To determine the intervals on which the function
\[ f(x) = \frac{|2x-1| \sqrt{5-x}}{\cos\left(\frac{\pi x}{6}\right) \cdot x \left(x^2 - \frac{1}{9}\right)} \]
is differentiable, we need to analyze the components of the function for points where they may become non-differentiable. This can occur at points where the function is undefined or where it can change behavior, especially with absolute values.
Step 1: Analyze the function for potential issues
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Absolute Value: The term \(|2x - 1|\) is non-differentiable at the point where \(2x - 1 = 0\), which gives: \[ x = \frac{1}{2} \]
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Square Root: The term \(\sqrt{5 - x}\) is defined for \(5 - x \geq 0\), or: \[ x \leq 5 \]
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Cosine Function: The term \(\cos\left(\frac{\pi x}{6}\right)\) is zero whenever: \[ \frac{\pi x}{6} = \frac{\pi}{2} + k\pi \quad \text{for } k \in \mathbb{Z} \] Solving for \(x\) gives: \[ x = 3 + 6k \] for \(k \in \mathbb{Z}\). The first few values are \(x = 3, 9, 15, \ldots\)
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Zero in Denominator: The function also has terms that can make the denominator zero. The term \(x^2 - \frac{1}{9} = 0\) gives: \[ x = \frac{1}{3} \quad \text{and} \quad x = -\frac{1}{3} \] Thus, the function is undefined at \(x = -\frac{1}{3}\) and \(x = \frac{1}{3}\).
Step 2: Summary of Non-differentiable Points
Thus, the function \(f(x)\) has potential issues at:
- \(x = \frac{1}{2}\) (due to absolute value)
- \(x = 3, 9, \ldots\) (due to cosine function being zero)
- \(x = -\frac{1}{3}, \frac{1}{3}\) (denominator is zero)
Step 3: Determine the Intervals
Given these points, \(f(x)\) will be differentiable on intervals excluding these critical points. Therefore, we can express the intervals on which \(f(x)\) is differentiable as:
Intervals: \[ (-\infty, -\frac{1}{3}), , \left(-\frac{1}{3}, \frac{1}{3}\right), , \left(\frac{1}{3}, \frac{1}{2}\right), , \left(\frac{1}{2}, 3\right), , (3, 5) \]
Conclusion
The function \(f(x)\) is differentiable in the intervals:
- \((-∞, -\frac{1}{3})\)
- \((- \frac{1}{3}, \frac{1}{3})\)
- \((\frac{1}{3}, \frac{1}{2})\)
- \((\frac{1}{2}, 3)\)
- \((3, 5)\)
Remember to avoid points of non-differentiability when considering the differentiability of \(f(x)\).