Suppose
f
′
′
(
x
)
=
−
18
x
+
4
e
−
2
x
\displaystyle {f}{''}{\left({x}\right)}=-{18}{x}+{4}{e}^{{-{2}{x}}}
f
′′
(
x
)
=
−
18
x
+
4
e
−
2
x
,
f
′
(
0
)
=
7
\displaystyle {f}'{\left({0}\right)}={7}
f
′
(
0
)
=
7
, and
f
(
0
)
=
5
\displaystyle {f{{\left({0}\right)}}}={5}
f
(
0
)
=
5
.
Then
f
(
x
)
=
\displaystyle {f{{\left({x}\right)}}}=
f
(
x
)
=
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Question 6 Part 1 of 2
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