Let f(x) be continuous and differentiable function for all reals and f(x + y) = f(x) - 3xy + ff(y). If `lim_(h to 0)(f(h))/(h) = 7`, then the value of f'(x) is

Options

A. `-3x`
B. 7
C. `-3x + 7`
D. `2f(x) + 7`

Answer

To solve the problem, we start with the given functional equation and the limit condition.

### Step 1: Analyze the functional equation
We have the equation:
\[ f(x + y) = f(x) - 3xy + f(y) \]
This suggests that \( f(x) \) might be a polynomial function.

### Step 2: Differentiate the functional equation
We will differentiate both sides of the equation with respect to \( y \):
\[ \frac{d}{dy}[f(x + y)] = \frac{d}{dy}[f(x) - 3xy + f(y)] \]

Using the chain rule on the left side:
\[ f'(x + y) \]

On the right side, we differentiate term by term:
- The derivative of \( f(x) \) with respect to \( y \) is \( 0 \) since \( f(x) \) is constant with respect to \( y \).
- The derivative of \( -3xy \) with respect to \( y \) is \( -3x \).
- The derivative of \( f(y) \) with respect to \( y \) is \( f'(y) \).

Thus, we have:
\[ f'(x + y) = -3x + f'(y) \]

### Step 3: Set \( y = 0 \)
Now, we will set \( y = 0 \) in the differentiated equation:
\[ f'(x + 0) = -3x + f'(0) \]
This simplifies to:
\[ f'(x) = -3x + f'(0) \]

### Step 4: Identify \( f'(0) \)
Let \( f'(0) = c \), where \( c \) is a constant. Thus, we can express \( f'(x) \) as:
\[ f'(x) = -3x + c \]

### Step 5: Use the limit condition
We are given:
\[ \lim_{h \to 0} \frac{f(h)}{h} = 7 \]
This limit represents \( f'(0) \):
\[ f'(0) = 7 \]

### Step 6: Substitute \( c \) back into the derivative
Now substituting \( c = 7 \) into our expression for \( f'(x) \):
\[ f'(x) = -3x + 7 \]

### Conclusion
Thus, the value of \( f'(x) \) is:
\[ \boxed{-3x + 7} \]