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Lebl-Basic-Analysis-I
· Problem 4.4.6
The Derivative
Mathematics · inverse-function-theorem-counterexample
Problem
Let
f
(
x
)
:
=
x
+
2
x
2
sin
(
1
/
x
)
f(x) := x + 2x^2 \sin(1/x)
f
(
x
)
:=
x
+
2
x
2
sin
(
1/
x
)
for
x
≠
0
x \neq 0
x
=
0
and
f
(
0
)
:
=
0
f(0) := 0
f
(
0
)
:=
0
. Show that
f
f
f
is differentiable at all
x
x
x
, that
f
′
(
0
)
>
0
f'(0) > 0
f
′
(
0
)
>
0
, but that
f
f
f
is not invertible on any open interval containing the origin.
Related theory
Smooth Manifolds And The Inverse Function Theorem
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