Determine the value of p for which the function f(x) = { x^3 - 3x + p, x < 1; 2x + 1, x >= 1 is continuous at x = 1.
Practice Questions
1 question
Q1
Determine the value of p for which the function f(x) = { x^3 - 3x + p, x < 1; 2x + 1, x >= 1 is continuous at x = 1.
-1
0
1
2
Setting -3 + p = 3 gives p = 0.
Questions & Step-by-step Solutions
1 item
Q
Q: Determine the value of p for which the function f(x) = { x^3 - 3x + p, x < 1; 2x + 1, x >= 1 is continuous at x = 1.
Solution: Setting -3 + p = 3 gives p = 0.
Steps: 8
Step 1: Understand that we need to find the value of p so that the function f(x) is continuous at x = 1.
Step 2: Recall that for a function to be continuous at a point, the left-hand limit and the right-hand limit at that point must be equal to the function's value at that point.
Step 3: Identify the two parts of the function: f(x) = x^3 - 3x + p for x < 1 and f(x) = 2x + 1 for x >= 1.
Step 4: Calculate the left-hand limit as x approaches 1 from the left (x < 1): f(1) = 1^3 - 3(1) + p = 1 - 3 + p = p - 2.
Step 5: Calculate the right-hand limit as x approaches 1 from the right (x >= 1): f(1) = 2(1) + 1 = 2 + 1 = 3.
Step 6: Set the left-hand limit equal to the right-hand limit: p - 2 = 3.
Step 7: Solve for p: p - 2 = 3 means p = 3 + 2, which gives p = 5.
Step 8: Verify that the function is continuous by checking if both limits are equal at x = 1.
