The greatest, the least values of the function ,`f(x) =2-sqrt(1+2x+x^(2)), x in [-2,1]` are respectively

Options

A. `2,1`
B. `2, -1`
C. `2,0`
D. ` -2,3`

Answer

To find the greatest and least values of the function \( f(x) = 2 - \sqrt{1 + 2x + x^2} \) for \( x \) in the interval \([-2, 1]\), we can follow these steps:

### Step 1: Rewrite the function
We can simplify the expression under the square root:
\[ 1 + 2x + x^2 = (x + 1)^2 \]
Thus, we can rewrite the function as:
\[ f(x) = 2 - \sqrt{(x + 1)^2} \]
Since the square root function yields non-negative values, we have:
\[ \sqrt{(x + 1)^2} = |x + 1| \]
So, the function becomes:
\[ f(x) = 2 - |x + 1| \]

### Step 2: Analyze the absolute value
The expression \( |x + 1| \) can be split into two cases based on the value of \( x \):
1. If \( x + 1 \geq 0 \) (i.e., \( x \geq -1 \)), then \( |x + 1| = x + 1 \).
2. If \( x + 1 < 0 \) (i.e., \( x < -1 \)), then \( |x + 1| = -(x + 1) = -x - 1 \).

### Step 3: Define the function in both cases
1. For \( x \geq -1 \):
\[ f(x) = 2 - (x + 1) = 1 - x \]
2. For \( x < -1 \):
\[ f(x) = 2 - (-x - 1) = 3 + x \]

### Step 4: Evaluate the function at critical points and endpoints
Now, we need to evaluate \( f(x) \) at the endpoints of the interval and at the critical point \( x = -1 \).

1. At \( x = -2 \):
\[ f(-2) = 3 + (-2) = 1 \]
2. At \( x = -1 \):
\[ f(-1) = 1 - (-1) = 2 \]
3. At \( x = 1 \):
\[ f(1) = 1 - 1 = 0 \]

### Step 5: Determine the greatest and least values
Now we compare the values obtained:
- \( f(-2) = 1 \)
- \( f(-1) = 2 \)
- \( f(1) = 0 \)

From these evaluations, we find:
- The **greatest value** of \( f(x) \) is \( 2 \) (at \( x = -1 \)).
- The **least value** of \( f(x) \) is \( 0 \) (at \( x = 1 \)).

### Final Answer
The greatest and least values of the function \( f(x) \) are \( 2 \) and \( 0 \) respectively.