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 $x > 0.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free