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Mathematics
Problem
For the curve
x
=
t
2
+
1
x = t^{2}+1
x
=
t
2
+
1
,
y
=
t
3
−
3
t
y = t^{3}-3t
y
=
t
3
−
3
t
, consider the open interval
(
−
3
,
3
)
(-3,3)
(
−
3
,
3
)
of the parameter
t
t
t
. On which part of this interval is the curve concave upward? a)
(
−
3
,
0
)
(-3,0)
(
−
3
,
0
)
only b)
(
0
,
3
)
(0,3)
(
0
,
3
)
only c)
(
−
3
,
0
)
∪
(
0
,
3
)
(-3,0)\,\cup\,(0,3)
(
−
3
,
0
)
∪
(
0
,
3
)
d) Nowhere in
(
−
3
,
3
)
(-3,3)
(
−
3
,
3
)
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