If `f(x)=" min "{(sqrt(9-x^(2)), sqrt(1+x^(2)))}, AA, x in [-3, 3]`
then the number of point(s) where `f(x)` is non - differentiable is/are

Options

A. 4
B. 3
C. 2
D. 0

Answer

To find the number of points where the function \( f(x) = \min\left(\sqrt{9 - x^2}, \sqrt{1 + x^2}\right) \) is non-differentiable for \( x \in [-3, 3] \), we will analyze the two functions involved in the minimum operation.

### Step 1: Identify the two functions
We have:
1. \( g_1(x) = \sqrt{9 - x^2} \)
2. \( g_2(x) = \sqrt{1 + x^2} \)

### Step 2: Determine the domain of each function
- The function \( g_1(x) \) is defined for \( -3 \leq x \leq 3 \) because \( 9 - x^2 \geq 0 \).
- The function \( g_2(x) \) is defined for all real numbers since \( 1 + x^2 \) is always positive.

### Step 3: Find the points of intersection
To find where \( g_1(x) \) and \( g_2(x) \) intersect, we set them equal to each other:
\[ \sqrt{9 - x^2} = \sqrt{1 + x^2} \]
Squaring both sides gives:
\[ 9 - x^2 = 1 + x^2 \]
Rearranging this leads to:
\[ 9 - 1 = 2x^2 \implies 8 = 2x^2 \implies x^2 = 4 \implies x = \pm 2 \]

### Step 4: Evaluate the function at the intersection points
Now we need to evaluate \( f(x) \) at the intersection points \( x = -2 \) and \( x = 2 \):
- For \( x = -2 \):
\[ f(-2) = \min\left(\sqrt{9 - (-2)^2}, \sqrt{1 + (-2)^2}\right) = \min\left(\sqrt{5}, \sqrt{5}\right) = \sqrt{5} \]
- For \( x = 2 \):
\[ f(2) = \min\left(\sqrt{9 - (2)^2}, \sqrt{1 + (2)^2}\right) = \min\left(\sqrt{5}, \sqrt{5}\right) = \sqrt{5} \]

### Step 5: Check endpoints and other critical points
Next, we check the endpoints of the interval and the behavior of the function:
- At \( x = -3 \):
\[ f(-3) = \sqrt{9 - (-3)^2} = 0 \]
- At \( x = 3 \):
\[ f(3) = \sqrt{9 - (3)^2} = 0 \]

### Step 6: Analyze differentiability
The function \( f(x) \) is non-differentiable at points where the two functions intersect because the minimum function can switch from one function to another, leading to a change in the slope.

The points of non-differentiability are:
1. \( x = -3 \) (endpoint)
2. \( x = -2 \) (intersection)
3. \( x = 2 \) (intersection)
4. \( x = 3 \) (endpoint)

### Conclusion
Thus, the total number of points where \( f(x) \) is non-differentiable is **4**.