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I'm looking at

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

I'm having the feeling (based on the application where I am using it), that $f(x)$ should be strictly positive for $x > 0$. Indeed, Wolfram Alpha plots it as such, with a global minimum of ($f(0)x=0$).

However, I fail to show this. It is trivial for $x \geq 1$, but what for $x < 1$?

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    $\begingroup$ It seems trivial to show $x> 0 \implies f'(x) > 0$. What have you tried? $\endgroup$
    – Macavity
    Feb 24, 2016 at 12:01

4 Answers 4

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$f'(x)=x \, e^x > 0$ for $x>0$, so $f$ is strictly increasing on $[0,\infty)$.

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Since $$ e^x\ge 1+x $$ and thus also $$ e^{-x}\ge 1-x $$ one gets $$ f(x)=e^x·(e^{-x}-(1-x))\ge 0. $$

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    $\begingroup$ +^I just love it when all you ever wanted to know about the exponential is the inequality $e^x\ge 1+x$ $\endgroup$
    – Hagen von Eitzen
    Feb 24, 2016 at 20:47
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For $x\in(0,1)$, the inequality $e^x (x-1)+1 > 0$ is equivalent to: $$ e^x < \frac{1}{1-x} \tag{1}$$ or to: $$ 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots < 1+x+x^2+x^3+\ldots \tag{2} $$ that is trivial.

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    $\begingroup$ If confused by the last step, Jack converts $e^x$ to its Taylor series and the $\dfrac{1}{1-x}$ to its geometric series. $\endgroup$
    – snoram
    Feb 24, 2016 at 16:18
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$$f^{'}(x)=e^{x}x$$ the derivative is positive for $x>1$ and thus it is increasing. I think you are having doubts about $x<1$. So you have two answers.

  1. range of (0,1) which is increasing, since the derivative is positive.

  2. From$ (-\infty,0)$ which is decreasing since $ e^{x}x$ becomes negative.

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