Find the greatest value of $f (x) = x e^{- x}$ on $[0, \infty] .$

0

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1

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e

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\frac{1}{e}

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Solution

The correct option is D $\frac{1}{e}$
$f (x) = x e^{- x}$
$\Rightarrow f^{'} (x) = e^{- x} - x e^{- x} = e^{- x} (1 - x) = 0$ for max. or min.
$\Rightarrow x = 1$
Now $f^{''} (x) = - e^{- x} - e^{- x} + x e^{- x} = (x - 2) e^{- x}$
Clearly at $x = 1$ $f^{''} (x) < 0$
Hence $f_{m a x} = \frac{1}{e}$

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