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 .)

  • 2
    $\begingroup$ You are correct. $\endgroup$ – tomi Dec 10 '19 at 22:26

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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