The function f(x) = { x^2, x < 0; 2x + 1, x >= 0 } is continuous at which point?
Practice Questions
1 question
Q1
The function f(x) = { x^2, x < 0; 2x + 1, x >= 0 } is continuous at which point?
x = -1
x = 0
x = 1
x = 2
To check continuity at x = 0, we find f(0) = 1 and limit as x approaches 0 is also 1.
Questions & Step-by-step Solutions
1 item
Q
Q: The function f(x) = { x^2, x < 0; 2x + 1, x >= 0 } is continuous at which point?
Solution: To check continuity at x = 0, we find f(0) = 1 and limit as x approaches 0 is also 1.
Steps: 8
Step 1: Identify the function f(x) which is defined in two parts: f(x) = x^2 for x < 0 and f(x) = 2x + 1 for x >= 0.
Step 2: To check if the function is continuous at x = 0, we need to find f(0).
Step 3: Calculate f(0) using the second part of the function since 0 is greater than or equal to 0: f(0) = 2(0) + 1 = 1.
Step 4: Next, we need to find the limit of f(x) as x approaches 0 from both sides.
Step 5: Calculate the left-hand limit (as x approaches 0 from the left): limit as x approaches 0- of f(x) = limit as x approaches 0- of x^2 = 0.
Step 6: Calculate the right-hand limit (as x approaches 0 from the right): limit as x approaches 0+ of f(x) = limit as x approaches 0+ of (2x + 1) = 1.
Step 7: Compare the left-hand limit, right-hand limit, and f(0): left-hand limit = 0, right-hand limit = 1, and f(0) = 1.
Step 8: Since the left-hand limit does not equal the right-hand limit, the function is not continuous at x = 0.
