For which value of p is the function f(x) = { x^2 + p, x < 0; 3x - 1, x >= 0 } continuous at x = 0?
Practice Questions
1 question
Q1
For which value of p is the function f(x) = { x^2 + p, x < 0; 3x - 1, x >= 0 } continuous at x = 0?
-1
0
1
2
Setting the two pieces equal at x = 0 gives us p = -1.
Questions & Step-by-step Solutions
1 item
Q
Q: For which value of p is the function f(x) = { x^2 + p, x < 0; 3x - 1, x >= 0 } continuous at x = 0?
Solution: Setting the two pieces equal at x = 0 gives us p = -1.
Steps: 6
Step 1: Identify the two pieces of the function f(x). The function is defined as f(x) = x^2 + p for x < 0 and f(x) = 3x - 1 for x >= 0.
Step 2: Find the value of f(0) using the second piece of the function since 0 is included in that piece. Calculate f(0) = 3(0) - 1 = -1.
Step 3: To ensure the function is continuous at x = 0, the limit of f(x) as x approaches 0 from the left (x < 0) must equal f(0).
Step 4: Calculate the limit of f(x) as x approaches 0 from the left. This means using the first piece: limit as x approaches 0 from the left of f(x) = x^2 + p = 0^2 + p = p.
Step 5: Set the limit from the left equal to f(0): p = -1.
Step 6: Conclude that for the function to be continuous at x = 0, p must equal -1.
