if `f:[1,oo)->[2,oo)` is given by `f(x)=x+1/x` then `f^-1(x)` equals to : a) `(x+sqrt(x^2-4))/2` b) `x/(1+x^2)` c) `(x-sqrt(x^2-4))/2` d) `1+sqrt(x^2-4)`

Options

A. `((x+sqrt(x^(2)-4)))/(2)`
B. `(x)/(1+x^(2))`
C. `((x-sqrt(x^(2)-4)))/(2)`
D. `1+sqrt(x^(2)-4)`

Explanation

To find the inverse function \( f^{-1}(x) \) for the function \( f(x) = x + \frac{1}{x} \), defined on the domain \([1, \infty)\) and mapping to the range \([2, \infty)\), we will follow these steps:

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

### 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 rearranges to:
\[ x^2 - yx + 1 = 0 \]

### Step 3: Apply the quadratic formula
Now we can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -y \), and \( c = 1 \). Plugging these values into the formula gives:
\[ x = \frac{y \pm \sqrt{(-y)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{y \pm \sqrt{y^2 - 4}}{2} \]

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

### Step 5: Write the inverse function
Thus, we can express the inverse function as:
\[ f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2} \]

### Conclusion
The correct answer is:
\[ \text{a) } f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2} \]

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