Step 5: Solve for a. Subtract 1 from both sides: -a = 0. Then multiply both sides by -1: a = 0.
Step 6: Conclude that the function f(x) is continuous at x = 1 when a = 0.
