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Problem 27 Medium Difficulty
Sketch the graph of a function that satisfies all of the given conditions
$ f'(0) = f'(2) = f'(4) = 0 $,
$ f'(x) > 0 $ if $ x < 0 $ or $ 2 < x < 4 $,
$ f'(x) < 0 $ if $ 0 < x < 2 $ or $ x > 4 $,
$ f"(x) > 0 $ if $ 1 < x < 3 $, $ f"(x) < 0 $ if $ x < 1 $ or $ x > 3 $
Answer
$f^{\prime}(0)=f^{\prime}(2)=f^{\prime}(4)=0 \Rightarrow$ horizontal tangents at $x=0,2,4$
$f^{\prime}(x)>0$ if $x<0$ or $2<x<4 \Rightarrow f$ is increasing on $(-\infty, 0)$ and (2,4)
$f^{\prime}(x)<0$ if $0<x<2$ or $x>4 \Rightarrow f$ is decreasing on (0,2) and $(4, \infty)$
$f^{\prime \prime}(x)>0$ if $1<x<3 \Rightarrow f$ is concave upward on (1,3)
$f^{\prime \prime}(x)<0$ if $x<1$ or $x>3 \Rightarrow f$ is concave downward on $(-\infty, 1)$
and $(3, \infty) .$ There are inflection points when $x=1$ and 3
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Top Calculus 2 / BC Educators
Heather Z.
Oregon State University
Kayleah T.
Harvey Mudd College
Kristen K.
University of Michigan - Ann Arbor
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Video Transcript
Okay, so we're giving a bunch of conditions and we're gonna try and use these to sketch a pretty accurate graph of our function F. Of X. And so the first thing I'm gonna do is try and figure out if these points at 02 and four are actually minimums are maximum values. So if we look at values less than zero, you can see that's here are function or the derivative, our function is greater than zero. So we're increasing and then we are decreasing for values between zero and two. So we go from increasing and decreasing. So at zero we have a local maximum. So I'm just going to draw a point, we'll put it here and it's going to be a maximum. And then for this point at two we can look at points or values of X that are less than two but greater than zero. And that's this part. You can see we're decreasing and the values greater than two but less than four were increasing. So we're going from decreasing increasing. So this is going to be a minimum value. And then if we look at values less than four but greater than two were increasing. And if we look at values that are greater than four um we are decreasing and I actually forgot a parentheses on this part. So we're going from increasing to decreasing. So this is another maximum at X. Is equal to four. So I'm just gonna put a point over here and now what we wanna do is make sure that we are doing the right thing cavity. So between one and three we have our second derivative being greater than zero. So that means that We have we are concave up between one and 3. So that would be I guess I'll put a point put this in blue at one and at three and then between negative infinity and one, we are concave down and from three to infinity were also concave down. So what I'm gonna do is I'm gonna draw this graph. So we're concave down before zero. So we have to be concave down and then this is a maximum and then we're concave down until There or wherever we come again until one. And then we start to be concave up And we're going and then we're concave down at three and we are decreasing from four to infinity. So this would be a good sketch of our graph. Or concave down before this point at one we have a maximum at X is equal to zero, a minimum and X equals two and then a maximum X is equal to four. And then we're concave up between one and three and then concave down again from three to infinity. So this is a good sketch of our graph given the conditions
Oregon State University
Top Calculus 2 / BC Educators
Grace H.
Numerade Educator
Heather Z.
Oregon State University
Kayleah T.
Harvey Mudd College
Kristen K.
University of Michigan - Ann Arbor
