If `f : R to R` is a function such that `f(x) = x^3 + x^2 f'(1) + x f '' (2) + f ' ' '(3) `for all `x in R,` then `f (2) -f (1)` is

Options

A. `f(0)`
B. `-f(0)`
C. `f'(0)`
D. `-f'(0)`

Answer

To solve the problem, we need to analyze the function given and compute \( f(2) - f(1) \).

### Step 1: Understand the function
We have:
\[ f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3) \]
This function is defined for all \( x \in \mathbb{R} \).

### Step 2: Differentiate the function
We will differentiate \( f(x) \) 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:
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)

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

### Step 3: Evaluate \( f'(1) \)
Now, we substitute \( x = 1 \) 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:
\[ f'(1) - 2f'(1) = 3 + f''(2) \]
\[ -f'(1) = 3 + f''(2) \]
Thus:
\[ f'(1) = -3 - f''(2) \] (Equation 1)

### Step 4: Differentiate again to find \( f''(x) \)
Now, we differentiate \( f'(x) \):
\[ f''(x) = \frac{d}{dx}(3x^2 + 2x f'(1) + f''(2)) \]
Calculating each term:
1. \( \frac{d}{dx}(3x^2) = 6x \)
2. \( \frac{d}{dx}(2x f'(1)) = 2f'(1) \)
3. \( \frac{d}{dx}(f''(2)) = 0 \)

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

### Step 5: Evaluate \( f''(2) \)
Substituting \( x = 2 \):
\[ f''(2) = 6(2) + 2f'(1) = 12 + 2f'(1) \] (Equation 2)

### Step 6: Substitute Equation 1 into Equation 2
Now we substitute \( f'(1) = -3 - f''(2) \) into \( f''(2) = 12 + 2f'(1) \):
\[ f''(2) = 12 + 2(-3 - f''(2)) \]
This simplifies to:
\[ f''(2) = 12 - 6 - 2f''(2) \]
\[ 3f''(2) = 6 \]
Thus:
\[ f''(2) = 2 \]

### Step 7: Substitute back to find \( f'(1) \)
Now substituting \( f''(2) = 2 \) back into Equation 1:
\[ f'(1) = -3 - 2 = -5 \]

### Step 8: Find \( f(x) \)
Now we can substitute \( f'(1) \) and \( f''(2) \) back into the original function:
\[ f(x) = x^3 + x^2(-5) + x(2) + 6 \]
This simplifies to:
\[ f(x) = x^3 - 5x^2 + 2x + 6 \]

### Step 9: Calculate \( f(1) \) and \( f(2) \)
Now we compute \( f(1) \) and \( f(2) \):
1. \( f(1) = 1^3 - 5(1^2) + 2(1) + 6 = 1 - 5 + 2 + 6 = 4 \)
2. \( f(2) = 2^3 - 5(2^2) + 2(2) + 6 = 8 - 20 + 4 + 6 = -2 \)

### Step 10: Find \( f(2) - f(1) \)
Finally, we compute:
\[ f(2) - f(1) = -2 - 4 = -6 \]

Thus, the final answer is:
\[ \boxed{-6} \]