What value of m makes the function f(x) = { 3x + 1, x < 2; mx + 4, x = 2; x^2 - 1, x > 2 continuous at x = 2?

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Q: What value of m makes the function f(x) = { 3x + 1, x < 2; mx + 4, x = 2; x^2 - 1, x > 2 continuous at x = 2?

Solution: Setting the left limit (3(2) + 1 = 7) equal to the right limit (2^2 - 1 = 3), we find m = 3.

Steps: 10

Step 1: Identify the function f(x) and the point where we want it to be continuous, which is x = 2.
Step 2: Determine the left limit as x approaches 2 from the left (x < 2). This is given by the function 3x + 1.
Step 3: Calculate the left limit by substituting x = 2 into 3x + 1: 3(2) + 1 = 6 + 1 = 7.
Step 4: Determine the right limit as x approaches 2 from the right (x > 2). This is given by the function x^2 - 1.
Step 5: Calculate the right limit by substituting x = 2 into x^2 - 1: 2^2 - 1 = 4 - 1 = 3.
Step 6: For the function to be continuous at x = 2, the left limit must equal the right limit. Set 7 (left limit) equal to 3 (right limit).
Step 7: Since the right limit is defined by the function mx + 4 at x = 2, we need to set mx + 4 equal to the left limit (7).
Step 8: Substitute x = 2 into mx + 4: 2m + 4 = 7.
Step 9: Solve for m by isolating it: 2m = 7 - 4, which simplifies to 2m = 3.
Step 10: Divide both sides by 2 to find m: m = 3/2.