Max, Min

$\begin{aligned} h(x) & = \begin{cases} \max \{ f(x), \ g(x)\} & = \max \{1-x, \ x^2+1 \} & \text{if } x\le 0 \\ \min \{ f(x), \ g(x)\} & = \min \{1+x, \ x^2+1\} & \text{if } x> 0 \end{cases} \end{aligned}$

Equating $f(x) = g(x)$ for $x \le 0$ and $x > 0$ ,

$\begin{aligned} x \le 0: \quad 1-x & = x^2 + 1 \\ x^2 + x & = 0 \\ x(x+1) & = 0 \end{aligned}$

$\implies f(-1) = g(-1) = 2$ , $f(0) = g(0) = 1$ , when $x<-1$ , $g(x) > f(x)$ and when $-1 < x < 0$ , $f(x) > g(x)$ .

Similarly,

$\begin{aligned} x > 0: \quad 1+x & = x^2 + 1 \\ x^2 - x & = 0 \\ x(x-1) & = 0 \end{aligned}$

$\implies f(0) = g(0) = 1$ , $f(1) = g(1) = 2$ , when $0 < x< 1$ , $g(x) < f(x)$ and when $x > 1$ , $f(x) < g(x)$ .

Therefore, $h(x) = \begin{cases} x^2 + 1 & \text{if } x < -1 \\ 1-x & \text{if } -1 < x < 0 \\ x^2 + 1 & \text{if } 0 < x < 1 \\ 1+x & \text{if } x > 1 \end{cases}$

There are three critical points at $x=-1,0,1$ .

Since $\begin{cases} x = -1 & \implies \begin{cases} f'(-1) = -1 \\ g'(-1) = -2 \end{cases} & \implies f'(-1) \ne g'(-1) & \text{Not differentiable} \\ x = 0 & \implies \begin{cases} f'(0) = -1 \\ g'(0) = 0 \end{cases} & \implies f'(0) \ne g'(0) & \text{Not differentiable} \\ x = 1 & \implies \begin{cases} f'(1) = 1 \\ g'(1) = 2 \end{cases} & \implies f'(1) \ne g'(1) & \text{Not differentiable} \end{cases}$

Therefore, there are $\boxed{3}$ points at which $h(x)$ is not differentiable.