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How do I use the mean value theorem to find $$ 1+x < e^x < 1+xe^x $$ for all x>0

I'm not really sure what function I can use or if I can use any function to show with MVT. I tried using $f(x)=x$ over $[1+x,1+xe^x]$ but in the end it shows 1=1 so how should I solve it?

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$$\frac{e^x-e^0}{x-0}=e^{\xi},\quad \xi \in (0,x), $$ therefore $$\frac{e^x-1}{x}\ge 1.$$ Now it's easy to see the inequality is strict for $x>0$.

Can you do the same for the other inequality?

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For the first inequality you could use $f(x)=e^x-(1+x)$. For the second use $g(x)=(1+xe^x)-e^x$.

Remark: There are alternatives, and the functions chosen above are not necessarily best. But they are the most mechanical.

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