how would i prove that e^x >= 1 + x for all x in [0,inf)?
where are your thoughts?
hint: expand out to see.
Define f(x) = e^x - x - 1. What can f'(x) tell us?
like morphism stated, now what u have to do is just find f'(x)= (e^x-x-1)', what is f'(x) now?,
After that look at what interval the function f(x) is always greater or equal to zero, using the first derivative test,can you do it? and the result surely should be for any x greater than zero. It is obvious that for x=0 the >= sing is valid.
Separate names with a comma.