Step-by-step Solution:

The function given is: \[ f(x) = |\cos x| + 3 \] The function consists of two parts: \[\] 1. \(\cos x\), which is a smooth and differentiable function. \[\] 2. The absolute value function \(|\cos x|\), which introduces potential points of non-differentiability. The key observation is that the absolute value function introduces non-differentiability at the points where the argument inside the absolute value changes sign. In our case, the points where \(\cos x = 0\) will be the points where \(|\cos x|\) might cause non-differentiability. \[\] We need to solve for \(x\) in the interval \([-\pi, \pi]\) such that \(\cos x = 0\). This happens at: \[ x = \pm \frac{\pi}{2} \] \[\] At \(x = \pm \frac{\pi}{2}\), \(\cos x = 0\), so \(|\cos x|\) \[\] changes from positive to negative or vice versa. The function \(f(x) = |\cos x| + 3\) will have a "corner" at these points, which means the function is not differentiable at \(x = \pm \frac{\pi}{2}\). \[\] The function \(f(x) = |\cos x| + 3\) is not differentiable at two points: \(x = -\frac{\pi}{2}\) and \(x = \frac{\pi}{2}\). Thus, the number of points where \(f(x)\) is not differentiable in the interval \([-\pi, \pi]\) is: 2