AP Calculus BC Quiz
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LIMITS AND CONTINUITY
Question 1
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A continuous function
m
m
m
on
[
−
2
,
1
]
[-2,1]
[
−
2
,
1
]
satisfies
m
(
−
2
)
=
4
m(-2)=4
m
(
−
2
)
=
4
and
m
(
1
)
=
−
5
m(1)=-5
m
(
1
)
=
−
5
; does IVT guarantee a solution to
m
(
x
)
=
0
m(x)=0
m
(
x
)
=
0
?
A
Yes, because
m
m
m
is continuous on
[
−
2
,
1
]
[-2,1]
[
−
2
,
1
]
and
0
0
0
lies between
4
4
4
and
−
5
-5
−
5
.
B
No, because
0
0
0
is not between
−
2
-2
−
2
and
1
1
1
.
C
Yes, because
m
(
−
2
)
m
(
1
)
<
0
m(-2)m(1)<0
m
(
−
2
)
m
(
1
)
<
0
even if
m
m
m
is discontinuous.
D
No, because IVT requires
m
m
m
to be decreasing.
E
Yes, because
m
m
m
has a minimum on
[
−
2
,
1
]
[-2,1]
[
−
2
,
1
]
.
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