Let `f:R-> R and g:R-> R` be respectively given by `f(x) = |x| +1 and g(x) = x^2 + 1`. Define `h:R-> R` by `h(x)={max{f(x), g(x)}, if xleq 0 and min{f(x), g(x)}, if x > 0`.The number of points at which `h(x)` is not differentiable is

Answer

To determine the number of points at which the function \( h(x) \) is not differentiable, we need to analyze the functions \( f(x) \) and \( g(x) \) and how they interact in the definition of \( h(x) \).

### Step 1: Define the Functions
We have:
- \( f(x) = |x| + 1 \)
- \( g(x) = x^2 + 1 \)

### Step 2: Analyze \( f(x) \) and \( g(x) \)
1. **For \( f(x) \)**:
- \( f(x) = |x| + 1 \) is continuous everywhere and has a corner point at \( x = 0 \) where the derivative changes.
- The derivative is:
- \( f'(x) = 1 \) for \( x > 0 \)
- \( f'(x) = -1 \) for \( x < 0 \)
- \( f'(0) \) is not defined (corner point).

2. **For \( g(x) \)**:
- \( g(x) = x^2 + 1 \) is a parabola that opens upwards and is differentiable everywhere.
- The derivative is:
- \( g'(x) = 2x \)

### Step 3: Define \( h(x) \)
The function \( h(x) \) is defined as:
- \( h(x) = \max\{f(x), g(x)\} \) if \( x \leq 0 \)
- \( h(x) = \min\{f(x), g(x)\} \) if \( x > 0 \)

### Step 4: Analyze \( h(x) \) for \( x \leq 0 \)
For \( x \leq 0 \):
- We need to find where \( f(x) \) and \( g(x) \) intersect.
- Set \( f(x) = g(x) \):
\[ |x| + 1 = x^2 + 1 \]
Simplifying gives:
\[ |x| = x^2 \]
For \( x \leq 0 \), \( |x| = -x \):
\[ -x = x^2 \implies x^2 + x = 0 \implies x(x + 1) = 0 \]
Thus, \( x = 0 \) or \( x = -1 \).

### Step 5: Analyze \( h(x) \) for \( x > 0 \)
For \( x > 0 \):
- Again, we find where \( f(x) \) and \( g(x) \) intersect:
\[ f(x) = g(x) \implies x + 1 = x^2 + 1 \]
Simplifying gives:
\[ x = x^2 \implies x^2 - x = 0 \implies x(x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

### Step 6: Points of Non-Differentiability
1. **At \( x = 0 \)**:
- \( h(x) \) switches from \( \max \) to \( \min \), and \( f'(0) \) is not defined.

2. **At \( x = -1 \)**:
- This is where \( f(x) \) and \( g(x) \) intersect for \( x \leq 0\), leading to a potential corner point.

3. **At \( x = 1 \)**:
- This is where \( f(x) \) and \( g(x) \) intersect for \( x > 0\), leading to another potential corner point.

### Conclusion
Thus, the points at which \( h(x) \) is not differentiable are \( x = -1 \), \( x = 0 \), and \( x = 1 \). Therefore, the number of points at which \( h(x) \) is not differentiable is **3**.