The function \[x{e^x}\] is increasing in
\[\left( { - 3,0} \right)\]
\[\left( {0,4} \right)\]
\[x > - 1\]
\[\left( {\dfrac{1}{e},\infty } \right)\]

Hint:
We know that for a function to be increasing its derivative is greater than zero. So we will find the derivative and will check, in which interval it’s positive and in which interval it’s negative.

Complete step by step solution:
Given the function is \[x{e^x}\] .
So
\[f(x) = x{e^x}\]
Taking its derivative with respect to x.
\[ \Rightarrow f'(x) = {e^x}.1 + x.{e^x} > 0\] \[ \to \left( {u.v = vu' + uv'} \right)\]
Taking exponential term common,
\[ \Rightarrow {e^x}\left( {1 + x} \right) > 0\]
\[ \Rightarrow 1 + x > 0\]
\[ \Rightarrow x > - 1\]

Thus option C is correct.

Note:
A function is said to be increasing when its derivative is greater than zero or positive. On the other hand a function is said to be decreasing when its derivative is less than zero or negative. we have used product rule of derivative here that is \[ \to \left( {u.v = vu' + uv'} \right)\]