Showing that the exponential expression $e^x (x-1) + 1$ is positive

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 $x < 1$?

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

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

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

• +^I just love it when all you ever wanted to know about the exponential is the inequality $e^x\ge 1+x$ Feb 24, 2016 at 20:47

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.

• If confused by the last step, Jack converts $e^x$ to its Taylor series and the $\dfrac{1}{1-x}$ to its geometric series. Feb 24, 2016 at 16:18

$$f^{'}(x)=e^{x}x$$ 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.