Question

The function f(x) = | cos(pi x) | is differentiable everywhere on the interval (0, 5/4) except at which value of x?

Answer 1

To determine where the function \( f(x) = |\cos(\pi x)| \) is differentiable on the interval \( (0, 5/4) \), we first need to identify points where \( \cos(\pi x) \) is zero, as these will be points where \( f(x) \) may not be differentiable due to the absolute value.

The expression \( \cos(\pi x) = 0 \) occurs at:
\[
\pi x = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z}
\]
This simplifies to:
\[
x = \frac{1}{2} + n
\]

Within the interval \( (0, 5/4) \), let's identify the values of \( n \):
- For \( n = 0 \): \( x = \frac{1}{2} \)
- For \( n = 1 \): \( x = \frac{3}{2} \) (this value is outside the interval \( (0, 5/4) \))

Thus, the only value we need to consider in the interval \( (0, 5/4) \) is \( x = \frac{1}{2} \).

At \( x = \frac{1}{2} \):
- \( \cos(\pi/2) = 0 \), which means \( f(x) \) changes behavior here because \( f\left(\frac{1}{2}\right) = |\cos(\pi/2)| = 0 \).

To check differentiability, we must observe the nature of \( f(x) \) around \( x = \frac{1}{2} \):
- For \( x < \frac{1}{2} \), \( \cos(\pi x) > 0 \), so \( f(x) = \cos(\pi x) \).
- For \( x > \frac{1}{2} \), \( \cos(\pi x) < 0 \), so \( f(x) = -\cos(\pi x) \).

Let's compute the derivatives from the left and right:
- From the left: \( f'(x) = -\pi\sin(\pi x) \) as \( x \to \frac{1}{2}^- \)
- From the right: \( f'(x) = \pi\sin(\pi x) \) as \( x \to \frac{1}{2}^+ \)

Evaluating the derivatives at \( x = \frac{1}{2} \):
- \( f'(\frac{1}{2}^-) = -\pi\sin(\pi \cdot \frac{1}{2}) = -\pi(1) = -\pi \)
- \( f'(\frac{1}{2}^+) = \pi\sin(\pi \cdot \frac{1}{2}) = \pi(1) = \pi \)

Since the left-hand and right-hand derivatives are not equal:
\[
-\pi \neq \pi
\]
\( f(x) \) is not differentiable at \( x = \frac{1}{2} \).

Thus, the function \( f(x) = |\cos(\pi x)| \) is differentiable everywhere on the interval \( (0, 5/4) \) except at:
\[
\boxed{\frac{1}{2}}
\]