Exam 3 solution

# Show your work clearly solution we need to find the x

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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 x-values, because there must be a point on the graph. We must confirm that the concavity changes at those x-values. For the answer to the first question, we know that polynomial: its domain is all real numbers. exists at those x-values because For the answer to the second question, we will investigate the sign of x-values sample number is a at sample x-values sign of conclusion about is concave up is concave down is concave up We see that does change sign at and x-values. 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 x-values.) Solution: Start by finding the critical values. Those are x-values 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 x-values that cause . Two x-values 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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