Minimum Sum of Distinct Real Roots of Derivatives of Polynomial

To find the minimum possible value of $(m_{f'} + m_{f''})$ for $f \in S$, we analyze the properties of the polynomial $f(x)$ and its derivatives using Rolle's Theorem.

Step 1: Definition of $f(x)$

Any polynomial $f \in S$ is given by: $f(x) = (x^2 - 1)^2 (a_0 + a_1 x + a_2 x^2 + a_3 x^3) = (x - 1)^2 (x + 1)^2 P(x)$ where $P(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$ for some $a_0, a_1, a_2, a_3 \in \mathbb{R}$.

Step 2: Minimum Number of Distinct Real Roots of $f'(x)$

From the structure of $f(x)$, we see that $x = 1$ and $x = -1$ are roots of $f(x)$ with multiplicity at least $2$. This implies:

1. $f(1) = 0$ and $f(-1) = 0$
2. $f'(1) = 0$ and $f'(-1) = 0$

Since $f(x)$ is continuous and differentiable everywhere, by Rolle's Theorem on the interval $[-1, 1]$, there exists at least one $c \in (-1, 1)$ such that: $f'(c) = 0$

Since $-1 < c < 1$, the roots $-1$, $c$, and $1$ are distinct. Therefore, $f'(x)$ must have at least $3$ distinct real roots. $\implies m_{f'} \ge 3$

Step 3: Minimum Number of Distinct Real Roots of $f''(x)$

The derivative $f'(x)$ has distinct real roots at $x = -1$, $x = c$, and $x = 1$ with $-1 < c < 1$.

Applying Rolle's Theorem to $f'(x)$:

1. On the interval $[-1, c]$, there exists at least one $d_1 \in (-1, c)$ such that $f''(d_1) = 0$.
2. On the interval $[c, 1]$, there exists at least one $d_2 \in (c, 1)$ such that $f''(d_2) = 0$.

Since $-1 < d_1 < c < d_2 < 1$, the roots $d_1$ and $d_2$ are distinct. Therefore, $f''(x)$ must have at least $2$ distinct real roots. $\implies m_{f''} \ge 2$

Thus, for any non-zero $f \in S$: $m_{f'} + m_{f''} \ge 3 + 2 = 5$

Step 4: Constructing a Polynomial to Attain the Minimum Value

Consider $a_0 = 1$ and $a_1 = a_2 = a_3 = 0$. Then $P(x) = 1$, which gives $f(x) \in S$: $f(x) = (x^2 - 1)^2 = x^4 - 2x^2 + 1$

Calculating its derivatives:

1. First derivative: $f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)$ The distinct real roots of $f'(x)$ are $x = -1, 0, 1$. $\implies m_{f'} = 3$

2. Second derivative: $f''(x) = 12x^2 - 4 = 4(3x^2 - 1)$ The distinct real roots of $f''(x)$ are $x = \pm \frac{1}{\sqrt{3}}$. $\implies m_{f''} = 2$

For this specific choice of $f(x)$, we have: $m_{f'} + m_{f''} = 3 + 2 = 5$

Conclusion

The minimum possible value of $(m_{f'} + m_{f''})$ for $f \in S$ is 5.