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


Given:- Function $f(x)=(x-1) e^{x}+1$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for $a l l x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$


(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

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

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left((\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+1\right)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{x}+(\mathrm{x}-1) \mathrm{e}^{x}$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}(1+\mathrm{x}-1)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \mathrm{e}^{\mathrm{x}}$

as given


$\Rightarrow e^{x}>0$

$\Rightarrow x e^{x}>0$

$\Rightarrow f^{\prime}(x)>0$

Hence, condition for $f(x)$ to be increasing

Thus $f(x)$ is increasing on interval $x>0$

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