The function f(x) = { 3x + 1, x < 1; 2, x = 1; x^2, x > 1 } is continuous at x = 1 if which condition holds?
Practice Questions
1 question
Q1
The function f(x) = { 3x + 1, x < 1; 2, x = 1; x^2, x > 1 } is continuous at x = 1 if which condition holds?
3 = 2
1 = 2
2 = 1
2 = 4
For continuity at x = 1, the left limit (3) must equal f(1) (2), which is not true.
Questions & Step-by-step Solutions
1 item
Q
Q: The function f(x) = { 3x + 1, x < 1; 2, x = 1; x^2, x > 1 } is continuous at x = 1 if which condition holds?
Solution: For continuity at x = 1, the left limit (3) must equal f(1) (2), which is not true.
Steps: 5
Step 1: Identify the function f(x) and its pieces: f(x) = { 3x + 1 for x < 1; 2 for x = 1; x^2 for x > 1 }.
Step 2: Find the value of f(1). Since x = 1, we use the second piece of the function, which gives us f(1) = 2.
Step 3: Calculate the left limit as x approaches 1 from the left (x < 1). We use the first piece of the function: limit as x approaches 1 from the left of (3x + 1) = 3(1) + 1 = 4.
Step 4: Calculate the right limit as x approaches 1 from the right (x > 1). We use the third piece of the function: limit as x approaches 1 from the right of (x^2) = 1^2 = 1.
Step 5: Check if the left limit (4) equals f(1) (2). Since 4 does not equal 2, the function is not continuous at x = 1.
