what is the number of connected components of given set? 
Consider the function $f: \Bbb{R} \to \Bbb{R}$ is given below:
  \begin{cases}|t/2|, & t <-2,\\ |t+3/2| + 1/2, & -2\le t<-1,\\|t^3|, & -1\le t< 1,\\|t-3/2|^2+3/4 ,& 1\le t <2\\|t/2|,& t\ge 2 \end{cases}
  What is the number of connected components of set $\{t\in \Bbb{R}: f \text{ is differentiable at } t\}$?

I think there  will be five connected components of the set $\{t\in \Bbb{R}: f \text{ is differentiable at } t\}$ as there are five disjoint nonempty sets.
 A: Infact if we check $f$ is differentiable on each of the intervals (*inclusive of the critical points). Hence the number of connected components is 6.
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The check:
Using definition of limit for calculating derivative viz existence of $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ and its value gives $f'(x)$.
By using the definition of limit  we get the following: 
For $t<-2$ , $f(t)=-t/2\quad$ $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\displaystyle \lim_{h\rightarrow 0}\frac{\frac{-(t+h)}{2}-\frac{-t}{2}}{h}=\frac{-1}{2}$ 
We rewrite $$\displaystyle f(t)=\begin {align}  \begin{cases}-(t+1), &-2\leq t\leq -3/2 \\ (t+2), &-3/2\leq t<-1\\-t^3, &-1\leq t\leq 0 \\ t^3,& 0\leq t<1\\(t-3/2)^2+3/4, & 1\leq t<2\\t/2,& t\geq 2\end{cases}\end{align}$$
Calculating derivative using limit as above we get 
$$f'(t)=\begin{cases}-1, &-2<t< -3/2 \\ 1,& -3/2<t<-1\\-3t^2,& -1< t\leq 0 \\ 3t^2, &0\leq t<1\\2t-3, &1\leq t<2 \\1/2, & t\geq 2\end{cases}$$
We see that the derivative of $f$ does not exist at the points $x=-2,-3/2,-1,1,2$. Thus the intervals where derivatives exist are $(-\infty,-2),(-2,-3/2),(-3/2,-1),(-1,1),(1,2),(2,\infty).$ Hence the number of connected components are 6.
*Sorry I wanted to write exclusive of critical points. 
