Prove that e^x >= 1+x for all x in [0,inf).
My solution so far:
I let f(x) = e^x - 1 - x, now if I can prove f is increasing on [0,inf), then it will work, but I have difficulties trying to do that...
Let f(x)=e^x-x-1
This function is differenciable apply the increasing/decreasing theorem,
f'(x)=e^x-1
Now if it true that e^x-1>0 on x>0.
Yes, why?
Because, it is equivalent to saying,
e^x>1 and e^0=1 thus if x>0 then e^x > 1 because e^x is an increasing function. Q.E.D.