(5)

Step 1:

The 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.

$\"\"$

$\"\"$ and $\"\"$ .

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

Solution :

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

(6)

Step 1:

The function is $\"\"$ .

$\"\"$ .

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

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

\ \

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

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 $\"\"$ .

(7)

Step 1:

The function is $\"\"$ .

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

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

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

Find the local maximum and local minimum.

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

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

$\"\"$

Local maximum is $\"\"$ .

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

$\"\"$

Local minimum is $\"\"$ .

Solution :

Local maximum is $\"\"$ .

Local minimum is $\"\"$ .