Problem source

Find the three values of $x$ for which the derivative of $x^2 \e^{-x^2}$ is zero. Given that $a$ and $b$ are distinct positive numbers, find a polynomial $\P(x)$ such that the derivative of $\P(x)\e^{-x^2}$ is zero for $x=0$, $x=\pm a$ and $x=\pm b\,$, but for no other values of $x$.

Solution source

\begin{align*}
&& y &= x^2e^{-x^2} \\
\Rightarrow && y' &= 2xe^{-x^2} +x^2 \cdot (-2x)e^{-x^2} \\
&&&= e^{-x^2}(2x-2x^3) \\
&&&= 2e^{-x^2}x(1-x^2)
\end{align*}

Therefore the derivative is zero iff $x = 0, \pm 1$

\begin{align*}
&& y &= \P(x) e^{-x^2} \\
\Rightarrow && y' &= e^{-x^2} (\P'(x)-2x\P(x))
\end{align*}

Therefore we want $\P'(x) - 2x\P(x) = Kx(x^2-a^2)(x^2-b^2)$

Since this has degree $5$, we should look at polynomials degree $4$ for $\P$. We can also immediately see that $0$ is a root of $\P'(x)$, so $\P(x) = a_4x^4+a_3x^3+a_2x^2+a_0$. WLOG $a_4 = 1$ and $K = -2$, so 

\begin{align*}
&& -2(x^5-(a^2+b^2)x^3+a^2b^2x) &= 4x^3+3a_3x^2+2a_2x- 2x(x^4+a_3x^3+a_2x^2+a_0) \\
&&&= -2x^5-2a_3 x^4+(4-2a_2)x^3+(2a_2-2a_0)x \\
\Rightarrow && a_3 &= 0 \\
&& a^2+b^2 &= 2-a_2 \\
\Rightarrow && a_2 &= 2-a^2-b^2 \\
&& a^2b^2 &= a_0-a_2 \\
\Rightarrow && a_0 &= a^2b^2 + 2-a^2-b^2 \\
\Rightarrow && \P(x) &= x^4+(2-a^2-b^2)x^2+(a^2-1)(b^2-1)x
\end{align*}