Does anyone know how to prove that for all x>0,
e^x > 1+x
using just alevel maths and further maths knowledge. Ive seen proofs which consider f(x) = e^x-x-1 and show that it is an increasing function etc... But they then go on to use the mean-value theorem which i am not familiar with.
[Btw i am writing this from memory so could anyone clarify if it is either of these or could you have both really]:
e^x> 1+x for x>0
or e^x> or = to 1+x, for x> or equal to 0 (with equality at x=0).
Also, is this true for all x (so now considering negatives; on a quick google search people specify x>0 but e^x gets small very quick for as x gets more negative but is always greater than 0 so i can see it clearly applying for x<-1.What about -1<x<0 then.
I can see in the increasing fucntion proof that f(x) is only increasing for x>0 so i get that restriction but is there another proof which doesnt require that, because i thought it would be true for all x.