The range of `f(x) =x+ (1)/(x) ` is

Options

A. `[-2,2]`
B. `[2,oo)`
C. `(-oo,-2]`
D. `(-oo,-2] uu [2,oo)`

Explanation

To find the range of the function \( f(x) = x + \frac{1}{x} \), follow these steps:

### Step 1: Set up the equation
Let \( y = f(x) \). So, we have:
\[ y = x + \frac{1}{x} \]

### Step 2: Rearrange the equation
Multiply both sides by \( x \) to get rid of the fraction:
\[ yx = x^2 + 1 \]

### Step 3: Form a quadratic equation
Rearrange the equation to form a standard quadratic equation:
\[ x^2 - yx + 1 = 0 \]

### Step 4: Analyze the quadratic equation
For \( x \) to be real, the discriminant of the quadratic equation must be non-negative. The discriminant \( \Delta \) of the quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \Delta = b^2 - 4ac \]

For our equation \( x^2 - yx + 1 = 0 \), \( a = 1 \), \( b = -y \), and \( c = 1 \). So, the discriminant is:
\[ \Delta = (-y)^2 - 4 \cdot 1 \cdot 1 \]
\[ \Delta = y^2 - 4 \]

### Step 5: Ensure the discriminant is non-negative
For \( x \) to be real, we need:
\[ y^2 - 4 \geq 0 \]
\[ y^2 \geq 4 \]
\[ |y| \geq 2 \]

This means:
\[ y \leq -2 \quad \text{or} \quad y \geq 2 \]

### Step 6: Write the range
The range of \( f(x) = x + \frac{1}{x} \) is:
\[ (-\infty, -2] \cup [2, \infty) \]