Step 1: Define the functions f(x) and g(x)

The given functions are: $f(x) = |x| + 1$ $g(x) = x^2 + 1$

We can write $f(x)$ as a piecewise function: $f(x) = \left\{ \begin{matrix} -x + 1, & \text{if } x < 0 \\ x + 1, & \text{if } x \ge 0 \end{matrix} \right.$ The function $h(x)$ is defined as: $h(x) = \left\{ \begin{matrix} \max\{ f(x), g(x) \}, & \text{if } x \le 0 \\ \min\{ f(x), g(x) \}, & \text{if } x > 0 \end{matrix} \right.$

Step 2: Analyze h(x) for x ≤ 0

For $x \le 0$, we have $f(x) = -x + 1$ and $g(x) = x^2 + 1$. We need to find $h(x) = \max\{-x + 1, x^2 + 1\}$. First, let's find the points where $f(x) = g(x)$ for $x \le 0$: $-x + 1 = x^2 + 1$ $x^2 + x = 0$ $x(x + 1) = 0$ This gives $x = 0$ and $x = -1$.

Now, we compare $f(x)$ and $g(x)$ in the intervals $(-\infty, -1]$ and $[-1, 0]$.

• For $x \in (-\infty, -1)$, let's take $x = -2$. $f(-2) = -(-2) + 1 = 3$ $g(-2) = (-2)^2 + 1 = 5$ So, $g(x) > f(x)$ for $x < -1$.
• For $x \in (-1, 0)$, let's take $x = -0.5$. $f(-0.5) = -(-0.5) + 1 = 1.5$ $g(-0.5) = (-0.5)^2 + 1 = 1.25$ So, $f(x) > g(x)$ for $-1 < x < 0$.

Therefore, for $x \le 0$, the definition of $h(x)$ is: $h(x) = \left\{ \begin{matrix} g(x) = x^2 + 1, & \text{if } x \le -1 \\ f(x) = -x + 1, & \text{if } -1 < x \le 0 \end{matrix} \right.$

Step 3: Analyze h(x) for x > 0

For $x > 0$, we have $f(x) = x + 1$ and $g(x) = x^2 + 1$. We need to find $h(x) = \min\{x + 1, x^2 + 1\}$. First, let's find the points where $f(x) = g(x)$ for $x > 0$: $x + 1 = x^2 + 1$ $x^2 - x = 0$ $x(x - 1) = 0$ This gives $x = 0$ and $x = 1$. Since we are considering $x > 0$, the intersection point is $x=1$.

Now, we compare $f(x)$ and $g(x)$ in the intervals $(0, 1]$ and $[1, \infty)$.

• For $x \in (0, 1)$, let's take $x = 0.5$. $f(0.5) = 0.5 + 1 = 1.5$ $g(0.5) = (0.5)^2 + 1 = 1.25$ So, $g(x) < f(x)$ for $0 < x < 1$.
• For $x > 1$, let's take $x = 2$. $f(2) = 2 + 1 = 3$ $g(2) = 2^2 + 1 = 5$ So, $f(x) < g(x)$ for $x > 1$.

Therefore, for $x > 0$, the definition of $h(x)$ is: $h(x) = \left\{ \begin{matrix} g(x) = x^2 + 1, & \text{if } 0 < x \le 1 \\ f(x) = x + 1, & \text{if } x > 1 \end{matrix} \right.$

Step 4: Combine to get the full definition of h(x)

Combining the pieces, we get the complete definition of $h(x)$: $h(x) = \left\{ \begin{matrix} x^2 + 1, & \text{if } x \le -1 \\ -x + 1, & \text{if } -1 < x \le 0 \\ x^2 + 1, & \text{if } 0 < x \le 1 \\ x + 1, & \text{if } x > 1 \end{matrix} \right.$

Step 5: Check for differentiability

The potential points of non-differentiability are where the definition of the function changes: $x = -1$, $x = 0$, and $x = 1$. A function is not differentiable at a point if the left-hand derivative (LHD) is not equal to the right-hand derivative (RHD).

Let's find the derivative of $h(x)$ for each interval: $h'(x) = \left\{ \begin{matrix} 2x, & \text{if } x < -1 \\ -1, & \text{if } -1 < x < 0 \\ 2x, & \text{if } 0 < x < 1 \\ 1, & \text{if } x > 1 \end{matrix} \right.$ Now we check the LHD and RHD at the critical points.

At x = -1: LHD = $\lim_{x \to -1^-} h'(x) = \lim_{x \to -1^-} (2x) = 2(-1) = -2$ RHD = $\lim_{x \to -1^+} h'(x) = \lim_{x \to -1^+} (-1) = -1$ Since LHD $\neq$ RHD, $h(x)$ is not differentiable at $x = -1$.

At x = 0: LHD = $\lim_{x \to 0^-} h'(x) = \lim_{x \to 0^-} (-1) = -1$ RHD = $\lim_{x \to 0^+} h'(x) = \lim_{x \to 0^+} (2x) = 2(0) = 0$ Since LHD $\neq$ RHD, $h(x)$ is not differentiable at $x = 0$.

At x = 1: LHD = $\lim_{x \to 1^-} h'(x) = \lim_{x \to 1^-} (2x) = 2(1) = 2$ RHD = $\lim_{x \to 1^+} h'(x) = \lim_{x \to 1^+} (1) = 1$ Since LHD $\neq$ RHD, $h(x)$ is not differentiable at $x = 1$.

Step 6: Conclusion

The function $h(x)$ is not differentiable at three points: $x = -1, 0, 1$. The number of points at which $h(x)$ is not differentiable is 3.