Mathematics - Continuity - Differentiability Question with Solution | TestHub

The function is $f(x)=\min \{|x^2-4|,|x^2-2x|\}$. Let $g(x)=|x^2-4|$ and $h(x)=|x^2-2x|$.$g(x)$ is not differentiable at $x=\pm 2$. $h(x)$ is not differentiable at $x=0,2$.

The points where $g(x)=h(x)$ are found by solving $x^2-4=\pm (x^2-2x)$. This yields $x=2$ (from $x^2-4=x^2-2x$) and $x=-1,2$ (from $x^2-4=-(x^2-2x)$). Potential points of non-differentiability are $x=-2,-1,0,2$. By analyzing the function's definition in intervals:

1. At $x=-2$: $f(x)$ changes from $x^2-4$ to $4-x^2$. Left derivative is $-4$, right derivative is $4$. Not differentiable.

2. At $x=-1$: $f(x)$ is $4-x^2$ on both sides. Left derivative is $2$, right derivative is $2$. Differentiable.

3. At $x=0$: $f(x)$ changes from $4-x^2$ to $-x^2+2x$. Left derivative is $0$, right derivative is $2$. Not differentiable.

4. At $x=2$: $f(x)$ changes from $-x^2+2x$ to $x^2-2x$. Left derivative is $-2$, right derivative is $2$. Not differentiable. The points where $f(x)$ is not differentiable are $x=-2,0,2$. Thus, $N=3$.