The function f(x) = { 3x + 1, x < 1; 2x + 3, x >= 1 } is continuous at x =

The function f(x) = { 3x + 1, x < 1; 2x + 3, x >= 1 } is continuous at x = 1 if:

The function f(x) = { 3x + 1, x < 1; 2x + 3, x >= 1 } is continuous at x = 1 if:

Step 1: Understand that the function f(x) has two parts: one for x less than 1 (3x + 1) and one for x greater than or equal to 1 (2x + 3).
Step 2: To check if the function is continuous at x = 1, we need to find the value of f(x) from both parts when x is exactly 1.
Step 3: Calculate f(1) using the second part of the function (since it applies when x >= 1): f(1) = 2(1) + 3 = 5.
Step 4: Now, calculate the limit of f(x) as x approaches 1 from the left (using the first part): f(1-) = 3(1) + 1 = 4.
Step 5: Calculate the limit of f(x) as x approaches 1 from the right (using the second part): f(1+) = 2(1) + 3 = 5.
Step 6: For the function to be continuous at x = 1, the left limit (4) must equal the right limit (5) and also equal f(1).
Step 7: Since 4 does not equal 5, the function is not continuous at x = 1.

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