Prove that e^(x+1)=x+2, where x is a real number

Question

Prove that e^(x+1)>=x+2, where x is a real number

turingtest · Answer

induction?
how far have you gotten?

anonymous · Answer

i was thinking more like the first derivate tabel

turingtest · Answer

ok, well, take the first derivative of both sides and what do you get?

anonymous · Answer

(x+1)e^x>=1

turingtest · Answer

not quite, you are trying to use d/dx(x^n) on d/dx(e^x)

anonymous · Answer

yeah you're right induction is the only key

anonymous · Answer

sorry for my mistake

turingtest · Answer

there may be a calc way to do it... but it's not jumping out at me

anonymous · Answer

now in my country its 00:25 am so..
i'm not filled with ideas

turingtest · Answer

let's see if I have something

turingtest · Answer

actually this proof is easier and can be done with calculus

turingtest · Answer

Simply show that the left function is concave up for all x, that both functions have the same first derivative at x=-1, and that the two graphs touch at that point. This means that the slope of e^(x+1) is less than that of x+2 for x<-1, and greater for x>-1, hence f(x)=e^(x+1) is always above f(x)=x+2, except where they touch at x=-1.

anonymous · Answer

done it

turingtest · Answer

cool :)