(a)

Thu function is $\"\"$ .

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

$\"\"$

Find the critical points.

Since $\"\"$ is a polynomial it is continuous at all the point.

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

Equate $\"\"$ to zero.

$\"\"$

The critical point is $\"\"$ .

The domain of $\"\"$ is $\"\"$ .

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

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

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

The function is increasing on the interval $\"\"$ .

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

(b)

Find the local maximum and local minimum.

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

Substitute $\"\"$ in $\"\"$ .

$\"\"$

Local minimum is $\"\"$ .

(c)

$\"\"$ .

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

$\"\"$

Find the inflection points.

Equate $\"\"$ to zero.

$\"\"$

The domain of $\"\"$ is $\"\"$ .

There is no inflection point, since $\"\"$ is not in the domian.

Consider the test intervals are $\"\"$ .

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

\ Interval \	Test Value	Sign of $\"\"$	Concavity
$\"\"$	$\"\"$	$\"\"$	\ Concave Down \

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

(d)

Graph :

Graph the function $\"\"$ :

$\"\"$

(a)

Increasing on the interval $\"\"$ .

Decreasing on the interval $\"\"$ .

(b)

Local maximum is $\"\"$ .

(c)

Concave down in the interval $\"\"$ .

There is no inflection point.

(d)

Graph of the function $\"\"$ is

$\"\"$ .