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For which value of b is the function f(x) = { x^2 - 4, x < 2; bx + 2, x >=
For which value of b is the function f(x) = { x^2 - 4, x < 2; bx + 2, x >= 2 } continuous at x = 2?
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Q1
For which value of b is the function f(x) = { x^2 - 4, x < 2; bx + 2, x >= 2 } continuous at x = 2?
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Setting the two pieces equal at x = 2 gives us 0 = 2b + 2. Solving for b gives b = -1.
Questions & Step-by-step Solutions
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Q
Q: For which value of b is the function f(x) = { x^2 - 4, x < 2; bx + 2, x >= 2 } continuous at x = 2?
Solution:
Setting the two pieces equal at x = 2 gives us 0 = 2b + 2. Solving for b gives b = -1.
Steps: 5
Show Steps
Step 1: Identify the two pieces of the function f(x). The first piece is x^2 - 4 for x < 2, and the second piece is bx + 2 for x >= 2.
Step 2: Find the value of the first piece at x = 2. Substitute x = 2 into the first piece: f(2) = 2^2 - 4 = 4 - 4 = 0.
Step 3: Find the value of the second piece at x = 2. Substitute x = 2 into the second piece: f(2) = b(2) + 2 = 2b + 2.
Step 4: Set the two pieces equal to each other at x = 2 to ensure continuity: 0 = 2b + 2.
Step 5: Solve the equation 0 = 2b + 2 for b. Subtract 2 from both sides: -2 = 2b. Then divide both sides by 2: b = -1.
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