Find the Taylor polynomial of order 4(four) generated by f(x) =x^1/3 at x=8

The Taylor series is a particular way to approximate a function with a polynomial in the neighbourhood of a generic point
(
x
0
,
f
(
x
0
)
)
.
f
(
x
)
=
f
(
x
0
)
+
f
'
(
x
0
)
(
x
−
x
0
)
1
1
!
+
f
'
'
(
x
0
)
(
x
−
x
0
)
2
2
!
+
...
or, with the series notation:
f
(
x
)
=
+
∞
∑
i
=
0

f
(
i
)
(
x
0
)
(
x
−
x
0
)
i
i
!
.
So, let's calculate all the derivatives we need:
f
(
1
)
=
5
f
(
1
)
(
x
)
=
8
x
7
+
4
x
3
and
f
(
1
)
(
1
)
=
12
f
(
2
)
(
x
)
=
56
x
6
+
12
x
2
and
f
(
2
)
(
1
)
=
68
f
(
3
)
(
x
)
=
336
x
5
+
24
x
and3
)
1
)=360f(4)(
x)=
1680x
4+24
and f(4)(1)
=1704 so:
f(x)=5+12(x−1)11!+68(x−1)22!+360(x−1)33!+1704(x−1)44!
It's easy to understand that if you would calculate the Taylor's series until the 8th order you will obtain the function itself, because it is a polynomial!

D) pirates hope it ..

Idon’t know what the answer is i wish i could