School of Mathematics and Statistics
Carleton University
Math. 69.104
SOLUTIONS TO TEST 2

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Student Number: This test is out of a Total of 30.

PART I: Multiple Choice Questions
(Choose and CIRCLE only ONE answer)

1. [2 marks] Let . Which of the following expressions represents the value of ?

   (a) -1, , (c) 1, (d) This derivative does not exist.

2. [2 marks] Let . Then is equal to:

   (a) , (b) , (c) ,

3. [2 marks] Let . Then the function f is increasing on the interval:

   , (b) , (c) , (d) .

4. [2 marks] Let . Then f has points of inflection at:

   (a) no point whatsoever, , (c) x=0 only, (d) , only.

5. [2 marks] Answer TRUE or FALSE:

   If then, for each x, its derivative

   

   (a) , (b) FALSE

PART II: Show all work here.
No additional pages will be accepted

1. [5+5 marks] Evaluate the following integrals using any method:

   a) .

   Solution: Let . Then .

   

   Thus,

   

   b) Evaluate and find that antiderivative such that

   Solution Let . Then or . So,

   

   Since we want we get . Hence

   

2. [5+5 marks] Let .

   a) Determine all the intervals where f is increasing and decreasing.

   Solution:We know that .

   It follows that f is increasing if , that is, when |x| > 1, or is in either or .

   Similarly, f is decreasing when , which in this case means that |x| < 1, or x is in the interval (-1, 1). b) In what intervals is f concave up and concave down? Where, if any, is there a point of inflection?

   Solution: In this case, we don' t need the SDT of since Note that when 6x > 0 or, equivalently, when x > 0. So f is concave up in this case.

   Similarly, we can see that f is concave down when x < 0. This makes x= 0 a point of inflection!

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