Let $$f : \mathbb{R} \to \mathbb{R}$$ be a thrice differentiable function such that $$f(0) = 0, f(1) = 1, f(2) = -1, f(3) = 2$$ and $$f(4) = -2$$. Then, the minimum number of zeros of $$(3f'f'' + ff''')(x)$$ is _____

1. Identify the expression as a derivative

Let us define a helper function:

$$h(x) = f(x)f'(x)$$

Differentiating h(x) with respect to x gives:

$$h'(x) = (f'(x))^2 + f(x)f''(x)$$

Let this result be another function g(x):

$$g(x) = f(x)f''(x) + (f'(x))^2$$

Now, differentiate g(x) with respect to x:

$$g'(x) = f'(x)f''(x) + f(x)f'''(x) + 2f'(x)f''(x)$$

$$g'(x) = 3f'(x)f''(x) + f(x)f'''(x)$$

This matches the given expression exactly. Therefore, finding the minimum number of zeros of (3f'f'' + ff''')(x) is equivalent to finding the minimum number of zeros of g'(x).

2. Count the zeros of f(x)

We are given the following values for f(x):

$$f(0) = 0$$

$$f(1) = 1$$

$$f(2) = -1$$

$$f(3) = 2$$

$$f(4) = -2$$

By the Intermediate Value Theorem, since f(x) changes sign between the given intervals, it must cross zero at least once in each of those intervals:

Between x = 1 and x = 2 because f(1) > 0 and f(2) < 0. Let this zero be c_1 \in (1, 2).

Between x = 2 and x = 3 because f(2) < 0 and f(3) > 0. Let this zero be c_2 \in (2, 3).

Between x = 3 and x = 4 because f(3) > 0 and f(4) < 0. Let this zero be c_3 \in (3, 4).

Including f(0) = 0, the function f(x) has at least 4 distinct zeros:

$$x = 0, c_1, c_2, c_3$$

3. Count the zeros of h(x)

Since h(x) = f(x)f'(x), h(x) will be equal to zero whenever either f(x) = 0 or f'(x) = 0.

First, h(x) = 0 at the 4 zeros of f(x):

$$x = 0, c_1, c_2, c_3$$

Second, by Rolle's Theorem, between any two consecutive zeros of a differentiable function, its derivative must become zero at least once. Since f(x) has zeros at 0, c_1, c_2, c_3, its derivative f'(x) must have at least 3 zeros in between them:

One zero d_1 \in (0, c_1)

One zero d_2 \in (c_1, c_2)

One zero d_3 \in (c_2, c_3)

At these points, f'(d_1) = 0, f'(d_2) = 0, and f'(d_3) = 0, which makes h(x) = 0 as well.

Combining both sets of points, h(x) has at least 4 + 3 = 7 distinct zeros ordered as:

$$0 < d_1 < c_1 < d_2 < c_2 < d_3 < c_3$$

4. Apply Rolle's Theorem sequentially

Since h(x) has at least 7 distinct zeros, applying Rolle's Theorem tells us that its derivative h'(x) = g(x) must have at least:

$$7 - 1 = 6 \text{ zeros}$$

Applying Rolle's Theorem once more to g(x), since g(x) has at least 6 distinct zeros, its derivative g'(x) must have at least:

$$6 - 1 = 5 \text{ zeros}$$

Final Answer

The minimum number of zeros of (3f'f'' + ff''')(x) is 5.