Suppose: `f : R to S` is defined by
`f(x) = 1/(x^(2) + 2x + 2) AA x in R`, If f is a surjective function, then S is given by

Options

A. `[1, infty)`
B. `(1, infty)`
C. `[0,1]`
D. `(0,1]`

Answer

To determine the set \( S \) for the surjective function \( f: \mathbb{R} \to S \) defined by

\[ f(x) = \frac{1}{x^2 + 2x + 2} \]

we need to find the range of the function \( f(x) \).

### Step 1: Rewrite the function
First, we can rewrite the quadratic expression in the denominator:

\[ x^2 + 2x + 2 = (x+1)^2 + 1 \]

Thus, we can express \( f(x) \) as:

\[ f(x) = \frac{1}{(x+1)^2 + 1} \]

### Step 2: Analyze the function
Next, we analyze the function to find its range. The term \( (x+1)^2 \) is always non-negative for all \( x \in \mathbb{R} \), which means:

\[ (x+1)^2 \geq 0 \implies (x+1)^2 + 1 \geq 1 \]

This indicates that the minimum value of the denominator is 1.

### Step 3: Determine maximum and minimum values
Since the denominator \( (x+1)^2 + 1 \) achieves its minimum value of 1 when \( x = -1 \), we can find the maximum value of \( f(x) \):

\[ f(-1) = \frac{1}{1} = 1 \]

As \( x \) approaches \( \pm \infty \), \( (x+1)^2 \) becomes very large, leading to:

\[ f(x) \to \frac{1}{\infty} = 0 \]

### Step 4: Establish the range
From the above analysis, we conclude that:

- The maximum value of \( f(x) \) is 1 (achieved at \( x = -1 \)).
- The function approaches 0 but never actually reaches it as \( x \) approaches \( \pm \infty \).

Thus, the range of \( f(x) \) is:

\[ (0, 1] \]

### Step 5: Conclusion
Since \( f \) is surjective, the codomain \( S \) must equal the range of \( f(x) \). Therefore, we have:

\[ S = (0, 1] \]

### Final Answer
The set \( S \) is given by \( (0, 1] \).

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