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

Question

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 $x < 1$?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.