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

$\"\"$

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

$\"\"$

Step 2:

Fin the point of tangency.

Substitute $\"\"$ in the funtion. \ \

$\"\"$

Point of tangency is $\"\"$ .

Step 3:

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

Substitute $\"\"$ and slope $\"\"$ in the point slope form.

$\"\"$

Solution:

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

(8)

Step 1:

The function is $\"\"$ .

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

$\"\"$

Find the critical points. \ \

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

Equate $\"\"$ to zero. \ \

$\"\"$

Apply logarithm on each side.

$\"\"$

Critical points is $\"\"$ .

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

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

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

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

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

(a)

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

Hence the function $\"\"$ is decreasing at $\"\"$ and $\"\"$ .

(b)

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

Hence the function $\"\"$ is decreasing at $\"\"$ .

(c)

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

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

Solution: \ \

(a) The function $\"\"$ is decreasing at $\"\"$ .

(b) The function $\"\"$ is decreasing at $\"\"$ .