A differentiable function f (x) satisfies the condition `f(x+y) =f(x) + f(y) + xy` and `lim_(h to 0) 1/h f(h) = 3` then f is:

Options

A. linear
B. `f(x) =3x + x^(2)/2`
C. `f(x) = 3x + x^(2)`
D. none of these

Answer

To find the function \( f(x) \) that satisfies the given conditions, we can follow these steps:

### Step 1: Use the functional equation
We start with the functional equation given:
\[ f(x+y) = f(x) + f(y) + xy \]
Let's substitute \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0) = f(0) + f(0) + 0 \cdot 0 \]
This simplifies to:
\[ f(0) = 2f(0) \]
From this, we can deduce that:
\[ f(0) = 0 \]

### Step 2: Differentiate the functional equation
Next, we differentiate the functional equation with respect to \( y \):
\[ \frac{d}{dy} f(x+y) = \frac{d}{dy} \left( f(x) + f(y) + xy \right) \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ f'(x+y) = f'(y) + x \]
Now, let’s set \( y = 0 \):
\[ f'(x+0) = f'(0) + x \]
This simplifies to:
\[ f'(x) = f'(0) + x \]

### Step 3: Integrate to find \( f(x) \)
Now we can integrate \( f'(x) \):
\[ f(x) = \int (f'(0) + x) \, dx = f'(0)x + \frac{x^2}{2} + C \]
Since we know \( f(0) = 0 \), we can substitute \( x = 0 \) into our equation:
\[ f(0) = f'(0) \cdot 0 + \frac{0^2}{2} + C = 0 \implies C = 0 \]
Thus, we have:
\[ f(x) = f'(0)x + \frac{x^2}{2} \]

### Step 4: Use the limit condition
We are given the limit condition:
\[ \lim_{h \to 0} \frac{1}{h} f(h) = 3 \]
Substituting \( f(h) \):
\[ \lim_{h \to 0} \frac{1}{h} \left( f'(0)h + \frac{h^2}{2} \right) = 3 \]
This simplifies to:
\[ \lim_{h \to 0} \left( f'(0) + \frac{h}{2} \right) = 3 \]
As \( h \to 0 \), the term \( \frac{h}{2} \) goes to 0, so we have:
\[ f'(0) = 3 \]

### Step 5: Substitute \( f'(0) \) back into \( f(x) \)
Now substituting \( f'(0) = 3 \) back into the equation for \( f(x) \):
\[ f(x) = 3x + \frac{x^2}{2} \]

### Final Result
Thus, the function \( f(x) \) is:
\[ f(x) = 3x + \frac{x^2}{2} \]