Step 1 :  

(a)

Thu function .

Differentiate with respect to x:

Determination of critical points:

Since  is a polynomial it is continuous for all real numbers.

Thus, the critical points exist when .

Equate to zero:

The critical points are and 

Consider the test intervals are and

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

Thus, The function is increasing on the intervals and

And The function is decreasing on the interval

Step 2 :  

(b)

is changes its sign from positive to negative, hence f has a local maximum at .

Local maximum is  .

is changes its sign from negative to positive, hence f has a local minimum at .

Local minimum is .

Step 3 :  

(c)

Differentiate with respect to x:

Determination of concavity and inflection points : 

Equate to zero:

Thus, the inflection point is

Consider the test intervals as and .

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

Thus, the graph is concave up on the interval

The graph is concave down on the interval .

Inflection point :

Inflection point is

Step 4 : 

(d)

Graph of the function .

Solution :

(a)

Increasing on the intervals and .

Decreasing on the interval and .

(b)

Local maximum is  .

Local minimum is .

(c)

Concave up on the interval and .

Concave down on the interval .

Inflection points .