`f(x) =` maximum `{4, 1 + x^2, x^2-1) AA x in R`. Total number of points, where `f(x)` is non-differentiable,is equal to

Answer

To solve the problem, we need to analyze the function defined as:

\[ f(x) = \max \{ 4, 1 + x^2, x^2 - 1 \} \]

### Step 1: Identify the individual functions
The function consists of three parts:
1. \( f_1(x) = 4 \)
2. \( f_2(x) = 1 + x^2 \)
3. \( f_3(x) = x^2 - 1 \)

### Step 2: Find the points of intersection
To determine where the maximum function changes, we need to find the points where these functions intersect.

**1. Intersection of \( f_1(x) \) and \( f_2(x) \):**
Set \( 4 = 1 + x^2 \):
\[ x^2 = 3 \implies x = \pm \sqrt{3} \]

**2. Intersection of \( f_1(x) \) and \( f_3(x) \):**
Set \( 4 = x^2 - 1 \):
\[ x^2 = 5 \implies x = \pm \sqrt{5} \]

**3. Intersection of \( f_2(x) \) and \( f_3(x) \):**
Set \( 1 + x^2 = x^2 - 1 \):
\[ 1 = -1 \quad \text{(no solution)} \]

### Step 3: Determine the intervals
Next, we need to analyze the behavior of \( f(x) \) in the intervals defined by the intersection points:

- For \( x < -\sqrt{5} \): \( f(x) = 4 \)
- For \( -\sqrt{5} < x < -\sqrt{3} \): \( f(x) = 1 + x^2 \)
- For \( -\sqrt{3} < x < \sqrt{3} \): \( f(x) = 1 + x^2 \)
- For \( \sqrt{3} < x < \sqrt{5} \): \( f(x) = 4 \)
- For \( x > \sqrt{5} \): \( f(x) = 4 \)

### Step 4: Check for non-differentiability
The function \( f(x) \) is non-differentiable at points where the maximum function changes from one piece to another. This occurs at the intersection points:

- \( x = -\sqrt{5} \)
- \( x = -\sqrt{3} \)
- \( x = \sqrt{3} \)
- \( x = \sqrt{5} \)

### Conclusion
Thus, the total number of points where \( f(x) \) is non-differentiable is:

\[ \text{Total points} = 4 \]