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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 $x_0$

a group of graphs, the second has a check mark under it

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 .)

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    $\begingroup$ You are correct. $\endgroup$ – tomi Dec 10 '19 at 22:26
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It appears that you are correct.
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.

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