Question

which of the following intervals is the function f(x)=|2x-1|xsqrt5-x/cos(pix/6)x(x^2-1/9) differentiable

Answer 1

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

1. **Absolute Value**: The term \(|2x - 1|\) is non-differentiable at the point where \(2x - 1 = 0\), which gives:
\[
x = \frac{1}{2}
\]

2. **Square Root**: The term \(\sqrt{5 - x}\) is defined for \(5 - x \geq 0\), or:
\[
x \leq 5
\]

3. **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 \(x = 3, 9, 15, \ldots\)

4. **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 \(f(x)\).