If `f:[1, oo) rarr [2, oo)` is defined by `f(x)=x+1/x`, find `f^(-1)(x)`

Answer

To find the inverse of the function \( f(x) = x + \frac{1}{x} \) defined from \( [1, \infty) \) to \( [2, \infty) \), we will follow these steps:

### Step 1: Set up the equation
We start by setting \( y = f(x) \):
\[ y = x + \frac{1}{x} \]

### Step 2: Rearrange the equation
To find \( x \) in terms of \( y \), we can rearrange the equation:
\[ y = x + \frac{1}{x} \implies yx = x^2 + 1 \]
This can be rewritten as:
\[ x^2 - yx + 1 = 0 \]

### Step 3: Identify the quadratic equation
The equation \( x^2 - yx + 1 = 0 \) is a quadratic equation in \( x \) where:
- \( a = 1 \)
- \( b = -y \)
- \( c = 1 \)

### Step 4: Apply the quadratic formula
We can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting \( a \), \( b \), and \( c \):
\[ x = \frac{y \pm \sqrt{(-y)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{y \pm \sqrt{y^2 - 4}}{2} \]

### Step 5: Determine the correct sign
Since \( f(x) \) is increasing for \( x \geq 1 \), we will take the positive root:
\[ f^{-1}(y) = \frac{y + \sqrt{y^2 - 4}}{2} \]

### Step 6: Replace \( y \) with \( x \)
For the inverse function, we replace \( y \) with \( x \):
\[ f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2} \]

### Final Answer
Thus, the inverse function is:
\[ f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2}, \quad \text{for } x \in [2, \infty) \]

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