Number of points at which $g(x)$ is non-differentiable I could not type the question in latex correctly,so i am posting the pic of my question.

My problem is that in the first part,i think that number of points where function is non differentiable are $3$(i.e at $x=-2,x=0,x=1 )$ but i am not sure about the answer.Is my answer correct.In second part,i am convinced that option (C) is correct.g(x) is continuous and non-differentiable at $x=0$.

For the third part,both $f(x)$ and $g(x)$ are zero at $x=-2$.So how will we find its finite limit as $x$ tends to $-2$.Please help me.
Would rather discard the first answer I have posted, but instead of editing it, I am reposting a more detailed answer, hopefully removing any errors of commission or omission from the first one. Also. OP edited his question in the mean time.

I $\color{blue}{agree}$ with you on the first part - with the following explanation. 

We first start with the definition of $|x|$ which is used throughout.

$$
|x| = \left\{
        \begin{array}{ll}
            x &amp; \quad x \ge 0 \\
            -x &amp; \quad x \lt 0
        \end{array}
    \right.
$$

This is now used to simplify $f(x)$ successively

$$ f(x) = |x^2 - 3|x| + 2|$$
$$
 = \left\{
        \begin{array}{ll}
            |x^2 - 3x + 2| &amp; \quad x \ge 0 \\
            |x^2 + 3x + 2| &amp; \quad x \le 0
        \end{array}
    \right.
$$

$$
 = \left\{
        \begin{array}{ll}
            |(x-1)(x-2)| &amp; \quad x \ge 0 \\
            |(x+1)(x+2)| &amp; \quad x \le 0
        \end{array}
    \right.
$$

$$
 = \left\{
        \begin{array}{ll}
            (x-1)(x-2) &amp; \quad x \ge 2 \\
            (x-1)(2-x) &amp; \quad 1 \le x \lt 2 \\
            (x-1)(x-2) &amp; \quad 0 \le x \lt 1 \\
            (x+1)(x+2) &amp; \quad -1 \le x \lt 0 \\
            -(x+1)(x+2) &amp; \quad -2 \le x \lt -1 \\
            (x+1)(x+2) &amp; \quad x \lt -2
        \end{array}
    \right.
$$

Having done it so explicitly, it is easy to see that its roots $\pm 1$ and $\pm 2$

So now, the definition of $g$ becomes easy (there were intervals when maxima had to be found, but leaving that out for brevity).
$$
g(x) = \left\{
        \begin{array}{ll}
            0 &amp; \quad x \le -1 \\
            2 &amp; \quad -1 \le x \lt 1 \\
            0 &amp; \quad 1 \le x \lt 2 \\
            x^2 + 3x + 2 &amp; \quad  x \ge 2 
        \end{array}
    \right.
$$

So $-1, 1$ and $2$ are $\color{blue}{3}$ points of discontinuity for $g(x)$, and hence not differentiable. At all other places, it is smooth.

On the $\color{blue}{second}$ part, since $g(x) = 2$ around $0$, so it is both continuous and differentiable.

On the $\color{blue}{third}$ part, the left hand limit is 0, the right hand is infinity, so the limit does not exist.
I read the first part of the definition as 
$$\min_{t\in(-\infty,x]}f(t)$$
whenever $x&lt;-1$.  Since $f(t)$ is decreasing on $(-\infty,-2)$, the minimum on the interval is $f(x)$, so $g(x)=f(x)$ if $x&lt;-2$.  For $x\in[-2,-1]$, the interval $(-\infty,x)$ contains $-2$, so $g(x)=0$.