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Mathematics
Mathematics
Problem
Given that
f
′
(
2
)
=
6
f'(2)=6
f
′
(
2
)
=
6
and
f
′
(
1
)
=
4
f'(1)=4
f
′
(
1
)
=
4
, then
lim
h
→
0
f
(
2
h
+
2
+
h
2
)
−
f
(
2
)
f
(
h
−
h
2
+
1
)
−
f
(
1
)
=
\lim_{h \rightarrow 0} \frac{f(2h+2+h^2) - f(2)}{f(h-h^2+1) - f(1)} =
lim
h
→
0
f
(
h
−
h
2
+
1
)
−
f
(
1
)
f
(
2
h
+
2
+
h
2
)
−
f
(
2
)
=
(a) Does not exist (b) Is equal to
−
3
/
2
-3/2
−
3/2
(c) Is equal to
3
/
2
3/2
3/2
(d) Is equal to 3
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