WBJEE · Maths · Application of Derivatives
Let \(f(x)=x^3 e^{-3 x}, x > 0\). Then the maximum value of \(f(x)\) is
- A \(\mathrm{e}^{-3}\)
- B \(3 \mathrm{e}^{-3}\)
- C \(27 \mathrm{e}^{-9}\)
- D \(\infty\)
Answer & Solution
Correct Answer
(A) \(\mathrm{e}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Hints: } f(x)=x^3 \cdot e^{-3 x} \\ & =f^{\prime}(x)=3 x^2 e^{-3 x}+x^3 e^{-3 x}(-3) \\ & =x^2 3 e^{-3 x}[1-x]=0, x=1 \\ & \text { Maximum at } x=1 \\ & f(1)=e^{-3} \end{aligned}\)
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