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in Continuity and Differentiability by (48.6k points)
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Prove that \(f(x) = \begin{cases} x - 1, & \quad \text{when } 1 \text{ ≤ x < 2}\\ 2x - 3, & \quad \text{when } 2 \text{ ≤ x ≤ 3} \end{cases} \) is continuous at x = 2.

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by (49.9k points)
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Best answer

It is given that

\(f(x) = \begin{cases} x - 1, & \quad \text{when } 1 \text{ ≤ x < 2}\\ 2x - 3, & \quad \text{when } 2 \text{ ≤ x ≤ 3} \end{cases} \)

Consider left hand limit at x = 2

Therefore, f(x) is continuous at x = 2.

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