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

e^{- 3}

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3 e^{- 3}

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27 e^{- 9}

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\infty

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Solution

The correct option is A $e^{- 3}$
Given,
$f (x) = x^{3} e^{3 x}, x > 0$

On differentiating w.r.t $x$ we get
$f^{'} (x) = 3 x^{2} e^{- 3 x} + x^{3} e^{- 3 x} (- 3)$
$= x^{2} 3 e^{- 3 x} (1 - x)$
$= 3 e^{- 3 x} (x^{2} - x^{3})$

For maxima and minima, put $f^{'} (x) = 0$
$\Rightarrow 3 x^{2} 3 e^{- 3 x} (1 - x) = 0$
$\Rightarrow x = 0, 1$

Now,
$f^{''} (x) = 3 e^{- 3 x} (2 x - 3 x^{2}) - 9 e^{- 3 x} (x^{2} - x^{3})$
$= 3 e^{- 3 x} (3 x^{3} - 6 x^{2} + 2 x)$

At $x = 1$ , $f^{''} (1) = 3 e^{- 3 x} (- 1) < 0$ maxima.
$∴$ It is maximum at $x = 1$

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Let f(x)=x3e−3x,x>0. Then the maximum value of f(x) is

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