(10)

Step 1:

Thu function is and the interval is .

Differentiate on each side with respect to .

.

Differentiate on each side with respect to .

.

Equate to zero.

.

General solution for trigonometric equation is .

.

For ,

For ,

For ,

is not in the .

Consider the test intervals as ,,  , and .

\ Interval \	Test Value	Sign of	Concavity
			\ Concave Down \
			Concave Up
			Concave Down
			Concave Up
			Concave Down

The graph of the function is concave up on the interval and .

The graph of the function is concave down on the interval , and .

Solution :

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

11)

Thu function is .

Find the inflection points.

Inflection Point :

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

The curve changes concave down to concave up at and .

The curve changes concave up to concave down at and .

.

Substitute in the function.

Inflection point is .

Substitute in the function.

Inflection point is . 

Substitute in the function.  

Inflection point is . 

\ \

Substitute  in the function. 

Inflection point is .

Inflection points are , , and .

Find the critical points.

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

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

Thus, the critical points exist when .

Equate to zero.

and  

and  

The critical point are and .

Solution :

The critical point are and .

(13)

Step 1:

Thu function is .

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

The test intervals are , and .

Interval	Test Value	Sign of	Conclusion
		\ \	Increasing
		\ \	Decreasing
		\ \	\ Increasing \

The function is increasing on the intervals and .

The function is decreasing on the interval .

Solution :

The function is increasing on the intervals and .

The function is decreasing on the interval .

(14)

Step 1:

Find the local maximum and local minimum.

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

Substitute in .

Local maximum is .

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

Substitute in .

Local minimum is .

Solution : 

Local maximum is .

Local minimum is .

(15).

Step 1:

Thu function is .

.

Differentiate on each side with respect to .

.

Equate to zero.

Consider the test intervals as and .

\ Interval \	Test Value	Sign of	Concavity
			\ Concave Down \
			Concave Up

The graph of the function is concave up on the interval .

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

Solution :

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

(16)

Step 1:

Thu function is .

Find the inflection points.

Inflection Point :

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

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

Substitute in the function.

Inflection point is .

Solution :

Inflection point is .

(17)

Thu function is .

The function is increasing on the intervals and .

The function is decreasing on the interval .

Local maximum is .

Local minimum is .

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

Inflection point is .

Graph :

Graph the function :