Newton's Method will fail to approximate the solution to
f
(
x
)
=
0
\displaystyle {f{{\left({x}\right)}}}={0}
f
(
x
)
=
0
with initial guess
x
0
\displaystyle {x}_{{0}}
x
0
when:
f
′
(
x
0
)
=
0
\displaystyle {f}'{\left({x}_{{0}}\right)}={0}
f
′
(
x
0
)
=
0
.
f
(
x
)
\displaystyle {f{{\left({x}\right)}}}
f
(
x
)
does not have a root.
x
0
\displaystyle {x}_{{0}}
x
0
is not close to the solution of
f
(
x
)
=
0
\displaystyle {f{{\left({x}\right)}}}={0}
f
(
x
)
=
0
.
f
(
x
)
\displaystyle {f{{\left({x}\right)}}}
f
(
x
)
is not a polynomial.
f
(
x
)
\displaystyle {f{{\left({x}\right)}}}
f
(
x
)
is not differentiable at
x
0
\displaystyle {x}_{{0}}
x
0
.
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