How to prove that $f^{'}(x_{0})<0$ for some point $x_{0}$ implies the existence of $a$ and $b$ such that $af(b)$. I want to prove that $f^{'}(x_{0})<0$ for some point $x_{0}\in\mathbb{R}$ implies the existence of $a$ and $b$ in $\mathbb{R}$ such that $a<x_{0}<b$ and $f(a)>f(b)$. I think I should use $\epsilon-\delta$ definition for derivative i.e. for a given $\epsilon>0$ there exists $\delta>0$ such that $|x-x_{0}|<\delta\,\Rightarrow\,\left| \frac{f(x)-f(x_{0}}{x-x_{0}}-f^{'}(x_{0}) \right|<\epsilon$. However, I am stuck at this level. Is there a related $\textbf{Theorem}$ to solve this problem? 
 A: Your stated definition of the derivative is incomplete or incorrect, but you're on the right track. What you mean is that $f'(x_0) = c$ if, for any $\varepsilon > 0$, there exists $\delta > 0$ such that $| x - x_0 | < \delta \implies \left|  \frac{f(x) - f(x_0)}{x-x_0} - c \right| < \varepsilon$. If you know that $f'(x_0) < 0$, then set $c = f'(x_0)$, $\varepsilon = -c/2$, $a=x_0$, and $b = x_0 + \delta/2$ where $\delta$ is the one that goes with $\varepsilon$ in the derivative definition.
A: One easy way: Denote $\alpha=f'(x_0)<0$. We know for any $\epsilon>0$, there exists $\delta>0$ such that when $x\in (x_0-\delta,x_0+\delta)$, we have $-\epsilon<\frac{f(x)-f(x_0)}{x-x_0}-\alpha<\epsilon.$ Choose any $\epsilon$ such that $\epsilon+\alpha<0$. Then we can say for such a $\delta$ choice that
$$
-\epsilon+\alpha<\frac{f(x)-f(x_0)}{x-x_0}<\epsilon+\alpha<0.
$$
What does the fact that $f(x)-f(x_0)$ and $x-x_0$ have opposite signs tell you? Finish the proof by selecting points on the left and right halves of the delta neighborhood, respectively.
A: No need for a Theorem.
Since
$f'(x_0) = \lim_{ x \to x_o} { f(x) -f(x_0) \over x -x_0}  < 0$, there is some $\delta>0$
such that if $0<|x-x_0| < \delta $ then
${ f(x) -f(x_0) \over x -x_0}< {1 \over 2} f'(x_0)$.
Then if $b=x_0+{1 \over 2 } \delta$ we have $f(b)-f(x_0) < {1 \over 4} f'(x_0) \delta$
and if $a = x_0 -{1 \over 2}\delta$ we have $f(a)-f(x_0) > {1 \over 4} f'(x_0) \delta$.
In particular, $a<b$ and $f(a)>f(b)$.
