Let $f(x)=(x^2-1)^2P(x),$ where $P(x)=a_0+a_1x+a_2x^2+a_3x^3$ is any real polynomial of degree at most $3$.

We must find the minimum possible value of $m_{f'}+m_{f''},$ where $m_g$ denotes the number of distinct real roots of $g$.

1. First understand the structure of $f$

Since $f(x)=(x-1)^2(x+1)^2P(x),$ we see that $x=1$ and $x=-1$ are always roots of $f$, each of multiplicity at least $2$.

Therefore:

• $f'(1)=f'(-1)=0$,
• so $f'$ has at least two distinct real roots: $1$ and $-1$.

Hence, $m_{f'}\ge 2.$

2. Use Rolle's theorem on $f'$

Since $f'$ has distinct real roots at $-1$ and $1$, Rolle's theorem applied to $f'$ on $[-1,1]$ gives at least one real root of $f''$ in $(-1,1)$.

Thus, $m_{f''}\ge 1.$

So immediately, $m_{f'}+m_{f''}\ge 2+1=3.$

But this is not enough; we need the exact minimum.

3. Compute derivatives in a useful form

Let $Q(x)=(x^2-1)^2=x^4-2x^2+1.$ Then $f=QP.$ So $f'=Q'P+QP',$ with $Q'(x)=4x(x^2-1)=4x(x-1)(x+1).$ Hence $f'(x)=(x^2-1)\Big(4xP(x)+(x^2-1)P'(x)\Big).$

Therefore $f'$ always has factors $(x-1)$ and $(x+1)$, confirming the roots $\pm1$.

Let $R(x)=4xP(x)+(x^2-1)P'(x).$ Then $f'(x)=(x^2-1)R(x).$ Since $P$ has degree at most $3$, $R$ has degree at most $4$.

Now differentiate again: $f''(x)=\frac{d}{dx}\big((x^2-1)R(x)\big)=2xR(x)+(x^2-1)R'(x).$

In particular, evaluate at $x=\pm1$: $f''(1)=2R(1),\qquad f''(-1)=-2R(-1).$ But $R(1)=4P(1),\qquad R(-1)=-4P(-1),$ because $(x^2-1)P'(x)$ vanishes at $x=\pm1$. Thus $f''(1)=8P(1),\qquad f''(-1)=8P(-1).$

So:

• if $P(1)=0$, then $1$ is also a root of $f''$;
• if $P(-1)=0$, then $-1$ is also a root of $f''$.

4. Try to make the total as small as possible

To minimize $m_{f'}+m_{f''}$, we want:

1. as few real roots of $f'$ as possible,
2. as few real roots of $f''$ as possible.

Since $f'$ already has roots at $\pm1$, the best hope is $m_{f'}=2.$ Then $f''$ must have at least one real root between $-1$ and $1$. Could we make $m_{f''}=1?$ If yes, total would be $3$.

But there is a stronger obstruction.

5. Show that $f''$ must have at least 3 distinct real roots if $m_{f'}=2$

Assume we choose $P$ so that $f'$ has exactly the two real roots $-1$ and $1$. Then these are consecutive real roots of $f'$, so by Rolle there is at least one root of $f''$ in $(-1,1)$.

Now examine behavior outside this interval.

Because $f'(x)=(x^2-1)R(x),$ and $f'$ has no real roots other than $\pm1$, the polynomial $R(x)$ has no real roots. Therefore $R$ has constant sign on $\mathbb R$.

Hence for $|x|>1$, since $x^2-1>0$, the sign of $f'(x)$ is the same constant sign on both intervals $(-\infty,-1)$ and $(1,\infty)$. Inside $(-1,1)$, since $x^2-1<0$, the sign of $f'(x)$ is the opposite constant sign. Thus the sign pattern of $f'$ is:

• one constant sign on $(-\infty,-1)$,
• opposite sign on $(-1,1)$,
• original sign again on $(1,\infty)$.

So $f'$ changes sign at both $x=-1$ and $x=1$.

Therefore:

• on $(-\infty,-1)$, $f'$ must go from its limiting sign at $-\infty$ to $0$ at $-1$ with a sign change there, forcing an extremum of $f'$ in $(-\infty,-1)$, hence a root of $f''$ there;
• on $(-1,1)$, Rolle already gives one root of $f''$;
• on $(1,\infty)$, similarly there must be another root of $f''$.

Hence, $m_{f''}\ge 3.$

So if $m_{f'}=2$, then $m_{f'}+m_{f''}\ge 2+3=5.$

6. Could taking more roots for $f'$ reduce the sum?

If $m_{f'}\ge 3$, then at best $m_{f''}\ge 2$ by Rolle between consecutive roots of $f'$. So $m_{f'}+m_{f''}\ge 3+2=5.$

If $m_{f'}\ge 4$, the sum is even larger.

Therefore for all $f\in S$, $m_{f'}+m_{f''}\ge 5.$

7. Construct an example attaining 5

We now need an example with equality. Take $P(x)=1.$ Then $f(x)=(x^2-1)^2.$

Compute derivatives: $f'(x)=4x(x^2-1)=4x(x-1)(x+1).$ So the distinct real roots of $f'$ are $-1,0,1,$ thus $m_{f'}=3.$

Next, $f''(x)=12x^2-4=4(3x^2-1).$ Its distinct real roots are $x=\pm\frac{1}{\sqrt3},$ thus $m_{f''}=2.$

Therefore, $m_{f'}+m_{f''}=3+2=5.$

So the minimum possible value is $\boxed{5}.$

8. Comparison with stored answer

Stored correct answer: $5$.

Our derived answer matches it.