interval where function with larger derivative is larger Would like to show the following:
If $f(x_0)=g(x_0)$ and $f'(x_0) \ge g'(x_0)$, then $\exists \epsilon >0$ such that $f(x)\ge g(x)$ on $[x_0,\epsilon)$
Maybe something like Taylor series? Help appreciated.
 A: The 'strictness' of the relationship between $f(x_0)$ and $g(x_0)$ matters.
It is not true in general if we just have equality. Take $g(x) = 0$, $f(x) = -x^2$, and $x_0 = 0$. Then $f(x_0) = g(x_0) = 0$, $f'(x_0) = g'(x_0) = 0$, but $f(x) < g(x)$ for all $x\neq x_0$.
If $f'(x_0) > g'(x_0)$, then $\lim_{x \to x_0} { f(x)-g(x)-(f(x_0)-g(x_0)) \over x-x_0 } > 0$, and so for some $\delta>0$, if $|x-x_0| < \delta$, we have
${ f(x)-g(x)-(f(x_0)-g(x_0)) \over x-x_0 } > 0$. If we choose $x > x_0$, then this gives $f(x)-g(x) > f(x_0)-g(x_0)$ for $x_0 < x < x_0 + \delta$.
A: Let $h(x)=f(x)-g(x)$. Then $h(x_0)=0$, $h'(x_0)>0$. From $h'(x_0)>0$, there exists $h_0>0$ with 
$$
\frac{h(x_0+h)-h(x_0)}h>0, \ \ 0<h<h_0.
$$
As $h(x_0)=0$ and $h>0$, the inequality above tells us that $h(x_0+h)>0$ for all $0<h<h_0$. In other words, $h(x)>0$ for all $x\in(x_0,x_0+h_0)$. 
A: Assume that $h=f-g$. Let $h'(x_0)=l>0$,. Then, we can use definition of  $h'$.
For every $m$, there are $\epsilon$ such that for $|x-x_0|<\epsilon$ we have $|f-g-l|<m$. Now, choose $m=\frac{l}{2}$. Thus, there are $\epsilon$ such that for $|x-x_0|<\epsilon$ we have $0<\frac{l}{2}<f-g$.
