For the function f(x) = { x^2 + 1, x < 0; 2x + b, x = 0; 3 - x, x > 0 to be continuous at x = 0, what is b?
Practice Questions
1 question
Q1
For the function f(x) = { x^2 + 1, x < 0; 2x + b, x = 0; 3 - x, x > 0 to be continuous at x = 0, what is b?
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Setting the left limit (0 + 1 = 1) equal to the right limit (3 - 0 = 3), we find b = 1.
Questions & Step-by-step Solutions
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Q
Q: For the function f(x) = { x^2 + 1, x < 0; 2x + b, x = 0; 3 - x, x > 0 to be continuous at x = 0, what is b?
Solution: Setting the left limit (0 + 1 = 1) equal to the right limit (3 - 0 = 3), we find b = 1.
Steps: 6
Step 1: Understand that the function f(x) has different expressions depending on the value of x: x^2 + 1 for x < 0, 2x + b for x = 0, and 3 - x for x > 0.
Step 2: To find the value of b that makes the function continuous at x = 0, we need to ensure that the left limit (as x approaches 0 from the left) equals the right limit (as x approaches 0 from the right) and also equals the value of the function at x = 0.
Step 3: Calculate the left limit as x approaches 0 from the left (x < 0). This is f(x) = x^2 + 1. When x is very close to 0 (but negative), f(0) = 0^2 + 1 = 1.
Step 4: Calculate the right limit as x approaches 0 from the right (x > 0). This is f(x) = 3 - x. When x is very close to 0 (but positive), f(0) = 3 - 0 = 3.
Step 5: Set the left limit equal to the right limit: 1 (from the left) must equal 3 (from the right). This means we need to adjust the value of b in the middle expression (2x + b) so that it matches the left limit.
Step 6: Since we want the function to be continuous at x = 0, we set the value of the function at x = 0 (which is 2(0) + b = b) equal to the left limit (1). So, we have b = 1.
