Analyzing the Setup
We are given a thrice differentiable function f(x) with the following values:
f(0)=0, f(1)=1, f(2)=−1, f(3)=2, and f(4)=−2.
Our objective is to determine the minimum number of zeros for the expression (3f′f′′+ff′′′)(x).
The Detective Work
Recognizing the Pattern
At first glance, the expression (3f′f′′+ff′′′)(x) appears complex. However, we can identify a "parent" function by observing the structure of its derivatives. Let us define a function g(x)=f(x)f′(x).
Applying the product rule to
g(x), we find the first derivative:
g′(x)=(f′(x))2+f(x)f′′(x)
Differentiating once more to find
g′′(x), we apply the product rule to both terms:
g′′(x)=2f′(x)f′′(x)+[f′(x)f′′(x)+f(x)f′′′(x)]
Combining the like terms, we arrive at the elegant result:
g′′(x)=3f′(x)f′′(x)+f(x)f′′′(x)
Thus, the expression we are investigating is exactly the second derivative of g(x)=f(x)f′(x).
Mapping the Terrain
The Intermediate Value Theorem
We analyze the roots of f(x) using the Intermediate Value Theorem (IVT). We are given f(0)=0.
Between the given points, the function changes sign as follows:
1. In (1,2), f(x) changes from 1 to −1, implying at least one zero.
2. In (2,3), f(x) changes from −1 to 2, implying at least one zero.
3. In (3,4), f(x) changes from 2 to −2, implying at least one zero.
Including the known zero at x=0, f(x) has at least 4 distinct zeros.
The Chain Reaction
Rolle's Theorem
We now apply Rolle's Theorem to determine the zeros of the derivatives. If f(x) has 4 zeros, then f′(x) must have at least 4−1=3 zeros, located between the consecutive zeros of f(x).
Recall that g(x)=f(x)f′(x). The zeros of g(x) occur whenever f(x)=0 or f′(x)=0.
Since the zeros of f(x) and f′(x) are distinct, g(x) possesses at least 4+3=7 zeros.
The Final Descent
Reaching the Goal
We apply Rolle's Theorem iteratively to the function g(x). Since g(x) has at least 7 zeros, its first derivative g′(x) must have at least 7−1=6 zeros.
Applying Rolle's Theorem once more to g′(x), we conclude that g′′(x) must have at least 6−1=5 zeros.
The expression (3f′f′′+ff′′′)(x) is equivalent to g′′(x). Therefore, the minimum number of zeros for the given expression is 5.