Let `f: (-1,1)toR` be a function defind by f(x) =max. `{-absx,-sqrt(1-x^2)}`. If K is the set of all points at which f is not differentiable, then K has set of all points at which f is not differentiable, then K has exactly

Options

A. three elements
B. one element
C. five elements
D. two elements

Answer

To solve the problem, we need to analyze the function \( f(x) = \max\{-|x|, -\sqrt{1-x^2}\} \) defined on the interval \( (-1, 1) \). We will identify the points where this function is not differentiable.

### Step 1: Understand the components of the function
The function \( f(x) \) is defined as the maximum of two functions:
1. \( g_1(x) = -|x| \)
2. \( g_2(x) = -\sqrt{1 - x^2} \)

### Step 2: Analyze \( g_1(x) = -|x| \)
The function \( g_1(x) \) is a V-shaped graph that opens downwards. It has a vertex at the origin (0,0) and intersects the x-axis at \( x = -1 \) and \( x = 1 \). The function is linear and differentiable everywhere except at \( x = 0 \) where it has a sharp corner.

### Step 3: Analyze \( g_2(x) = -\sqrt{1 - x^2} \)
The function \( g_2(x) \) represents the lower half of a circle with radius 1 centered at the origin. It is differentiable everywhere in the interval \( (-1, 1) \) except at the endpoints \( x = -1 \) and \( x = 1 \).

### Step 4: Find the intersection points
To find the points where \( f(x) \) switches from one function to the other, we need to set \( g_1(x) = g_2(x) \):
\[ -|x| = -\sqrt{1 - x^2} \]
Squaring both sides to eliminate the square root gives:
\[ x^2 = 1 - x^2 \]
\[ 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \]

### Step 5: Identify points of non-differentiability
The function \( f(x) \) will not be differentiable at the following points:
1. \( x = 0 \) (corner point of \( g_1(x) \))
2. \( x = -\frac{1}{\sqrt{2}} \) (intersection point)
3. \( x = \frac{1}{\sqrt{2}} \) (intersection point)

### Conclusion
Thus, the set \( K \) of all points at which \( f \) is not differentiable contains exactly three points:
\[ K = \left\{ 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\} \]

### Final Answer
The set \( K \) has exactly 3 points.

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