Question
Show that $f(x) = (x - 1)e^x + 1$ is an increasing function for all $x > 0$.

Answer

$f(x) = (x - 1)e^x + 1 f'(x) = (x - 1)e^x + e^x= xe^x - e^x + e^x$
$= xe^x$
Given: x > 0
We know,
$e^x> 0$
$\Rightarrow xe^x > 0$
$\Rightarrow f'(x) > 0$, for all $x > 0$
So, f(x) is increasing on for all $x > 0$.

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