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This is my function $f(x)$enter image description here

What is the methodology for me determining if the sketch of the derivative has a horizontal asymptote? basically i want to be able to justify that it has a horizontal asymptote and not just because it looks like it does

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  • $\begingroup$ Show that $f(x)\to 0$ as $x\to\pm\infty$, and show the sequence if functions is non zero. $\endgroup$ – Pixel Jun 2 at 12:36
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By looking at your picture (with no values, or scale) it seems that for large positive $x$ and for large negative $x$ the slope is $0$. Thus, the value of the derivate for large positive $x$ and large negative $x$ is $0$. This suggest that the derivative of $f$, i.e $f'$ has the horizontal asymptot $y=0$.

Without additional info its hard to say much more. And it also could be false, since we don't know how much of $f$ is shown in the picture.

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  • $\begingroup$ thank you! if there was a scale, the idea would still be the same right? look at the function for a large negative and positive x $\endgroup$ – user130306 Jun 2 at 9:06
  • $\begingroup$ Yes. But if the function is only shown for lets say $x$ from $-0.1$ to $0.1$ then its hard to talk about large $x$ right? You can have a idea, but not much to back it up. @user130306 $\endgroup$ – Olba12 Jun 2 at 9:10
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In general, given a function $f(x)$, you can evaluate $$ \lim_{x \to \infty} f(x) \ \ \textrm{and} \ \ \lim_{x \to -\infty} f(x) $$ to get the horizontal asymptotes. There are at most two horizontal asymptotes since there are only two directions.

As pointed out by the previous answer, in your case, you have a graph instead of a explicit function. Therefore, you can't say much other than observation. The observation is that at the far left, and far right, the function get really close to zero! So your horizontal asymptote is zero.

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