<s>[INST] If $f(x)=\begin{cases} \cos (x^3 )& x<0 \\ \sin (x^3 )- |x^3-1| & x\ge 0 \end {cases}$, find the number of points where $f(|x|)$ is non differentiable Since we are dealing with $f(|x|)$, the $\cos (x^3)$ part can be ignored. (And $\cos$ is differentiable everywhere anyway)
For $\sin |x|^3$, the graph would break its smooth flow at $x=0$, so that is a potential point
$||x^3|-1|$ would be non differentiable at $x=\pm 1$, and that gives two more potential points
So there are a total of three points.
The given answer is 2 though (points are unknown)
Have I picked the right points ?
 [/INST] $$f(x)=\begin{cases}\cos(x^3 )& x<0 \\ \sin (x^3 )- |x^3-1| & x\geq 0 \end {cases}=\begin{cases}\cos(x^3 )& x<0 \\ \sin (x^3 )- 1+x^3 & 0\leq x<1\\ \sin(x^3)-x^3+1&x\geq1\end {cases}$$
$$f(|x|)=\begin{cases}\sin(x^3)-x^3+1&x\geq1\\\sin(x^3)+x^3-1&0\leq x<1\\-\sin(x^3)-x^3-1&-1\leq x<0\\-\sin(x^3)+x^3+1&x<-1\end {cases}$$
There are three points that lie on the seam between two pieces of a piecewise smooth function, and as such, we need to check if the function is continuous and differentiable only at the seams.
For the seam $x=1$, the function is continuous at $f(1)=\sin(1)$.  However, it is not differentiable, as the left handed derivative is $3\cos(1)+3$ and the right handed derivative is $3\cos(1)-3$.
For the seam $x=0$, the function is continuous at $f(0)=1$.  The left handed derivative is $0$ as is the right handed derivative, so the function is differentiable (at least for the first derivative) there as well.
For the seam $x=1$, the function is continuous at $f(1)=\sin(1)$.  However, it is not differentiable, as the left handed derivative is $-3\cos(1)-3$ and the right handed derivative is $-3\cos(1)+3$.
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