Show that f(x)=(x-1) e^x+1 is an increasing function for all x > … - Vidyadip

Question

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

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Answer

$\text { Given:- } f(x)=(x-1) e^x+1$
$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left((x-1) e^x+1\right)$
$=f^{\prime}(x)=e^x+(x-1) e^x$
$=f^{\prime}(x)=e^x(1+x-1)$
$=f^{\prime}(x)=x e^x$
as given
$x > 0$
$=e^x > 0$
$=x e^x > 0$
$=f^{\prime}(x) > 0$
Hence, the condition for $f(x)$ to be increasing
Thus, $f(x)$ is increasing for all $x > 0$

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