Differentiability

Question

Let $f(x) = |x - 1| + |x - 2| + \cos(\pi x)$. The number of points where $f(x)$ is non-differentiable in $[0, 3]$ is:

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The function contains modulus terms, which are generally non-differentiable at the points where the term inside becomes zero.

1. Points to check: $x = 1$ and $x = 2$.

2. At $x = 1$:
The term $|x - 1|$ has a 'corner' here (slope changes from -1 to 1). The derivative of the other terms, $|x - 2|$ and $\cos(\pi x)$, are well-defined and finite at $x = 1$. Specifically, $d/dx(\cos \pi x) = -\pi \sin(\pi) = 0$. The sum of a non-differentiable function and differentiable functions is non-differentiable. So, $f(x)$ is non-differentiable at $x = 1$.

3. At $x = 2$:
Similarly, $|x - 2|$ is non-differentiable at $x = 2$. The derivative of $|x - 1|$ is constant (1) nearby, and derivative of $\cos(\pi x)$ is $0$ at $x=2$. Thus, the sharp turn in $|x - 2|$ is not smoothed out. $f(x)$ is non-differentiable at $x = 2$.

Total points of non-differentiability: 2.

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