Find the value of a for which the function f(x) = { ax + 1, x < 2; 3x - 5, x >= 2 } is continuous at x = 2.
Practice Questions
1 question
Q1
Find the value of a for which the function f(x) = { ax + 1, x < 2; 3x - 5, x >= 2 } is continuous at x = 2.
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Setting the two pieces equal at x = 2 gives us 2a + 1 = 1. Solving for a gives a = 0.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the value of a for which the function f(x) = { ax + 1, x < 2; 3x - 5, x >= 2 } is continuous at x = 2.
Solution: Setting the two pieces equal at x = 2 gives us 2a + 1 = 1. Solving for a gives a = 0.
Steps: 7
Step 1: Understand that we have a piecewise function f(x) which has two parts: one for x < 2 and another for x >= 2.
Step 2: Identify the two parts of the function: f(x) = ax + 1 when x < 2 and f(x) = 3x - 5 when x >= 2.
Step 3: To find the value of a that makes the function continuous at x = 2, we need to make sure both parts of the function give the same output when x = 2.
Step 4: Calculate the output of the second part of the function at x = 2: f(2) = 3(2) - 5 = 6 - 5 = 1.
Step 5: Set the output of the first part of the function equal to 1 when x approaches 2 from the left: ax + 1 = 1. Since we are looking at x = 2, we substitute x with 2: 2a + 1 = 1.
Step 6: Solve the equation 2a + 1 = 1 for a. First, subtract 1 from both sides: 2a = 0. Then, divide both sides by 2: a = 0.
Step 7: Conclude that the value of a that makes the function continuous at x = 2 is a = 0.
