For which value of a is the function f(x) = { x^2, x < 1; ax + 1, x >= 1 } continuous at x = 1?
Practice Questions
1 question
Q1
For which value of a is the function f(x) = { x^2, x < 1; ax + 1, x >= 1 } continuous at x = 1?
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Setting the two pieces equal at x = 1 gives 1 = a(1) + 1, leading to a = 0.
Questions & Step-by-step Solutions
1 item
Q
Q: For which value of a is the function f(x) = { x^2, x < 1; ax + 1, x >= 1 } continuous at x = 1?
Solution: Setting the two pieces equal at x = 1 gives 1 = a(1) + 1, leading to a = 0.
Steps: 6
Step 1: Identify the two pieces of the function f(x). The first piece is x^2 for x < 1, and the second piece is ax + 1 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 when x = 1.
Step 3: Calculate the value of the first piece at x = 1. Since x < 1, we use the first piece: f(1) = 1^2 = 1.
Step 4: Calculate the value of the second piece at x = 1. Since x >= 1, we use the second piece: f(1) = a(1) + 1 = a + 1.
Step 5: Set the two values equal to each other: 1 = a + 1.
Step 6: Solve for a by subtracting 1 from both sides: 1 - 1 = a, which simplifies to a = 0.
