Which of the following intervals is the function f(x) = |2x-1| x sqrt5-x/cos(pi x / 6) x (x^2 - 1/9) differentiable

Answer 1

To determine the intervals where the function

\[ f(x) = |2x - 1| \cdot \sqrt{5 - x} \cdot \frac{1}{\cos\left(\frac{\pi x}{6}\right)} \cdot (x^2 - \frac{1}{9}) \]

is differentiable, we need to analyze the differentiability of each part of the function separately.

1. Absolute Value Function \( |2x - 1| \):

   • The absolute value function is not differentiable at the point where its argument is zero: \(2x - 1 = 0\) implies \(x = \frac{1}{2}\).
   • Thus, \(f(x)\) is not differentiable at \(x = \frac{1}{2}\).

2. Square Root Function \( \sqrt{5 - x} \):

   • The square root function is differentiable as long as its argument is positive. Therefore, we require: \[ 5 - x > 0 \implies x < 5 \]
   • The square root is not defined at \(x = 5\), so \(f(x)\) is not differentiable at \(x = 5\).

3. Cosine Function \(\frac{1}{\cos\left(\frac{\pi x}{6}\right)}\):

   • The cosine function is equal to zero at \( \frac{\pi x}{6} = \frac{\pi}{2} + k\pi \) for integers \(k\). This gives us points where \(x\) could cause issues: \[ x = 3 + 6k \quad (k \in \mathbb{Z}) \]
   • The first few points are \(x = 3\) (for \(k=0\)) and \(x = 9\) (for \(k=1\)). So \(f(x)\) is not differentiable at \(x = 3\).

4. Polynomial Function \(x^2 - \frac{1}{9}\):

   • This function is differentiable everywhere, so it does not introduce any additional restrictions.

In summary, the function \(f(x)\) is not differentiable at the following critical points:

• \(x = \frac{1}{2}\) from the absolute value component,
• \(x = 5\) from the square root component,
• \(x = 3\) from the cosine component.

Conclusion:

The function \(f(x)\) is differentiable on the intervals:

• \( (-\infty, \frac{1}{2}) \)
• \( \left(\frac{1}{2}, 3\right) \)
• \( (3, 5) \)
• \( (5, 9) \)
• \( (9, \infty) \)

These intervals exclude the points where the function is not differentiable.