Question

(a) Check the differentiability of \( f(x) = | \cos x | \) at \( x = \frac{\pi}{2} \).

Hint:

To check differentiability at a point, compute both the left-hand and right-hand derivatives. If these derivatives are not equal, the function is not differentiable at that point. For absolute value functions, pay attention to the behavior around critical points.

Answer 1

Step 1: Understand the behavior of \( f(x) \) near \( x = \frac{\pi}{2} \). The function \( f(x) = | \cos x | \) can be expressed as: \[ f(x) = \begin{cases} \cos x, & \text{if } \cos x \geq 0,
-\cos x, & \text{if } \cos x<0. \end{cases} \] At \( x = \frac{\pi}{2} \), \( \cos \left( \frac{\pi}{2} \right) = 0 \). To check differentiability, evaluate the left-hand derivative (LHD) and right-hand derivative (RHD). Step 2: Calculate the left-hand derivative (LHD). For \( x<\frac{\pi}{2} \), \( \cos x \geq 0 \), so: \[ f(x) = \cos x, \quad \text{and} \quad f'(x) = -\sin x. \] At \( x = \frac{\pi}{2} \), the left-hand derivative is: \[ \lim_{h \to 0^-} \frac{f\left( \frac{\pi}{2} + h \right) - f\left( \frac{\pi}{2} \right)}{h} = -\sin\left( \frac{\pi}{2} \right) = -1. \] Step 3: Calculate the right-hand derivative (RHD). For \( x>\frac{\pi}{2} \), \( \cos x<0 \), so: \[ f(x) = -\cos x, \quad \text{and} \quad f'(x) = \sin x. \] At \( x = \frac{\pi}{2} \), the right-hand derivative is: \[ \lim_{h \to 0^+} \frac{f\left( \frac{\pi}{2} + h \right) - f\left( \frac{\pi}{2} \right)}{h} = \sin\left( \frac{\pi}{2} \right) = 1. \] Step 4: Conclusion. Since the left-hand and right-hand derivatives are not equal (\(-1 \neq 1\)), the function \( f(x) = |\cos x| \) is not differentiable at \( x = \frac{\pi}{2} \). Final Answer: \[ \boxed{\text{The function } f(x) = |\cos x| \text{ is not differentiable at } x = \frac{\pi}{2}.} \]