Name:
Section:
Instructor:
Time: 75 minutes. You may not use calculators on this exam
Problem Points Score
1 14
2
3
4
∑
30
20
36
100 d sin
( x
) = cos
( x
)
, d x d d x d d x asin acsc
(
( x x
) =
1
√
1
− x
2
1
) = − x
√ x
2
,
−
1
, sin
(
2 x
) =
2 sin
( x
) cos
( x
) tan 2
( x
) +
1
= sec 2
( x
) d d cos
( x
) = − sin
( x
)
, tan
( x
) = sec 2
( x
)
, d x d x d d x d acos
( x
) = − asec
( x
) =
1
√
1
− x
2
1
√
,
, x x
2
−
1 d d x d atan
( x
) =
1
1
+ x
2
, sec
( x
) = sec
( x
) tan
( x
)
, d x d x
∫ ln x d x
= x ln x
− x
+
C
∫ sec
( x
) d x
= ln
∣ sec
( x
) + tan
( x
)∣ +
C
1
+ cos
(
2 x
)
1
− cos
(
2 x
) cos 2
( x
) = sin 2
( x
) =
2 2
Theorem (The Derivative Rule for Inverses) If f has an interval
I as domain and f ′
( x
) exists and is never zero on
I
, then f −
1 is differentiable at every point in its domain. The value of
( f −
1
)
′ at a point b in the domain of f
−
1 is the reciprocal of the value of f
′ at the point a
= f
−
1
( b
)
:
( f −
1
( b
) = f
′
( f
−
1
1
( b
))
.
)
′
A a b c d e
B a b c d e
C a b c d e
D a b c d e
E a b c d e
F a b c d e

1) a) Simplify cos
( arctan
( x
5
))
.
d b) Determine d x ln
(
√ x
5
( x
+
+
1
)
2
7
)
.

2) Evaluate the following integrals. Show your work.
a)
∫ t
⋅
3 t
2 d t b)
∫
√
− x
2
1
−
8 x
−
15 d x

3) Let f
( x
) =
2 x
− sin
( x
) cos
( x
)
.
a) What is the range of f
?
b) Explain why f has an inverse. (If you refer to a criterion or test, indicate why the test applies. If you refer to the graph of f you must show that your sketch of the graph is good enough!) c) Determine
( f −
1
)
′
(
2
π
)

4) The following multiple choice problems will be graded correct answer only. You do not need to show work, and no partial credit will be given. Record your answer in the answer block on the front page . Answers given on these pages will not be scored. You also may tear off these pages and do not need to hand them in.
It is strongly recommended that you work out the problems until the correct answer is uniquely determined and don’t just try to solve them by “intuition” or “guessing” – doing so is likely to result in a wrong pick.
Each correct anwer is worth 6 points, each incorrect answer is counted as
−
2 points. (Unanswered questions are 0 points, questions in which more than one answer is ticked are considered to have been answered wrongly.) You cannot get less than 0 points in this part.
A) a
If the function f is defined by f
( x
) = x
5
−
1 then the inverse f
− 1
√
1
/
5 x
+
1 b 1
/
√
5 x
+
1 c
√
5
√ x
−
1 d 5 x
−
1 e
√
5 x
+
1 is given by
B) a
The set of all points
( e t
, t
) where t is a real number is the graph of y
=
1
/ e x b e
1 / x c x e
1 / x d 1
/ ln
( x
) e ln
( x
)

C) Which of the following functions is a general solution to the differential equation y
′
+
2 y
= e
− x a d y
( t
) =
2 e y
( t
) =
C e
− x
− 2 x
+
C b y
( t
) =
C e
+ e
− x e y
( t
) = e
− x
− 2 x c
+ e y
( t
) = e
−
C x
− 2 x
+
C e
− x
D) The area of the region bounded by the lines x
=
0, x
=
2, and y
=
0 and the curve y
= e x
/ 2 is a
( e
−
1
)/
2 b e
−
1 c 2
( e
−
1
) d 2 e
−
1 e 2 e

E) a
If f
′
( x
) = − f
( x
) and f
(
1
) =
1 then f
( x
) =
1
/
2 e
− 2 x
+ 2 b e
− x
− 1 c e
1 − x d e
− x e
− e x
F) The number of bacteria in a culture is growing according to the function f
( t
) =
3000 e
2 t
/ 5 . At a certain time t
= t
1
Find the number present at t
= t
1 +
5 the number of bacteria present was 7,500.
a 1200 e
2 b 3000 e
2 c 7500 e
2 d 7500 e
5 e 15000
/
7 e
7