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Problem 26 Easy Difficulty

Sketch the graph of a function that satisfies all of the given conditions

Vertical asymptote $ x = 0 $, $ f'(x) > 0 $ if $ x < -2 $,
$ f'(x) < 0 $ if $ x > -2 (x \not= 0) $,
$ f"(x) < 0 $ if $ x < 0 $, $ f"(x) > 0 $ if $ x > 0 $

Answer

see sketch or work graph

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Video Transcript

Okay, so we're given four conditions that are graph has to meet you have this vertical ascent owed at X is equal to zero. We are increasing or are derivative is greater than zero when X is less than negative two were decreasing when X is greater than negative two. And our second derivative is less than zero for X being less than zero. And our second derivative being less than zero means that we are just concave down on the interval from negative infinity to zero. And so what I've drawn already is just the vertical ascent to X is equal to zero. And I put on the X axis this point negative too. Since that's the point where we go from increasing to decreasing. So I'm just gonna put a point up here as the point that we go from increasing to decreasing, which means that it is a local maximum. Or it could be an absolute maximum as well, but we're guaranteed to have a maximum at this point since we're going from increasing and decreasing. And so I'm going to draw a line that is concave down and increasing up until this point here just call that the point and now we're decreasing. Still concave down and we have this vertical ascent tote so we're never going to go past X is equal to zero. And now what we can do is we can actually just stop there since we don't have any conditions for X being greater than zero. So the important thing to remember is that we're decreasing until X is equal to negative two. And then we are sorry, we're increasing until X is equal to negative two. And then we start decreasing. And since we have this vertical ascent, oh, we're never going to go past zero. And we have to be always concave down which we are Yeah.

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