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Show that \[e ^{x} \ge 1+x\] for all real numbers x

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I have the idea that I could use the McLauren expansion for e^x \[e ^{x}=\sum_{n=0}^{\infty}\frac{ x ^{n} }{ n! }\]
So the two first expansions would be \[1+x\] but since I have to add the Lagrange reminder it would be proven bigger than 1+x
Would that be somewhat correct?

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Other answers:

\[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(n+1)}(\xi) }{(n+1)! }x ^{n+1}\] \[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(2)}(\xi) }{(2)! }x ^{2} \] \[f(x)=e ^{x}; f \prime(x)=e ^{x}; f \prime \prime(x)=e {^x}\] so that gives the remainder \[\frac{ e ^{\xi}x ^{2}}{2! }\] and \[e ^{x}=1+x+\frac{ e ^{\xi}x ^{2}}{2! }\ \ge 1+x\] Is this a genuine proof that\[e^{x}\ge 1+x\] ?
Oh forgot to write that \[\xi \in (0,x)\]
i think we can use by calculus principle...
differential calculus i meant
What about the Mclaurinexpansion I made, doesn't it prove the fact that it's bigger?
sorry i forgot about Mclaurinexpansion, i f by differential i got it
How did you show it using differentials?
e^x ≥ x + 1 e^x - 1 - x ≥ 0 let given f(x) = e^x - 1 - x, so f '(x) = e^x - 1 criticals points of f, hapended when saat f '(x) = 0 so, e^x - 1 = 0 get x = 0 to knowing the kind (max or min) of f, use the 2nd derivative so, f ''(x) = e^x for x=0, gives f ''(0) = 1 > 0 because f ''(0) > 0, therefore its kind is minimum the minimum value of f is f(0) = e^0 - 1 - 0 = 0 because f has the minimum value, is 0 so,, obviously f(x) ≥ 0 e^x - 1 - x ≥ 0 e^x ≥ x + 1 (proof) :)
Oh that's a good way to show it, pretty simple to, thank you RadEn! :)
very, welcome.... :)

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