Problem 16 Solution pg. 3/3
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
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