Unformatted text preview: Show your work clearly.
Solution: We need to find the x coordinates of the points on the graph where the concavity changes.
For that, we will study the sign of
. We first rewrite in better form for taking the derivative.
Then find and . We see that
at
and
. We suspect that these are the inflection points, but we have
to confirm two things: We must confirm that
exists at those xvalues, because there must be a point on the graph. We must confirm that the concavity changes at those xvalues.
For the answer to the first question, we know that
polynomial: its domain is all real numbers. exists at those xvalues because For the answer to the second question, we will investigate the sign of
xvalues sample number is a at sample xvalues sign of conclusion about
is concave up
is concave down
is concave up We see that
does change sign at
and
xvalues. Therefore, has inflection points at , so we conclude that
and
. changes concavity at those Find the absolute max and absolute min (if either of them exists) on the interval
.
Be sure to indicate if your result is a max or a min, and show how you know.
(You must use calculus and show all details clearly. No credit for just guessing xvalues.)
Solution: Start by finding the critical values. Those are xvalues that satisfy these two requirements: is undefined or
. (That is, x is a partition number for .) exists.
We will need to find . First we rewrite in a form that is better for taking the derivative. Then we find We see that is undefined at . Next, we find xvalues that cause . Two xvalues make this equation true:
and
.
Therefore, has partition numbers
and
and
.
Since
exists and
exists while
does not exist, we conclude that only
to be called critical values of . Of these two, only
is in the interval
.
We need to determine...
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 Fall '12
 barsamnian
 Math, Algebra, Critical Point, Derivative, Convex function

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