Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Look at this very recent post. –  André Nicolas Nov 8 '12 at 17:08
add comment

3 Answers

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 $x-x_0>0$.

share|improve this answer
add comment

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 $x_0$.

share|improve this answer
add comment

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 $${\rm sgn}\bigl(f(x)-f(x_0)\bigr)= {\rm sgn}(x-x_0)\qquad(0<|x-x_0|<\delta)\ .$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.