Find the range of the function `g(x)=(x^2+2x+3)/x.`

Options

A. `y in R`
B. `y in [-2sqrt3+2, 2sqrt3+2]`
C. `y in (-oo,2sqrt3+2] cup [2sqrt3 +2, oo)`
D. None of these

Answer

To find the range of the function \( g(x) = \frac{x^2 + 2x + 3}{x} \), we will follow these steps:

### Step 1: Rewrite the function
Let \( g(x) = y \). Therefore, we have:
\[ y = \frac{x^2 + 2x + 3}{x} \]
This can be rewritten as:
\[ y = x + 2 + \frac{3}{x} \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ yx = x^2 + 2x + 3 \]
This can be rearranged to:
\[ x^2 + (2 - y)x + 3 = 0 \]

### Step 3: Analyze the quadratic equation
For \( x \) to have real values, the discriminant \( D \) of the quadratic equation must be non-negative:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = 2 - y \), and \( c = 3 \). Therefore,
\[ D = (2 - y)^2 - 4 \cdot 1 \cdot 3 \geq 0 \]
This simplifies to:
\[ (2 - y)^2 - 12 \geq 0 \]

### Step 4: Solve the inequality
Now we solve the inequality:
\[ (2 - y)^2 \geq 12 \]
Taking the square root of both sides gives us:
\[ |2 - y| \geq 2\sqrt{3} \]
This leads us to two cases:
1. \( 2 - y \geq 2\sqrt{3} \)
2. \( 2 - y \leq -2\sqrt{3} \)

### Step 5: Solve each case
**For the first case:**
\[ 2 - y \geq 2\sqrt{3} \implies y \leq 2 - 2\sqrt{3} \]

**For the second case:**
\[ 2 - y \leq -2\sqrt{3} \implies y \geq 2 + 2\sqrt{3} \]

### Step 6: Combine the results
From the two cases, we find:
- \( y \leq 2 - 2\sqrt{3} \)
- \( y \geq 2 + 2\sqrt{3} \)

Thus, the range of \( g(x) \) is:
\[ (-\infty, 2 - 2\sqrt{3}] \cup [2 + 2\sqrt{3}, \infty) \]

### Final Answer
The range of the function \( g(x) = \frac{x^2 + 2x + 3}{x} \) is:
\[ (-\infty, 2 - 2\sqrt{3}] \cup [2 + 2\sqrt{3}, \infty) \]