The peak of a decaying pulse

For $x\ge 0$ the function $f(x)=x^3 e^{-x}$ rises, peaks, then decays to $0$ . What is its maximum value? Give a decimal to four places.

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Product rule: $f'(x)=e^{-x}(3x^2-x^3)=x^2 e^{-x}(3-x)$ .
The peak is at $x=3$ (not $x=0$ , where $f=0$ ); report $f(3)=27e^{-3}$ , not the location.

Answer

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1.3443 (± 0.001)

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Asked at: Game-Based Aptitude, ETF Market-Making