What is the value of q for which the function f(x) = { 5 - q, x < 1; 3x + 2,

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Question: What is the value of q for which the function f(x) = { 5 - q, x < 1; 3x + 2, x >= 1 } is continuous at x = 1?

Options:

Correct Answer: 2

Solution:

Setting the two pieces equal at x = 1: 5 - q = 3(1) + 2. Solving gives q = 0.

What is the value of q for which the function f(x) = { 5 - q, x < 1; 3x + 2, x >= 1 } is continuous at x = 1?

What is the value of q for which the function f(x) = { 5 - q, x < 1; 3x + 2, x >= 1 } is continuous at x = 1?

Step 1: Identify the two pieces of the function f(x). The first piece is 5 - q for x < 1, and the second piece is 3x + 2 for x >= 1.
Step 2: To find the value of q that makes the function continuous at x = 1, we need to set the two pieces equal to each other at x = 1.
Step 3: Substitute x = 1 into the second piece of the function: 3(1) + 2 = 3 + 2 = 5.
Step 4: Now set the first piece equal to this value: 5 - q = 5.
Step 5: Solve for q by subtracting 5 from both sides: -q = 5 - 5, which simplifies to -q = 0.
Step 6: Multiply both sides by -1 to find q: q = 0.

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