For which value of a is the function f(x) = { x^2 - a, x < 0; 2x + 1, x >=

₹0.0

Question: For which value of a is the function f(x) = { x^2 - a, x < 0; 2x + 1, x >= 0 } continuous at x = 0?

Options:

Correct Answer: -1

Solution:

Setting the two pieces equal at x = 0 gives -a = 1, so a = -1.

For which value of a is the function f(x) = { x^2 - a, x < 0; 2x + 1, x >= 0 } continuous at x = 0?

For which value of a is the function f(x) = { x^2 - a, x < 0; 2x + 1, x >= 0 } continuous at x = 0?

Correct Answer: -1

Step 1: Identify the two pieces of the function f(x). The function is defined as f(x) = x^2 - a for x < 0 and f(x) = 2x + 1 for x >= 0.
Step 2: Find the value of f(0) using the second piece of the function since 0 is included in x >= 0. Calculate f(0) = 2(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 is f(x) = x^2 - a. So, as x approaches 0, f(x) approaches 0^2 - a = -a.
Step 5: Set the limit from the left equal to f(0): -a = 1.
Step 6: Solve for a. Rearranging gives a = -1.

Piecewise Functions – Understanding how to evaluate and ensure continuity at a point for functions defined in pieces.
Continuity – The condition that the left-hand limit and right-hand limit at a point must equal the function's value at that point.