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So I have $f(x)=e^x-1-x$, MVT gives me $f'(c)={f(x)-f(0)\over{x-0}}={e^x-x-1\over x}$. So $f(x)=xf'(c)$. I got stuck at how exactly I can explain that $f(x)\ge 0\, \forall\, x\ge0$

I know that $f'(c)$ has the same sign as $c$ which has the same sign as $x$, thats why $f'(c)\ge0$ when $x\ge0$, which makes $f(x)\ge0$, but how do I show it nicely?

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up vote 3 down vote accepted

You know that for any $x\geqslant 0$ there exists $\xi\in[0,x]$ such that $\frac{f(x)-f(0}{x-0}=f'(\xi)$, or this all says that $\frac{e^x-1-x}{x}=f'(\xi)=e^\xi-1\geqslant 0$ (since $e^x-1$ is nonnegative on the non-negative reals) and so $e^x-1-x\geqslant 0$.

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How did you get from ${e^x}-1-x\over x$ to $e^\xi -1$? – Akaichan Dec 11 '12 at 4:39
Oh, never mind, I know why. Sorry! – Akaichan Dec 11 '12 at 4:41

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