Polynomial Function Satisfying Differential Relation

To find the required value, we first determine the polynomial function $f(x)$.

Step 1: Determine the degree of the polynomial

Let $f(x)$ be a real polynomial of degree $n$. Then:

• Degree of $f'(x)$ is $n - 1$
• Degree of $f''(x)$ is $n - 2$

Given the differential relation: $f(x) = f'(x) f''(x)$

Comparing the degrees on both sides: $\text{deg}(f) = \text{deg}(f') + \text{deg}(f'')$ $n = (n - 1) + (n - 2)$ $n = 2n - 3 \implies n = 3$

Thus, $f(x)$ is a polynomial of degree $3$.

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Step 2: Find the coefficients of $f(x)$

Let $f(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$.

Given $f(0) = 0$, we immediately get $d = 0$. So, $f(x) = ax^3 + bx^2 + cx$.

The derivatives are: $f'(x) = 3ax^2 + 2bx + c$ $f''(x) = 6ax + 2b$

Substitute $f(x)$, $f'(x)$, and $f''(x)$ into $f(x) = f'(x) f''(x)$: $ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b)$ $ax^3 + bx^2 + cx = 18a^2 x^3 + 18ab x^2 + (4b^2 + 6ac)x + 2bc$

Equating the coefficients of corresponding powers of $x$:

1. Coefficient of $x^3$: $a = 18a^2 \implies a = \frac{1}{18} \quad (\text{since } a \neq 0)$

2. Coefficient of $x$: $c = 4b^2 + 6ac$ Substitute $a = \frac{1}{18}$: $c = 4b^2 + 6\left(\frac{1}{18}\right)c \implies c = 4b^2 + \frac{c}{3} \implies \frac{2}{3}c = 4b^2 \implies c = 6b^2$

3. Constant term: $2bc = 0$ Substitute $c = 6b^2$: $2b(6b^2) = 0 \implies 12b^3 = 0 \implies b = 0$

Since $b = 0$, we have $c = 6(0)^2 = 0$.

Therefore, the function is: $f(x) = \frac{1}{18} x^3$

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Step 3: Evaluate the required terms

1. First derivative at $x = 2$: $f'(x) = \frac{1}{6} x^2 \implies f'(2) = \frac{1}{6}(2)^2 = \frac{4}{6} = \frac{2}{3}$

2. Second derivative at $x = 2$: $f''(x) = \frac{1}{3} x \implies f''(2) = \frac{1}{3}(2) = \frac{2}{3}$

3. Definite integral: $\int_0^2 f(x) \, dx = \int_0^2 \frac{1}{18} x^3 \, dx = \left[ \frac{x^4}{72} \right]_0^2 = \frac{16}{72} = \frac{2}{9}$

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Step 4: Calculate the final value

Substitute these values into the given expression: $36 \left( f'(2) + f''(2) + \int_0^2 f(x) \, dx \right) = 36 \left( \frac{2}{3} + \frac{2}{3} + \frac{2}{9} \right)$

$= 36 \left( \frac{6 + 6 + 2}{9} \right) = 36 \left( \frac{14}{9} \right) = 4 \times 14 = 56$

Thus, the correct option is C.