(12)

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Step 1:

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Thu function is \"\".

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

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\"\"

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\"\".

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Find the critical points.

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A critical number of a function \"\" is a number \"\" in the domain of \"\" such that either \"\" or \"\" does not exist.

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Since \"\" is a polynomial it is continuous for all values of \"\".

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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 point are \"\" and \"\".

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Solution :

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The critical point are \"\" and \"\".

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(13)

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Step 1:

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Thu function is \"\".

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The critical point are \"\" and \"\".             (From(12))

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

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Interval Test Value Sign of \"\"Conclusion
\"\"\"\" \

\"\"

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Increasing
\"\" \"\" \

\"\"

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Decreasing
\"\" \"\" \

\"\"

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Increasing

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The function is increasing on the intervals \"\" and \"\".

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

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Solution :

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The function is increasing on the intervals \"\" and \"\".

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

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(14)

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Step 1:

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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 positve 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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Substitute \"\" in \"\".

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\"\"

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Local minimum is \"\".

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Solution : 

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Local maximum is \"\".

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Local minimum is \"\".

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(15).

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Step 1:

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Thu function is \"\".

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\"\".

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

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\"\"

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\"\".

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Equate \"\" to zero.

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\"\"

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Consider the test intervals as \"\" and \"\".

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

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

Concave Down

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\"\"\"\" \"\"Concave Up
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The graph of the function is concave up on the interval \"\".

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

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Solution :

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The function is concave up on the interval \"\" and concave down on the interval \"\".

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(16)

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Step 1:

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Thu function is \"\".

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Find the inflection points.

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Inflection Point :

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Inflection point is a point on the curve at which the function changes from concave up to down or vice versa.

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The curve changes concave down to concave up at \"\".           (from (15)).

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Substitute \"\" in the function.

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\"\"

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Inflection point is \"\".

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Solution :

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Inflection point is \"\".

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(17)

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Thu function is \"\".

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The function is increasing on the intervals \"\" and \"\".

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

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Local maximum is \"\".

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Local minimum is \"\".

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The function is concave up on the interval \"\" and concave down on the interval \"\".

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Inflection point is \"\".

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Graph :

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Graph the function \"\" :

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\"\"