Let `f(x)=x+1/(2x+1/(2x+1/(2x+...…...oo)))`
Compute the value of `f(100)dotf^(prime)(100)`

Options

A. 0
B. 100
C. -1
D. 2

Answer

To solve the problem, we need to compute the value of \( f(100) \cdot f'(100) \) where

\[ f(x) = x + \frac{1}{2x + \frac{1}{2x + \frac{1}{2x + \ldots}}} \]

### Step 1: Define the function
We start by recognizing that the infinite nested fraction can be represented as \( f(x) \). Thus, we can write:

\[ f(x) = x + \frac{1}{f(x)} \]

### Step 2: Rearranging the equation
Rearranging the equation gives us:

\[ f(x) - x = \frac{1}{f(x)} \]

### Step 3: Cross-multiplying
Cross-multiplying gives us:

\[ (f(x) - x) f(x) = 1 \]

### Step 4: Expanding the equation
Expanding the left side, we have:

\[ f(x)^2 - x f(x) = 1 \]

### Step 5: Rearranging into standard form
Rearranging this into standard quadratic form, we get:

\[ f(x)^2 - x f(x) - 1 = 0 \]

### Step 6: Using the quadratic formula
We can solve this quadratic equation for \( f(x) \) using the quadratic formula:

\[ f(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 1, b = -x, c = -1 \). Thus:

\[ f(x) = \frac{x \pm \sqrt{x^2 + 4}}{2} \]

### Step 7: Choosing the correct root
Since \( f(x) \) must be positive for positive \( x \), we take the positive root:

\[ f(x) = \frac{x + \sqrt{x^2 + 4}}{2} \]

### Step 8: Differentiate \( f(x) \)
Now, we need to differentiate \( f(x) \):

\[ f'(x) = \frac{1}{2} \left( 1 + \frac{d}{dx}(\sqrt{x^2 + 4}) \right) \]

Using the chain rule on \( \sqrt{x^2 + 4} \):

\[ \frac{d}{dx}(\sqrt{x^2 + 4}) = \frac{1}{2\sqrt{x^2 + 4}} \cdot 2x = \frac{x}{\sqrt{x^2 + 4}} \]

Thus,

\[ f'(x) = \frac{1}{2} \left( 1 + \frac{x}{\sqrt{x^2 + 4}} \right) \]

### Step 9: Compute \( f(100) \) and \( f'(100) \)
Now we compute \( f(100) \):

\[ f(100) = \frac{100 + \sqrt{100^2 + 4}}{2} = \frac{100 + \sqrt{10000 + 4}}{2} = \frac{100 + \sqrt{10004}}{2} \]

And \( f'(100) \):

\[ f'(100) = \frac{1}{2} \left( 1 + \frac{100}{\sqrt{100^2 + 4}} \right) = \frac{1}{2} \left( 1 + \frac{100}{\sqrt{10004}} \right) \]

### Step 10: Compute \( f(100) \cdot f'(100) \)
Now, we compute \( f(100) \cdot f'(100) \):

From our earlier derivation, we know:

\[ f(x) \cdot f'(x) = x \]

Thus,

\[ f(100) \cdot f'(100) = 100 \]

### Final Answer
The value of \( f(100) \cdot f'(100) \) is

\[ \boxed{100} \]