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

Answer

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

### Step 1: Rewrite the function
Let \( f(x) = y \). Then we have:
\[ y = \frac{x^2 + 2x + 3}{x} \]
Multiplying both sides by \( x \) (assuming \( x \neq 0 \)):
\[ yx = x^2 + 2x + 3 \]
Rearranging gives us:
\[ x^2 + (2 - y)x + 3 = 0 \]

### Step 2: Identify the discriminant
For \( x \) to have real solutions, the discriminant of this quadratic equation must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = 2 - y \), and \( c = 3 \). Thus,
\[ D = (2 - y)^2 - 4 \cdot 1 \cdot 3 \]
Simplifying this:
\[ D = (2 - y)^2 - 12 \]

### Step 3: Set the discriminant greater than or equal to zero
To ensure real solutions for \( x \):
\[ (2 - y)^2 - 12 \geq 0 \]
This simplifies to:
\[ (2 - y)^2 \geq 12 \]

### Step 4: Solve the inequality
Taking square roots on both sides gives:
\[ |2 - y| \geq \sqrt{12} \]
This implies:
\[ 2 - y \geq 2\sqrt{3} \quad \text{or} \quad 2 - y \leq -2\sqrt{3} \]
From the first inequality:
\[ -y \geq 2\sqrt{3} - 2 \quad \Rightarrow \quad y \leq 2 - 2\sqrt{3} \]
From the second inequality:
\[ -y \leq -2\sqrt{3} - 2 \quad \Rightarrow \quad y \geq 2 + 2\sqrt{3} \]

### Step 5: Write the range
Thus, the range of \( f(x) \) is:
\[ y \in (-\infty, 2 - 2\sqrt{3}] \cup [2 + 2\sqrt{3}, \infty) \]

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