💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Sketch the graph of a function that satisfies all of the given conditions$f'(5) = 0$, $f'(x) < 0$ when $x < 5$,$f'(x) > 0$ when $x > 5$, $f"(2) = 0$, $f"(8) = 0$,$f"(x) < 0$ when $x < 2$ or $x > 8$,$f"(x) > 0$ for $2 < x < 8$, $\displaystyle \lim_{x\to\infty} f(x) = 3$, $\displaystyle \lim_{x\to-\infty} f(x) = 3$

## $f^{\prime}(5)=0 \Rightarrow$ horizontal tangent at $x=5$$f^{\prime}(x)<0 when x<5 \Rightarrow f is decreasing on (-\infty, 5)$$f^{\prime}(x)>0$ when $x>5 \Rightarrow f$ is increasing on $(5, \infty)$$f^{\prime \prime}(2)=0, \quad f^{\prime \prime}(8)=0, \quad f^{\prime \prime}(x)<0 when x<2 or x>8 f^{\prime \prime}(x)>0 for 2<x<8 \Rightarrow f is concave upward on (2,8) and concave downward on (-\infty, 2) and (8, \infty)There are inflection points at x=2 and x=8$$\lim _{x \rightarrow \infty} f(x)=3, \quad \lim _{x \rightarrow-\infty} f(x)=3 \Rightarrow y=3$ is a horizontal asymptote.

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##### Top Calculus 2 / BC Educators  ##### Heather Z.

Oregon State University  ##### Kristen K.

University of Michigan - Ann Arbor

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So in this problem we are given a bunch of different conditions and the first one is that f prime of five, this is equal to zero. So this is a potential maximum minimum value. And if we look we are decreasing from negative infinity to five and increasing from five to infinity. So we go from decreasing to increasing, which means that we have a local minimum at x is equal to five. So I'm gonna let this be five and I'm just gonna put a point just say, and we'll put it here for now, maybe we'll have to change it, but that's going to be a minimum. And we also know that we are decreasing from negative infinity to five and then increasing from 5 to infinity. And we also know that we also know that The second derivative at two is equal to the second derivative at eight, which is equal to zero. So these are potential inflection points. Um and That we are concave down from negative infinity to to and from 8 to infinity. And then we're concave up from 2 to 8. And so that does mean that these points at two and eight are inflection point. Since we're going from concave down to concave up at two and from concave up to concave down at eight. So at two and eight we have inflection points. So I'm just going to put Say this is eight, put a point, put it over here and then appoint over here and they'll say this is too. And those two will be inflection points. And then the last thing that we want to look at is that the limit as X goes to infinity is equal to the limit as X goes to negative infinity Which is equal to three. So that means that we're going to have a horizontal assam tote at three right here. And so if we look at this um function again, we see that we are decreasing from negative infinity to five. And then we're increasing mm hmm. from 5 to infinity. So we're going to have and we also know that at five we have this minimum value. So we're going to have a minimum value at five. But we want to be increasing from five to infinity. So we want this minimum value to be below this um this horizontal assam tote. So this being at zero will actually work here. And so what we want to make sure is that we're decreasing from negative infinity to five. And we are concave down from negative infinity to to So we're concave down in decreasing from negative infinity. See I'm gonna write this or draw this green from negative infinity two, two. And then we start to be concave up and we have this minimum value. We're concave up until eight and then we're concave down again and we're going towards this horizontal assented at three on both sides. So we want to make sure that we are concave down from negative infinity to two, and then we're concave up from 2 to 8. We have this minimum value at five and we are concave down again from eight to infinity, with both sides Of our graph going towards this point that Y is equal to three. Oregon State University

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##### Top Calculus 2 / BC Educators  ##### Heather Z.

Oregon State University  ##### Kristen K.

University of Michigan - Ann Arbor

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