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.

Answer 2

Given \( f: \mathbb{R} \to \mathbb{R} \), thrice differentiable, with data:
\[ f(0)=0, f(1)=1, f(2)=-1, f(3)=2, f(4)=-2 \] We want to find the minimum number of zeros of \( g(x) = 3f'(x)^2 + f''(x) \).
Step 1: Use Rolle's theorem on \( f \)
As \( f(x) \) has values:
\[ f(0) = 0, f(1) = 1, f(2) = -1, f(3) = 2, f(4) = -2, \] there are changes of signs, so there exist at least 4 points where \( f(x) \) crosses some values.
There must be at least 4 points \( c_1, c_2, c_3, c_4 \) in \( (0,4) \) where \( f'(c_i) = 0 \) by Rolle's theorem, because \( f \) passes through these given values.
Step 2: Consider zeros of \( g(x) = 3 f'^2 + f''(x) \)
Note, zeros of \( g \) correspond to solutions of:
\[ 3 (f'(x))^2 + f''(x) = 0 \] Rewrite as: \[ f''(x) = -3 (f'(x))^2 \]
Differentiate \( g(x) \) to find its zeros, or observe the pattern that zeros of \( f' \) lead to zeros of \( g \).
There are at least 4 zeros of \( f'(x) \). At each zero \( f'(x) = 0 \), \( g(x) = f''(x) \). Using Rolle's theorem on \( g \) and \( f' \), there must be at least one zero between each zero of \( f' \) for \( g \). This means at least 5 zeros.
Step 3: Use Rolle's theorem on \( g(x) \)
Given \( g(x) = 3f'^2 + f''(x) \), between 4 zeros of \( f' \), there will be at least 3 zeros of \( f''(x) \), and via linearity argument, \( g(x) \) is continuous and differentiable.
So the minimal number of zeros of \( g \) is 5.
Final answer: minimum number of zeros of \( (3f'^2 + f'')(x) \) is 5