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...

Printable View

- Mar 18th 2007, 06:21 PMtttcomraderFunction with e problem
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... - Mar 18th 2007, 06:58 PMThePerfectHacker
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. - Mar 18th 2007, 07:00 PMtttcomrader
oh, man, this is easy. I was trying to do this in a much harder way...

Thanks!