# Graph of a function $f$, around a point $x_0$, with $f'(x_0)<0$

Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be a continuous function. Let $$x_0$$ be some point in which $$f$$ has derivative s.t $$f'(x_0)<0$$.

How does the function graph looks around $$x_0$$

Well, in my opinion, it is only the second graph, because it is strictly decreasing around that point.

Will you help me? thx a lot .)

• You are correct. – tomi Dec 10 '19 at 22:26

The correct answer cannot be the first, sixth, seventh, or eighth graphs. The reason for this is that when there is a horizontal asymptote, by definition the $$y$$ component of the graph is unchanging. In other words, if you zoom in very closely, you can see that at that point that a tangent line to that point will have a slope of $$0$$. Thus $$f'(c) = 0$$, not $$f'(c) < 0$$.
The forth graph might not even be defined at $$c$$, but if it is defined it falls under the previous category. Lastly the third and fifth graphs cannot be the answer as the $$y$$ component is increasing as $$x$$ increases. Thus the tangent line to the point would have a positive slope, so $$f'(c) > 0$$, not $$f'(c) < 0$$. Thus the only possible solution is the second graph.