Problem 16 – Solution: 1, 2, 3

4. Filling the empty spaces in VT.

4.1 The signs for

From point 2.3 of this presentation we can see that:

The denominator is positive on the defined interval (it is equal with ) therefore the sign of  is dependent on the sign of its numerator.

when .

would have been a minimum if it was in a defined domain.

When 

When 

4.2 Input and on row :

• for when is “”

• for when is “”

5. Graphing, on Cartesian Coordinates System the function.

5.1 Drawing the oblique asymptotes  calculated at point 3.1 in this presentation.

5.2 Drawing all set points :

and ,  and with approximation the points where or is or .

5.3 Graphing the function (Fig. 5).

Problem 16 – Solution: 1, 2, 3