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Monotonically Increasing Functions

The nth der...

Question

The $n$ th derivative of $x e^{x}$ vanishes when

x = 0

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x = - 1

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x = - n

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x = n

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Solution

The correct option is C $x = - n$
$f (x) = x e^{x}$
$f^{'} (x) = e^{x} + x e^{x}$
$f^{''} (x) = e^{x} + e^{x} + x e^{x}$
$f^{'''} (x) = 2 e^{x} + e^{x} + x e^{x} = 3 e^{x} + x e^{x}$
$\dots$
$\dots$
$f^{n} (x) = n e^{x} + x e^{x}$ .
Now, $f^{n} (x) = 0$
or $n e^{x} + x e^{x} = 0$ or $x = - n$

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