For a polynomial $g(x)$ with real coefficients, let $m_{g}$ denote the number of distinct real roots

Question

For a polynomial $g(x)$ with real coefficients, let $m_{g}$ denote the number of distinct real roots of $g(x)$. Suppose $S$ is the set of polynomials with real coefficients defined by $S=$ $\left\{\left(x^{2}-1\right)^{2}\left(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}\right): a_{0}, a_{1}, a_{2}, a_{3} \in R\right\}$. For a polynomial $f$, let $f^{\prime}$ and $f^{\prime \prime}$ denote its first and second order derivatives, respectively. Then the minimum possible value of $\left(m_{f^{\prime}}+m_{f}{ }^{\prime \prime}\right)$, where $f \in S$, is

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(5) : Let $f(x)=\left(x^{2}-1ight)^{2} P(x)$, where\ $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}$\ As $f(-1)=f(1)=0$\ $	herefore \quad \exists r \in(-1,1)$ such that $f^{\prime}(r)=0$\ (By Rolle's theorem)\ Again $f^{\prime}(x)=0$ has two obvious roots -1 and 1 Thus, $f^{\prime}(x)=0$ has 3 roots i.e.,\ $-1,1, r ;-1<r<1$\ Also, $f^{\prime \prime}(x)=0$ will have at least two roots, say $s$ and $t$ such that $-1<s<r<t<1$\ (By Rolle's theorem)\ Hence, $\min \left(m_{f^{\prime}}ight)=3, \min \left(m_{f^{\prime \prime}}ight)=2$\ The minimum possible value of $m_{f^{\prime}}+m_{f^{\prime \prime}}$ is 5 .\