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Calculus 1 Final Exam Answer Key
1. (5 pts) A B C D E2. (5 pts) A B C D E3. (5 pts) A B C D E4. (5 pts) A B C D E5. (5 pts) A B C D E6. (5 pts) A B C D E7. (5 pts) A B C D E8. (5 pts) A B C D E9. (15 pts)
12
x
+ 3
10. (15 pts) 11. (15 pts)
1/(4
π
)
cm/s12. (15 pts)
300
m
×
600
m
9
Calculus 1 Final Exam Solutions
1. C. Find the limit.
lim
x
→
3
g
(
x
)
lim
x
→
3
x
−
4
3
−
4
−
1
Find
f
(
−
1)
.
f
(
−
1) = 3(
−
1)
2
f
(
−
1) = 3(1)
f
(
−
1) = 3
lim
x
→
3
f
[
g
(
x
)] = 3
2. B. We need to make sure that the piece-wise function is continuous through each domain and at each “break” point (the transition between functions).
x
−
3
is continuous over all of
x
.
10
The piece-wise function is continuous at the transition
x
=
−
1
since
x
≤ −
1
for
x
−
3
.
1
x
is discontinuous at
x
= 0
which is in the interval
−
1 <
x
< 2
The piece-wise function is discontinuous at
x
= 2
since
−
1 <
x
< 2
for
1
x
and
x
> 2
for
x
+ 2
.
x
+ 2
is continuous for all
x
> 2
Therefore the piece-wise function is discontinuous at
x
= 0
and
x
= 2
.3. A. Yes, there’s a zero in the interval
[0,1]
. Find the limit at the endpoints by plugging the endpoints into the polynomial.
f
(0) = 0
3
+ 2(0)
−
1 =
−
1
f
(1) = 1
3
+ 2(1)
−
1 = 2
Since
f
(0) < 0
and
f
(1) > 0
, the Intermediate Value Theorem states that there must be some
x
-value in
[0,1]
where
f
(
x
) = 0
. This means that there is a zero in the interval
[0,1]
.4. C. Use the chain rule to find the derivative:
f
′
(
g
(
x
))
g
′
(
x
)
“derivative of the outside times the derivative of the inside”
f
(
x
) = 4
−
x
2
11

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