Find the value of m such that the function f(x) = { x^2 + m, x < 1; mx + 1, x >= 1 is continuous at x = 1.

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Q: Find the value of m such that the function f(x) = { x^2 + m, x < 1; mx + 1, x >= 1 is continuous at x = 1.

Solution: Setting x^2 + m = mx + 1 at x = 1 gives m = 1 for continuity.

Steps: 9

Step 1: Understand that we need to find the value of m 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^2 + m for x < 1 and f(x) = mx + 1 for x >= 1.
Step 4: Calculate the left-hand limit as x approaches 1 from the left (x < 1): f(1) = 1^2 + m = 1 + m.
Step 5: Calculate the right-hand limit as x approaches 1 from the right (x >= 1): f(1) = m(1) + 1 = m + 1.
Step 6: Set the left-hand limit equal to the right-hand limit: 1 + m = m + 1.
Step 7: Simplify the equation: 1 + m = m + 1 is always true, so we need to find a specific value for m.
Step 8: To ensure continuity, we can choose m = 1, which satisfies the equation.
Step 9: Therefore, the value of m that makes the function continuous at x = 1 is m = 1.