(7)

Step 1:

The function is $\"\"$ .

Find the tangent line equation at $\"\"$ .

Slope of the tangent line is the first derivative of the function at $\"\"$ .

$\"\"$

Apply derivative with respect to $\"\"$ .

$\"\"$

Slope of the tangent at $\"\"$ .

$\"\"$

Step 2:

Fin the point of tangency.

$\"\"$

So the point of tangency is $\"\"$ .

Step 3:

Slope point form of the equation is $\"\"$ .

Substitute point $\"\"$ and slope $\"\"$ .

$\"\"$

Solution:

The tangent line equation at $\"\"$ is $\"\"$ .

(8)

Step 1:

The function is $\"\"$ .

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

$\"\"$

Find the critical points.

Since it is a polynomial it is continuous at all the point.

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

Equate $\"\"$ to zero.

$\"\"$

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

The test intervals are $\"\"$ .

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

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

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

$\"\"$ in the interval of $\"\"$ .

So the function $\"\"$ is increasing at $\"\"$ .