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(a)

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The function is \"\".

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Differentiate \"\" on each side with respect to \"\".

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\"\"

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Find the critical points.

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Since it is a polynomial it is continuous at all the point.

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Thus, the critical points exist when \"\".

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Equate \"\" to zero.

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\"\"

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\"\" and \"\"

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\"\" and \"\".

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The critical points are \"\" and \"\".

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The test intervals are \"\", \"\" and \"\".

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
Interval Test Value Sign of \"\"Conclusion
\"\" \"\" \"\"Increasing
\"\" \"\" \"\" Decreasing
\"\" \"\"\"\"Increasing
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Therefore the function is increasing on the intervals \"\" and \"\".

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The function is decreasing on the interval \"\".

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(b)

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Find the local maximum and local minimum.

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The function \"\" has a local maximum at \"\", because \"\" changes its sign from positive to negative.

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Substitute \"\" in \"\".

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\"\"

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Local maximum is \"\".

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The function \"\" has a local minimum at \"\", because \"\" changes its sign from negative to positive.

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\"\"

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Local minimum is \"\".

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(c)

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\"\".

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Differentiate \"\" on each side with respect to \"\".

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\"\"

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Find the inflection points.

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Equate \"\" to zero.

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\"\"

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The inflection point is \"\".

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Substitute \"\" in \"\".

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\"\"

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The test intervals are \"\" and \"\".

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
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Interval

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Test Value Sign of \"\"Concavity
\"\"\"\" \"\"Down
\"\" \"\"\"\" \

Up

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The graph is concave up on the interval \"\".

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The graph is concave down on the interval \"\".

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The inflection point is \"\".

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(a)

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Increasing on the intervals \"\" and \"\".

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Decreasing on the interval \"\".

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(b)

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Local maximum is  \"\".

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Local minimum is \"\"

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(c)

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Concave up on the interval \"\".

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Concave down on the interval \"\".

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Inflection point is \"\".