Let `f(x)=1/2[f(xy)+f(x/y)] " for " x,y in R^(+)` such that f(1)=0,f'(1)=2.`
f'(3) is equal to

Options

A. `1/3`
B. `2/3`
C. `1/2`
D. `1/4`

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript:

### Step 1: Understand the Functional Equation
We are given the functional equation:
\[ f(x) = \frac{1}{2} [f(xy) + f\left(\frac{x}{y}\right)] \]
for \( x, y \in \mathbb{R}^+ \).

### Step 2: Simplify the Functional Equation
By multiplying both sides of the equation by 2, we can rewrite it as:
\[ 2f(x) = f(xy) + f\left(\frac{x}{y}\right) \]

### Step 3: Assume a Form for \( f(x) \)
To solve this functional equation, we can assume a logarithmic form for \( f(x) \):
\[ f(x) = k \log x \]
where \( k \) is a constant to be determined.

### Step 4: Substitute the Assumed Form into the Functional Equation
Substituting \( f(x) = k \log x \) into our simplified functional equation:
\[ 2(k \log x) = k \log(xy) + k \log\left(\frac{x}{y}\right) \]
Using the properties of logarithms, we can simplify the right-hand side:
\[ k \log(xy) = k (\log x + \log y) \quad \text{and} \quad k \log\left(\frac{x}{y}\right) = k (\log x - \log y) \]
Thus, we have:
\[ k \log(xy) + k \log\left(\frac{x}{y}\right) = k (\log x + \log y + \log x - \log y) = 2k \log x \]
This shows that both sides are equal, confirming our assumption.

### Step 5: Use Given Conditions to Find \( k \)
We are given two conditions:
1. \( f(1) = 0 \)
2. \( f'(1) = 2 \)

Using the first condition:
\[ f(1) = k \log(1) = k \cdot 0 = 0 \]
This condition is satisfied for any \( k \).

Now using the second condition, we first find the derivative:
\[ f'(x) = \frac{d}{dx}(k \log x) = \frac{k}{x} \]
Setting \( x = 1 \):
\[ f'(1) = \frac{k}{1} = k \]
Given \( f'(1) = 2 \), we have:
\[ k = 2 \]

### Step 6: Find \( f'(3) \)
Now we can find \( f'(3) \):
\[ f'(x) = \frac{2}{x} \]
Thus,
\[ f'(3) = \frac{2}{3} \]

### Final Answer
\[ f'(3) = \frac{2}{3} \]

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