Question on the proof of $e^x>1+x$ for $x>0$ 
Show that $e^x>1+x$ for $x>0$
Proof: Set $f(x)=e^x-(1+x)$. Show that $f(x)$ is always positive. We know that $f(0)=e^0-(1+0)=0$ and $f'(x)=e^x-1$. When $x$ is positive, $f'(x)$ is positive because $e^x>1$
We know that if $f'(x)>0$ on an interval, then $f(x)$ is increasing on that interval, so we can conclude that $f(x)>f(0)$ for $x>0$ and thus:
$$e^x-(1+x)>0 \iff e^x>1+x$$

Question: How do we know that $f(x)$ is positive just by $e^x>1$?
Additional question, how do we prove $f'(x)$ is positive/negative over an interval $[a,b]$?

For $x>0$, we know $e^x > 1$, so $f'(x) > 0$.  So, $f(x)$ is increasing, in particular $f(x) > f(0) = 0$, so $f(x)$ is positive. 
