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white (taw933) – Review 1 – benzvi – (55600)
1
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Below is the graph oF a Function
f
.
2
4
6
−
2
−
4
−
6
2
4
−
2
−
4
Use this graph to determine all oF the values
oF
x
on (
−
7
,
7) at which
f
is discontinuous.
1.
none oF the other answers
2.
no values oF
x
3.
x
=
−
1
,
1
correct
4.
x
= 1
5.
x
=
−
1
Explanation:
Since
f
(
x
) is defned everywhere on (
−
7
,
7),
the Function
f
will be discontinuous at a point
x
0
in (
−
7
,
7) iF and only iF
lim
x
→
x
0
f
(
x
)
n
=
f
(
x
0
)
or iF
lim
x
→
x
0
−
f
(
x
)
n
=
lim
x
→
x
0
+
f
(
x
)
.
As the graph shows, the only possible candi
dates For
x
0
are
x
0
= 1 and
x
0
=
−
1. But at
x
0
= 1,
f
(1) =
−
4
n
= lim
x
→
1
f
(
x
) = 0
,
while at
x
0
=
−
1,
lim
x
→−
1
−
f
(
x
) =
−
4
n
=
lim
x
→−
1+
f
(
x
) = 0
.
Consequently, on (
−
7
,
7) the Function
f
is
discontinuous only at
x
=
−
1
,
1
.
002
10.0 points
±ind all values oF
x
at which the Function
f
defned by
f
(
x
) =
x
−
7
x
2
−
5
is not continuous?
1.
x
=
√
5
2.
x
= 7
3.
x
= 7
,
−
√
5
,
√
5
4.
no values oF
x
5.
x
=
−
√
5
6.
x
=
−
√
5
,
√
5
correct
Explanation:
Because
f
is a rational Function it will Fail
to be continuous only at zeros oF the denomi
nator,
i.e.
, at the solutions oF
x
2
= 5
.
Consequently,
f
Fails to be continuous only at
x
=
−
√
5
,
√
5
.
003
10.0 points
Below is the graph oF a Function
f
.
2
4
6
−
2
−
4
−
6
2
4
6
8
−
2
−
4
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View Full Documentwhite (taw933) – Review 1 – benzvi – (55600)
2
Use the graph to determine all the values of
x
in (
−
6
,
6) at which
f
is not diFerentiable.
1.
x
=
−
3
2.
x
=
−
3
,
2
3.
x
=
−
1
,
2
4.
x
=
−
3
,
−
1
,
2
correct
5.
x
=
−
3
,
−
1
Explanation:
The graph shows that
f
has a removable
discontinuity at
x
=
−
3 and a jump disconti
inuity at
x
= 2, so
f
will not be diFerentiable
at these points. On the other hand, at
x
=
−
1
the graph is continuous but has a ‘corner’, so
it will not be diFerentiable at this point also.
Thus, on (
−
6
,
6) the function
f
will fail to be
diFerentiable at the points
x
=
−
3
,
−
1
,
2
.
004
10.0 points
After
t
seconds the displacement,
s
(
t
), of a
particle moving rightwards along the
t
axis is
given (in feet) by
s
(
t
) =
1
2
t
3
.
±ind the average velocity of the particle over
the time interval [1
,
1 +
h
].
1.
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 Spring '08
 schultz
 Differential Calculus

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