The function `f(x) = min {|x|, sqrt(1-x^(2))}, -1lt x lt 1` possesses

Options

A. Local minimum at x= -1
B. Local maximum at x = 0
C. Global minimum at x = 1
D. Global maximum at `x = pm 1/sqrt2`

Answer

To solve the problem, we need to analyze the function \( f(x) = \min \{ |x|, \sqrt{1 - x^2} \} \) over the interval \( -1 < x < 1 \).

### Step 1: Understand the Components of the Function
The function consists of two parts:
1. \( |x| \) - This is the absolute value function, which is linear and V-shaped.
2. \( \sqrt{1 - x^2} \) - This represents the upper half of a circle with radius 1, centered at the origin.

### Step 2: Find Points of Intersection
To find where these two functions intersect, we set them equal to each other:
\[ |x| = \sqrt{1 - x^2} \]
Squaring both sides gives:
\[ x^2 = 1 - x^2 \]
Rearranging this, we get:
\[ 2x^2 = 1 \quad \Rightarrow \quad x^2 = \frac{1}{2} \quad \Rightarrow \quad x = \pm \frac{1}{\sqrt{2}} \]

### Step 3: Analyze the Function in the Interval
We need to analyze the behavior of \( f(x) \) in the intervals defined by the points of intersection:
- For \( -1 < x < -\frac{1}{\sqrt{2}} \): Here, \( |x| > \sqrt{1 - x^2} \), so \( f(x) = \sqrt{1 - x^2} \).
- For \( -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \): Here, \( |x| < \sqrt{1 - x^2} \), so \( f(x) = |x| \).
- For \( \frac{1}{\sqrt{2}} < x < 1 \): Here, \( |x| > \sqrt{1 - x^2} \), so \( f(x) = \sqrt{1 - x^2} \).

### Step 4: Evaluate the Function at Critical Points
We evaluate \( f(x) \) at the critical points:
1. At \( x = -1 \): \( f(-1) = \min\{1, 0\} = 0 \)
2. At \( x = 0 \): \( f(0) = \min\{0, 1\} = 0 \)
3. At \( x = \frac{1}{\sqrt{2}} \): \( f\left(\frac{1}{\sqrt{2}}\right) = \min\left\{\frac{1}{\sqrt{2}}, 0\right\} = \frac{1}{\sqrt{2}} \)
4. At \( x = -\frac{1}{\sqrt{2}} \): \( f\left(-\frac{1}{\sqrt{2}}\right) = \min\left\{\frac{1}{\sqrt{2}}, 0\right\} = \frac{1}{\sqrt{2}} \)

### Step 5: Determine the Minimum and Maximum Values
From the evaluations:
- The minimum value of \( f(x) \) occurs at \( x = 0 \) and is \( 0 \).
- The maximum value occurs at \( x = \pm \frac{1}{\sqrt{2}} \) and is \( \frac{1}{\sqrt{2}} \).

### Conclusion
The function \( f(x) \) possesses:
- A minimum at \( x = 0 \) with \( f(0) = 0 \).
- A maximum at \( x = \pm \frac{1}{\sqrt{2}} \) with \( f\left(\pm \frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \).