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Demonstrate:
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Demonstrate what?
\[e ^{x}\ge x+1\]
Hmm, well the power series representation of e^x is \[e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\] or you could just plug in a number to show that I guess. How exactly do you want to demonstrate it?
This problem represents an application to derivative functions so..
If you show that they're the same at x=0 then take the derivative of both you can compare their slopes. Since e^x has a much steeper slope than x+1 we know that it must be greater because it is increasing faster and the other function can't catch up to it.
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