Question

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

Answer 1

Correct Answer: 5

Given that \( f(x) \) is a thrice differentiable function with specified values at \( x = 0, 1, 2, 3, \) and \( 4 \). The oscillatory nature of these values implies multiple sign changes, suggesting roots within the interval \([0, 4]\).

1. The data indicates \( f(x) \) has at least 4 roots in \([0, 4]\).
2. By Rolle’s Theorem, \( f'(x) \) must have at least 3 roots, given that \( f(x) \) has at least 4 roots.
3. The product \( f'(x)f(x) \) combines roots from \( f(x) \) and \( f'(x) \), resulting in at least 7 sign changes.

\[ (3f'f'' + ff''')(x) = \left((ff'' + (f')^2)(x)\right)' \]

\[ \left((ff'') + (f')^2\right)(x) = \left((ff')(x)\right)' \]

\[ \therefore (3f'f'' + ff''')(x) = \left(f(x) \cdot f'(x)\right)'' \]

\[ \text{min. roots of } f(x) \to 4 \] \[ \therefore \text{min. roots of } f'(x) \to 3 \] \[ \therefore \text{min. roots of } (f(x) \cdot f'(x)) \to 7 \] \[ \therefore \text{min. roots of } (f(x) \cdot f'(x))'' \to 5 \]

Consequently, the expression \( (3f'f'' + ff''')(x) = (f'(x) \cdot f(x))'' \) must have a minimum of 5 roots due to its oscillatory behavior and the order of differentiation.

The minimum number of roots for \( (3f'f'' + ff''')(x) \) is 5.