答案：(a) $\dfrac{dy}{dx} = e^{-x}(2x^2-11x+7)$, $\dfrac{d^2y}{dx^2}=e^{-x}(-2x^2+15x-18)$ (b) 同意
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$\begin{array}{rcl}
y & = & (7x-2x^2)e^{-x} \\
\dfrac{dy}{dx} & = & (7x-2x^2) (-e^{-x}) + e^{-x}(7-4x) \\
\dfrac{dy}{dx} & = &e^{-x}(2x^2-7x-4x+7) \\
\dfrac{dy}{dx} & = & e^{-x}(2x^2-11x+7)
\end{array}$$\begin{array}{rcl}
\dfrac{dy}{dx} & = & e^{-x}(2x^2-11x+7) \\
\dfrac{d^2y}{dx^2} & = & e^{-x}(4x-11)+(2x^2-11x+7)(-e^{-x}) \\
\dfrac{d^2y}{dx^2} & = & e^{-x}(-2x^2+11x-7+4x-11) \\
\dfrac{d^2y}{dx^2} & = & e^{-x}(-2x^2+15x-18) \\
\dfrac{d^2y}{dx^2} & = & -e^{-x}(2x-3)(x-6)
\end{array}$ - 對於圖像的拐點，
$\begin{array}{rcl}
\dfrac{d^2y}{dx^2} & = & 0 \\
-e^{-x}(2x-3)(x-6) & = & 0
\end{array}$$\therefore x=\dfrac{3}{2}$ 或 $x=6$。
$\begin{array}{|l|c|c|c|c|c|} \hline
x & x< \dfrac{3}{2} & x=\dfrac{3}{2} & \dfrac{3}{2} < x < 6 & x=6 & x > 6 \\ \hline
\dfrac{d^2y}{dx^2} & -ve & 0 & +ve & 0 & -ve \\ \hline
y& \text{向下凹} & \text{拐點} & \text{向上凹} & \text{拐點} & \text{向下凹} \\ \hline
\end{array}$所以，該圖像有兩個拐點。
由此，同意該宣稱。
