Let `f:RtoR` be a function such that `f(x)=x^(3)+x^(2)f'(1)+xf''(2)+f'''(3)` for `x in R`
What is `f'(1)` is equal to ?

Options

A. A) `-5`
B. B) `-1`
C. C) `1`
D. D) `0`

Answer

To find \( f'(1) \) for the function defined as \( f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3) \), we will follow these steps:

### Step 1: Differentiate \( f(x) \)
We start by differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2 f'(1)) + \frac{d}{dx}(x f''(2)) + \frac{d}{dx}(f'''(3)) \]
Calculating each term:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( x^2 f'(1) \) is \( 2x f'(1) \) (since \( f'(1) \) is a constant).
- The derivative of \( x f''(2) \) is \( f''(2) \) (since \( f''(2) \) is also a constant).
- The derivative of \( f'''(3) \) is \( 0 \) (as it is a constant).

Thus, we have:
\[ f'(x) = 3x^2 + 2x f'(1) + f''(2) \]

### Step 2: Evaluate \( f'(1) \)
Now we substitute \( x = 1 \) into the expression for \( f'(x) \):
\[ f'(1) = 3(1)^2 + 2(1) f'(1) + f''(2) \]
This simplifies to:
\[ f'(1) = 3 + 2 f'(1) + f''(2) \]

### Step 3: Rearrange the equation
We can rearrange the equation to isolate \( f'(1) \):
\[ f'(1) - 2 f'(1) = 3 + f''(2) \]
\[ -f'(1) = 3 + f''(2) \]
\[ f'(1) = -3 - f''(2) \]

### Step 4: Find \( f''(2) \)
Next, we need to find \( f''(x) \) by differentiating \( f'(x) \):
\[ f''(x) = \frac{d}{dx}(3x^2 + 2x f'(1) + f''(2)) \]
Calculating each term:
- The derivative of \( 3x^2 \) is \( 6x \).
- The derivative of \( 2x f'(1) \) is \( 2 f'(1) \) (since \( f'(1) \) is constant).
- The derivative of \( f''(2) \) is \( 0 \).

Thus, we have:
\[ f''(x) = 6x + 2 f'(1) \]

Now, substituting \( x = 2 \) into \( f''(x) \):
\[ f''(2) = 6(2) + 2 f'(1) = 12 + 2 f'(1) \]

### Step 5: Substitute \( f''(2) \) back into the equation for \( f'(1) \)
Now we substitute \( f''(2) \) back into the equation we derived for \( f'(1) \):
\[ f'(1) = -3 - (12 + 2 f'(1)) \]
This simplifies to:
\[ f'(1) = -3 - 12 - 2 f'(1) \]
\[ f'(1) + 2 f'(1) = -15 \]
\[ 3 f'(1) = -15 \]
\[ f'(1) = -5 \]

### Final Answer
Thus, the value of \( f'(1) \) is:
\[ \boxed{-5} \]