For a polynomial $$g(x)$$ with real coefficients, let mg denote the number of distinct real roots of
$$g(x)$$. Suppose 𝑆 is the set of polynomials with real coefficients defined by

$$S = \left\{(x^{2} − 1)^{2}(a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3}) ∶ a_{0}, a_{1}, a_{2}, a_{3} \epsilon R\right\}$$.

For a polynomial f, let $$f{'}$$ and $$f {'}{'}$$ denote its first and second order derivatives, respectively. Then the minimum possible value of $$(m_{f{'}} + m_{f{'}{'}})$$, where $$f \epsilon S$$, is _____

Let $$f(x)= (x^{2}-1)^{2}\,P(x)$$ where $$P(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}$$ and the coefficients are real.

Step 1: Mandatory real zeros of $$f$$, $$f'$$.
Because of the factor $$(x^{2}-1)^{2}$$ we have $$f(-1)=f(1)=0$$, and the multiplicity of each root is $$2$$. For a root of multiplicity $$k\;(\ge 2)$$ of $$f$$, the derivative $$f'$$ has the same root with multiplicity $$k-1$$. Therefore $$x=-1$$ and $$x=1$$ are simple (multiplicity 1) zeros of $$f'$$. Thus $$f'$$ already has at least the two distinct real roots $$-1$$ and $$1$$.

Step 2: One more root of $$f'$$ inside $$(-1,1)$$.
Since $$f(-1)=f(1)=0$$ and $$-1\neq 1$$, Rolle’s theorem guarantees the existence of some $$c\in(-1,1)$$ with $$f'(c)=0$$. This root $$c$$ is different from $$\pm1$$, hence
$$m_{f'}\ge 3\;.$$ So no matter how we choose the coefficients of $$P(x)$$, $$f'$$ must possess at least three distinct real zeros.

Step 3: At least two real zeros of $$f''$$.
Arrange the three distinct real roots of $$f'$$ in ascending order: $$-1 \lt c \lt 1.$$ Apply Rolle’s theorem again, now to $$f'$$ on the intervals $$[-1,c]$$ and $$[c,1]$$. Because $$f'(-1)=f'(c)=0$$, there exists $$\alpha\in(-1,c)$$ with $$f''(\alpha)=0$$; similarly, $$f'(c)=f'(1)=0$$ gives a $$\beta\in(c,1)$$ with $$f''(\beta)=0$$. These $$\alpha,\beta$$ are distinct, so
$$m_{f''}\ge 2.$$ Thus every polynomial $$f\in S$$ satisfies $$m_{f'}+m_{f''}\ge 3+2 = 5.$$ This establishes the lower bound.

Step 4: A polynomial that attains the bound.
Choose the simplest possibility $$P(x)=1$$, i.e. $$f(x)=(x^{2}-1)^{2}$$.

Compute the derivatives:
$$f'(x)=4x(x^{2}-1)=4x^{3}-4x,$$
so $$f'(x)=0$$ at $$x=-1,\,0,\,1.$$ Hence $$m_{f'}=3.$$

Next,
$$f''(x)=12x^{2}-4=4(3x^{2}-1),$$
and $$f''(x)=0$$ at $$x=\pm\frac1{\sqrt3}.$$ Therefore $$m_{f''}=2.$$

This example achieves $$m_{f'}+m_{f''}=3+2=5,$$ exactly the lower bound found in Step 3. Hence the minimum possible value of $$(m_{f'}+m_{f''})$$ for polynomials in the set $$S$$ is $$\boxed{5}$$.