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Discuss the continuity of 'f' in [0, 2],...

Discuss the continuity of 'f' in [0, 2], where `f(x) = {{:(|4x - 5|[x],"for",x gt 1),([cos pi x],"for",x le 1):}`, where [x] is greastest integer not greater than x.

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To discuss the continuity of the function \( f \) defined as: \[ f(x) = \begin{cases} |4x - 5| \cdot [x] & \text{for } x > 1 \\ \cos(\pi x) & \text{for } x \leq 1 \end{cases} \] on the interval \([0, 2]\), we need to check the continuity at all points in this interval, particularly at the point where the definition of \( f(x) \) changes, which is at \( x = 1 \). ### Step 1: Check continuity for \( x < 1 \) For \( x \leq 1 \), the function is defined as \( f(x) = \cos(\pi x) \). 1. **Evaluate \( f(0) \)**: \[ f(0) = \cos(0) = 1 \] 2. **Evaluate \( f(1) \)**: \[ f(1) = \cos(\pi \cdot 1) = \cos(\pi) = -1 \] ### Step 2: Check continuity for \( x > 1 \) For \( x > 1 \), the function is defined as \( f(x) = |4x - 5| \cdot [x] \). 1. **Evaluate \( f(2) \)**: \[ f(2) = |4 \cdot 2 - 5| \cdot [2] = |8 - 5| \cdot 2 = 3 \cdot 2 = 6 \] ### Step 3: Check continuity at \( x = 1 \) To check continuity at \( x = 1 \), we need to find the left-hand limit, the right-hand limit, and the value of the function at that point. 1. **Left-hand limit as \( x \) approaches 1**: \[ \lim_{x \to 1^-} f(x) = f(1) = -1 \] 2. **Right-hand limit as \( x \) approaches 1**: \[ \lim_{x \to 1^+} f(x) = |4 \cdot 1 - 5| \cdot [1] = |4 - 5| \cdot 1 = 1 \cdot 1 = 1 \] 3. **Value of the function at \( x = 1 \)**: \[ f(1) = -1 \] ### Step 4: Conclusion about continuity For \( f \) to be continuous at \( x = 1 \), the following must hold: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \] However, we found: - \( \lim_{x \to 1^-} f(x) = -1 \) - \( \lim_{x \to 1^+} f(x) = 1 \) - \( f(1) = -1 \) Since the left-hand limit and right-hand limit are not equal, \( f(x) \) is not continuous at \( x = 1 \). Thus, the function \( f(x) \) is continuous on the intervals \([0, 1)\) and \((1, 2]\) but not continuous at \( x = 1 \). ### Final Answer The function \( f \) is continuous on the interval \([0, 2]\) except at \( x = 1 \). ---
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