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Using second derivative graph to find certain parts of original graph

jlmccart

New member
Joined
Oct 9, 2016
Messages
1
I am attaching a picture of the problem, but first some background information.
I know that the graph is increasing if the first derivative is above the x-axis and decreasing if the first derivative is below the x-axis. That said I would like to make sure my intervals are correct for part a.
Second, I know that the graph of f must have a local min or max if the derivative graph is 0 and it is a min if it goes from neg to pos, and a max if it goes from pos to neg. Otherwise if it is 0 and does not cross the x-axis it is a saddle point (Like x = -2 is).
Third, I know that f is concave up if the SECOND derivative is pos and concave down if the SECOND derivative is neg. This is where I am stuck because I don't remember (and more importantly cant find a good example with this first derivative graph) on what to do. I do know that if the tangent line is zero on the first derivative then it is an inflection point, but I don't seem to understand how to determine if it is concave up or down on the original function. (There are two inflection points x = -2 and x = 0)
Finally, can anyone tell me what is happening on x = 2 for the original graph because I do not know how I utilize that point discontinuity to draw a continuous graph. (Unless does it mean that the slope is vertical thus undefined and it changes from increasing to decreasing)?
Thanks and here is the picture

< link to objectionable page removed >
 
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Otis

Full Member
Joined
Apr 22, 2015
Messages
959
... I don't seem to understand how to determine if it is concave up or down on the original function ...

Can you sketch a rough graph of the second derivative (i.e., graph the slopes of the first derivative). You need to think about where the second derivative is positive (above x-axis) and where it is negative (below x-axis), to answer specific questions about concavity.

... does [the point discontinuity] mean that the slope is vertical thus undefined and it changes from increasing to decreasing ...

Yes -- not that the slope is vertical (that would be a cusp), but that it changes radically from a positive value to a negative value (similar to how absolute-value graphs have an abrupt slope change, at the "pointy", only your graph won't be linear). Consider actual values of the given slopes , on each side, as x approaches 2.

Try to reason out a rough sketch of the original graph near x=2, to match the slope values that you read from the given graph of the first derivative. You can also consider the concavity on each side, from your graph of the second derivative.

The original graph will be continuous, but it's not a smooth transition there! :)
 
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