# Suppose $f'(x_0)$ exists and is positive, then there exists $x_1 > x_0$ such that $f(x) > f(x_0)$ for all $x \in (x_0,x_1)$.

Suppose $f'(x_0)$ exists and is positive, then there exists $x_1 > x_0$ such that $f(x) > f(x_0)$ for all $x \in (x_0,x_1)$.

What I have done so far: Since $f(x_0) > 0$ and exists, then $\lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0) > 0$. Then let $x_1 > x_0$. So $f'(x_0)(x_1-x_0)=x_1*f'(x_0) - x_0*f'(x_0) = x_1 * \lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0} - x_0 * \lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}$

I'm not sure whether I am going about this in the right direction and whether I can just deal with the inside part of the limit. Let me know if I'm doing this right so far. Thanks in advance.

• Look at this very recent post. – André Nicolas Nov 8 '12 at 17:08

Use the $\epsilon$-$\delta$ definition of the limit: For $\epsilon:=f'(x_0)>0$ there exists a $\delta>0$ such that $|x-x_0|<\delta$ (and $x\ne x_0$) implies $\left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)\right|<\epsilon$. With our choice of $\epsilon$, the latter inequality implies $\frac{f(x)-f(x_0)}{x-x_0}>0$, hence $f(x)-f(x_0)>0$ if additionally $x-x_0>0$.
Since $\lim_{x \rightarrow x_0} \frac{f(x)-f(x_0)}{x-x_0}=f'(x_0) > 0$, for $x$ sufficiently close to $x_0$ you must have $\frac{f(x)-f(x_0)}{x-x_0}\geq \frac{f'(x_0)}{2}$, which give $f(x)-f(x_0) \geq \frac{f'(x_0)}{2} (x-x_0)$ when $x>x_0$ is close to $x_0$.
The following is a little bit different from the other answers, but points to an understanding of $f'(x_0)$ which is applicable also in a multidimensional setting:
The fact $$f'(x_0)=\lim_{x\to x_0}{f(x)-f(x_0)\over x-x_0}=A>0$$ means: There is a small deviation $r(x)$ with $\lim_{x\to x_0} r(x)=0$ such that $${f(x)-f(x_0)\over x-x_0}=A+r(x)\qquad(x\ne x_0)\ .$$ Multiplying this with $x-x_0\ne0$ we get $$f(x)-f(x_0)=\bigl(A+r(x)\bigr)(x-x_0)\qquad(x\ne x_0)\ ,$$ and taking the sign on both sides gives $${\rm sgn}\bigl(f(x)-f(x_0)\bigr)={\rm sgn}\bigl(A+r(x)\bigr)\ {\rm sgn}(x-x_0)\ .$$ As $A>0$ and $\lim_{x\to x_0} r(x)=0$ we have $A+r(x)>0$ for all $x$ sufficiently near $x_0$. It follows that there is a $\delta>0$ such that $${\rm sgn}\bigl(f(x)-f(x_0)\bigr)= {\rm sgn}(x-x_0)\qquad(0<|x-x_0|<\delta)\ .$$