Determine the value of p for which the function f(x) = { x^2 - 1, x < 1; 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^2 - 1, x < 1; p, x = 1; 2x + 1, x > 1 is continuous at x = 1.
0
1
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3
Setting the left limit (1 - 1 = 0) equal to the right limit (2(1) + 1 = 3), we find p = 2.
Questions & Step-by-step Solutions
1 item
Q
Q: Determine the value of p for which the function f(x) = { x^2 - 1, x < 1; p, x = 1; 2x + 1, x > 1 is continuous at x = 1.
Solution: Setting the left limit (1 - 1 = 0) equal to the right limit (2(1) + 1 = 3), we find p = 2.
Steps: 6
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 limit, right limit, and the function value at that point must all be equal.
Step 3: Calculate the left limit as x approaches 1 from the left (x < 1). This is given by the function x^2 - 1. So, we find the limit: lim (x -> 1-) f(x) = 1^2 - 1 = 0.
Step 4: Calculate the right limit as x approaches 1 from the right (x > 1). This is given by the function 2x + 1. So, we find the limit: lim (x -> 1+) f(x) = 2(1) + 1 = 3.
Step 5: Set the left limit equal to the right limit to find p: 0 (left limit) must equal p (function value at x = 1) and also equal 3 (right limit).
Step 6: Since the left limit (0) does not equal the right limit (3), we need to set p equal to the right limit to make the function continuous. Therefore, p = 3.
