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This problem is related to the application of the Mean Value Theorem (MVT) in calculus. Let's go step by step. ### Part (i): Show that there exists \( c_1 \in (0,1) \) such that \( f'(c_1) = 1 \). The conditions given are: - \( f \) is continuous on the closed interval \([0,2]\) and differentiable on the open interval \( (0,2) \). - \( f(0) = 0 \), \( f(1) = 1 \), and \( f(2) = 1 \). According to the Mean Value Theorem (MVT), for any function \( f \) that is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a point \( c \in (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] Consider the interval _______ ___ _______ _____ ______.
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