The function $x e^{x}$ is increasing in

(-3,0)

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(0,4)

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x > -1

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(1 / e, \infty)

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Solution

The correct option is C x > -1
$\begin{matrix} f (x) = x e^{x} f^{^{'}} (x) > 0 f o r i n c r e a s i n g f u n c t i o n = e^{x} + x e^{x} = e^{x} (x + 1) > 0 x > - 1 H e n c e, o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

The function f(x) = $x e^{x}$ is strictly increasing in the interval

The interval in which the function $f (x) = x e^{x}$ is strictly increasing is

f (x) = \frac{x}{e^{x} - 1} + \frac{x}{2} + 1

f (x) = x \frac{e^{x} + 1}{e^{x} - 1}

f (x) = x \frac{e^{x} + 1}{e^{x} - 1}

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