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

We are given that \( f(x) \) is a thrice differentiable function with specific values at points \( x = 0, 1, 2, 3, \) and \( 4 \). Based on these values, it appears that \( f(x) \) oscillates, implying multiple sign changes, which indicate roots within the interval.

1. The given values suggest that \( f(x) \) has at least 4 roots within \([0, 4]\).
2. By Rolle’s Theorem, since \( f(x) \) has at least 4 roots, its first derivative \( f'(x) \) must have at least 3 roots.
3. Similarly, \( f'(x)f(x) \), which involves both \( f(x) \) and \( f'(x) \), will have more roots due to the combination of roots from \( f(x) \) and \( f'(x) \). This product results in at least 7 changes in sign.

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

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

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

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

Thus, the expression \( (3f'f'' + ff''')(x) = (f'(x) \cdot f(x))'' \) must have at least 5 roots, given the oscillatory behavior and the higher order of differentiation.

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