Let `f(x)="min"{sqrt(4-x^(2)),sqrt(1+x^(2))}AA,x in [-2, 2]` then the number of points where `f(x)` is non - differentiable is

Options

A. 1
B. 0
C. 4
D. 2

Answer

To determine the number of points where the function \( f(x) = \min\{\sqrt{4 - x^2}, \sqrt{1 + x^2}\} \) is non-differentiable for \( x \in [-2, 2] \), we will analyze the two functions involved and their intersection points.

### Step 1: Define the functions
Let:
- \( g(x) = \sqrt{4 - x^2} \)
- \( h(x) = \sqrt{1 + x^2} \)

### Step 2: Find the intersection points
To find where \( f(x) \) is non-differentiable, we need to find the points where \( g(x) = h(x) \).

Set the two functions equal to each other:
\[ \sqrt{4 - x^2} = \sqrt{1 + x^2} \]

### Step 3: Square both sides
Squaring both sides to eliminate the square roots gives:
\[ 4 - x^2 = 1 + x^2 \]

### Step 4: Rearrange the equation
Rearranging the equation results in:
\[ 4 - 1 = x^2 + x^2 \]
\[ 3 = 2x^2 \]
\[ x^2 = \frac{3}{2} \]

### Step 5: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \pm \sqrt{\frac{3}{2}} \approx \pm 1.2247 \]

### Step 6: Check the endpoints
We also need to check the endpoints of the interval \( [-2, 2] \):
- At \( x = -2 \): \( g(-2) = \sqrt{4 - 4} = 0 \) and \( h(-2) = \sqrt{1 + 4} = \sqrt{5} \)
- At \( x = 2 \): \( g(2) = \sqrt{4 - 4} = 0 \) and \( h(2) = \sqrt{1 + 4} = \sqrt{5} \)

### Step 7: Identify non-differentiable points
The function \( f(x) \) will be non-differentiable at the points where \( g(x) \) and \( h(x) \) intersect, as well as at the endpoints where the functions switch from one to the other.

### Step 8: Count the points
The points where \( g(x) \) and \( h(x) \) intersect are:
1. \( x = \sqrt{\frac{3}{2}} \)
2. \( x = -\sqrt{\frac{3}{2}} \)

Additionally, we have the endpoints:
3. \( x = -2 \)
4. \( x = 2 \)

Thus, there are a total of **4 points** where \( f(x) \) is non-differentiable.

### Final Answer
The number of points where \( f(x) \) is non-differentiable is **4**.

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