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Mathematics
Problem
A twice–differentiable function
h
h
h
satisfies
h
′
′
(
t
)
=
3
t
2
−
2
t
+
1
h''(t)=3t^{2}-2t+1
h
′′
(
t
)
=
3
t
2
−
2
t
+
1
for all real
t
t
t
. In addition,
h
′
(
0
)
=
4
h'(0)=4
h
′
(
0
)
=
4
and
h
(
0
)
=
−
1
h(0)=-1
h
(
0
)
=
−
1
. What is the value of
h
(
2
)
h(2)
h
(
2
)
? a)
25
3
\displaystyle \tfrac{25}{3}
3
25
b)
23
3
\displaystyle \tfrac{23}{3}
3
23
c)
31
3
\displaystyle \tfrac{31}{3}
3
31
d)
11
3
\displaystyle \tfrac{11}{3}
3
11
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