Let f: R → R and g: R → R be respectively given by f(x)=|x|+1 and g(x…

According to the given information we have function $f(x)=|x|+1$ and $g(x)=x^2+1$ where $f: R \rightarrow R$ and $g: R \rightarrow R$

So for function $f(x)=|x|+1=\left\{\begin{array}{c}f(x)=x+1 \text { for } x \leq 0 \\ f(x)=-x+1 \text { for } x<0\end{array}\right.$

For the function $g(x)=x^2+1$ we know that for $x \leq 0$ and for $x<0$ the function $g(x)=x^2+1$

Finding the meeting point for function f(x) and function g(x)

When x < 0

Function f(x) = -x + 1 and function $g(x)=x^2+1$

We know that at the meeting point of two function f(x) = g(x)

Therefore $x^2+1=-x+1$

$\Rightarrow x^2+x=1-1$

$\Rightarrow x(x+1)=0$

Since x can't be 0 therefore x = -1

For x = -1, y = 2

Therefore the function f(x) and function g(x) for x < 0 will meet at (-1, 2)

For $x ≥ 0$

Function $f(x)=x+1$ and function $g(x)=x^2+1$

We know that at the meeting point f(x) = g(x)

Therefore $x+1=x^2+1$

$\Rightarrow x^2-x=1-1$

$\Rightarrow x(x-1)=0$

Therefore x = 0, 1

For x = 0, y = 1

And for x = 1, y = 2

Therefore function f(x) and function g(x) for $x \leq 0$ will meet at (0, 1) and (1, 2)

The graphical representation of the above equation is given below

For function $h(x)$ we know that $h(x)=\left\{\begin{array}{l}\max \{f(x), g(x)\} \text { if } x ≤ 0 \\ \min \{f(x), g(x)\} \text { if } x>0\end{array}\right.$

Here green shows the representation of h(x) function

We know that for $x ≤ 0$ the function h(x) is maximum so in the above graph

For x < -1 function g(x) is showing the maximum value

For -1 < x < 0 function f(x) is showing the maximum value

Also for x > 0 the function h(x) is minimum do by the above graph

For 0 < x < 1 function g(x) is showing the minimum value

And for x > 1 function f(x) is showing the minimum value

So by the above statement we can say that function h(x) is not differentiable at (0, 1), (-1, 2) and (1, 2)

Hence the number of points at which h(x) is not differentiable at 3 points.