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Using the second derivative test on $y=3xe^{-x}$, the stationary point is:
Aa minimum at $x=1$, $y=\frac{3}{e}$
Ba maximum at $x=0$, $y=0$
Ca maximum at $x=1$, $y=\frac{3}{e}$
Da point of inflexion at $x=1$
Answer & Solution
Correct answer: C. a maximum at $x=1$, $y=\frac{3}{e}$
1. Differentiate by the product rule: $\dfrac{dy}{dx}=3e^{-x}-3xe^{-x}=3e^{-x}(1-x)$.
2. Setting $\dfrac{dy}{dx}=0$ and noting $e^{-x}\neq 0$ gives $x=1$.
3. Then $y=3(1)e^{-1}=\dfrac{3}{e}$.
4. Second derivative: $\dfrac{d^2y}{dx^2}=-6e^{-x}+3xe^{-x}$. At $x=1$ this is $-6e^{-1}+3e^{-1}=-3e^{-1}<0$.
5. A negative second derivative means a maximum. Writing $y'=3xe^{-x}+3e^{-x}$ loses the minus sign and moves the stationary point.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 10: Differentiation_
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