The range of `f(x) =2+2x -x^(2) ` is :

Options

A. `(-oo, oo)`
B. `(-oo, 1)`
C. `(-oo, 3]`
D. `(0, oo)`

Explanation

To find the range of the function \( f(x) = 2 + 2x - x^2 \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = 2 + 2x - x^2 \]
We can rearrange it to emphasize the quadratic nature:
\[ f(x) = -x^2 + 2x + 2 \]

### Step 2: Complete the square
Next, we will complete the square for the quadratic expression. We can factor out the negative sign:
\[ f(x) = -(x^2 - 2x) + 2 \]
Now, we complete the square for \( x^2 - 2x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Substituting this back into the function:
\[ f(x) = -((x - 1)^2 - 1) + 2 \]
This simplifies to:
\[ f(x) = - (x - 1)^2 + 1 + 2 = - (x - 1)^2 + 3 \]

### Step 3: Analyze the function
The expression \( -(x - 1)^2 \) is always less than or equal to 0 because a square is always non-negative. Therefore:
\[ -(x - 1)^2 \leq 0 \]
Adding 3 to both sides gives:
\[ -(x - 1)^2 + 3 \leq 3 \]
This implies:
\[ f(x) \leq 3 \]

### Step 4: Determine the minimum value
The maximum value of \( f(x) \) occurs when \( (x - 1)^2 = 0 \), which is when \( x = 1 \):
\[ f(1) = -0 + 3 = 3 \]
As \( (x - 1)^2 \) increases, \( f(x) \) decreases without bound. Thus, the minimum value of \( f(x) \) approaches negative infinity.

### Step 5: Conclusion
The range of the function \( f(x) \) is:
\[ (-\infty, 3] \]