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) equal to :

Options

A. `-2`
B. `-1`
C. 0
D. 4

Answer

To solve the problem, we need to evaluate the function \( f(1) \) given the expression for \( f(x) \):

\[ f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3) \]

### Step 1: Differentiate \( f(x) \)

First, we differentiate \( f(x) \) to find \( f'(x) \):

\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2 f'(1)) + \frac{d}{dx}(x f''(2)) + \frac{d}{dx}(f'''(3)) \]

Calculating each term:

1. \( \frac{d}{dx}(x^3) = 3x^2 \)
2. \( \frac{d}{dx}(x^2 f'(1)) = 2x f'(1) \) (since \( f'(1) \) is a constant)
3. \( \frac{d}{dx}(x f''(2)) = f''(2) \) (since \( f''(2) \) is a constant)
4. \( \frac{d}{dx}(f'''(3)) = 0 \) (since \( f'''(3) \) is a constant)

So, we have:

\[ f'(x) = 3x^2 + 2x f'(1) + f''(2) \]

### Step 2: Differentiate \( f'(x) \) to find \( f''(x) \)

Now, we differentiate \( f'(x) \) to find \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x f'(1)) + \frac{d}{dx}(f''(2)) \]

Calculating each term:

1. \( \frac{d}{dx}(3x^2) = 6x \)
2. \( \frac{d}{dx}(2x f'(1)) = 2 f'(1) \) (since \( f'(1) \) is a constant)
3. \( \frac{d}{dx}(f''(2)) = 0 \) (since \( f''(2) \) is a constant)

So, we have:

\[ f''(x) = 6x + 2f'(1) \]

### Step 3: Differentiate \( f''(x) \) to find \( f'''(x) \)

Now, we differentiate \( f''(x) \) to find \( f'''(x) \):

\[ f'''(x) = \frac{d}{dx}(6x) + \frac{d}{dx}(2f'(1)) \]

Calculating each term:

1. \( \frac{d}{dx}(6x) = 6 \)
2. \( \frac{d}{dx}(2f'(1)) = 0 \) (since \( f'(1) \) is a constant)

So, we have:

\[ f'''(x) = 6 \]

### Step 4: Substitute values to find \( f(1) \)

Now we can substitute \( x = 1 \) into \( f(x) \):

\[ f(1) = 1^3 + 1^2 f'(1) + 1 f''(2) + f'''(3) \]

Calculating each term:

1. \( 1^3 = 1 \)
2. \( 1^2 f'(1) = f'(1) \)
3. \( 1 f''(2) = f''(2) \)
4. \( f'''(3) = 6 \)

Now we need to find \( f'(1) \) and \( f''(2) \):

### Step 5: Find \( f'(1) \)

Substituting \( x = 1 \) into \( f'(x) \):

\[ f'(1) = 3(1^2) + 2(1)f'(1) + f''(2) \]

This simplifies to:

\[ f'(1) = 3 + 2f'(1) + f''(2) \]

Rearranging gives:

\[ f'(1) - 2f'(1) = 3 + f''(2) \implies -f'(1) = 3 + f''(2) \implies f'(1) = -3 - f''(2) \]

### Step 6: Find \( f''(2) \)

Substituting \( x = 2 \) into \( f''(x) \):

\[ f''(2) = 6(2) + 2f'(1) = 12 + 2f'(1) \]

### Step 7: Substitute \( f'(1) \) into \( f''(2) \)

Substituting \( f'(1) = -3 - f''(2) \) into \( f''(2) \):

\[ f''(2) = 12 + 2(-3 - f''(2)) \]

This simplifies to:

\[ f''(2) = 12 - 6 - 2f''(2) \implies 3f''(2) = 6 \implies f''(2) = 2 \]

### Step 8: Substitute \( f''(2) \) back to find \( f'(1) \)

Now we can find \( f'(1) \):

\[ f'(1) = -3 - f''(2) = -3 - 2 = -5 \]

### Step 9: Substitute \( f'(1) \) and \( f''(2) \) back to find \( f(1) \)

Now we can substitute \( f'(1) \) and \( f''(2) \) back into \( f(1) \):

\[ f(1) = 1 + (-5) + 2 + 6 = 1 - 5 + 2 + 6 = 4 \]

### Final Answer

Thus, the value of \( f(1) \) is:

\[ \boxed{4} \]