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

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.