Let f(x) = x^3 e^{-3x}, x 0. Then the maxim | vedclass

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(A) Given the function $f(x) = x^3 e^{-3x}$ for $x > 0$. To find the maximum value, we first find the derivative $f'(x)$ using the product rule: $f'(x

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$e^{-3}$

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(A) Given the function $f(x) = x^3 e^{-3x}$ for $x > 0$. To find the maximum value, we first find the derivative $f'(x)$ using the product rule: $f'(x) = \frac{d}{dx}(x^3) \cdot e^{-3x} + x^3 \cdot \frac{d}{dx}(e^{-3x})$ $f'(x) = 3x^2 e^{-3x} + x^3 (-3 e^{-3x})$ $f'(x) = 3x^2 e^{-3x} (1 - x)$ Setting $f'(x) = 0$ for critical points: $3x^2 e^{-3x} (1 - x) = 0$ Since $x > 0$ and $e^{-3x} 
eq 0$, we have $1 - x = 0$, which gives $x = 1$. To confirm it is a maximum, we check the sign of $f'(x)$ around $x = 1$. For $x  0$ and for $x > 1$, $f'(x) Thus, $f(x)$ has a local maximum at $x = 1$. The maximum value is $f(1) = (1)^3 e^{-3(1)} = e^{-3}$.

Let $f(x) = x^3 e^{-3x}, x > 0$. Then the maximum value of $f(x)$ is

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