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Inverse Function Theorem and Implicit Function
Theorem
Dr. M. Asaduzzaman
Professor
Dept. of Mathematics
University of Rajshahi
Email: [email protected]
1
Contents
Euclidean Spaces
Normed Linear Spaces
Metric Induced From Norms
Open Sets in ℝn
Continuous Functions
Directional and Partial Derivatives
Derivatives
Jacobian and Regular Maps
Inverse Function Theorem
Implicit Function Theorem
2
Euclidean Spaces
Definition: A nonempty set F with addition and multiplication defined in F is called a
field if the following conditions hold:
i) x+y, xy  F,  x, y F
ii) x+y=y+x and xy=yx  x, y F
iii) (x+y)+z=x+(y+z) and (xy)z=x(yz)  x, y, zF
iv)  two elements 1 and 0 in F such that x+0=x and x.1=x  xF
v)  xF  an element –xF such that x+(-x)=0 and for every nonzero xF
 an element x-1F such that xx-1=1
vi) x(y+z)=xy+xz,  x, y, z F.
Example 1. ℝ, the set of all real numbers is a field under usual addition and
multiplication. This field is called the real field.
Example 2. C, the set of all complex numbers is a field under addition and
multiplication of complex numbers. This field is called the complex field.
3
Euclidean Spaces
Definition: A nonempty set V with addition and scalar multiplication defined in V is
called a vector space over the field F if  x, y, zV and  , F the following
conditions hold:
i) x+y, x V
ii) x+y=y+x
iii) (x+y)+z=x+(y+z)
iv)  an element 0 in V such that x+0=x
v)  xV  an element –xV such that x+(-x)=0
vi) (x+y)=x+y and (+)x=x+x
vii) ()x=(x) and (+)x=x+x
viii) 1x=x
The elements of V are called vectors and those of F are called scalars.
4
Euclidean Spaces
Example: Every field is a vector space over itself.
Example: Let n be a positive integer. Then the set
ℝn={(x1,x2,,xn): xiℝ, 1≤i ≤n} is a vector space over the real field if we define
addition and scalar multiplication component wise.
Definition: We define the so-called “inner product “ (or scalar product) of two
vectors x=(x1,x2,,xn), y=(y1,y2,,yn) in ℝn by
xy=x1y1+x2y2++xnyn
and the norm of x by
|x|=(x12+x22++xn2)1/2
then the vector space with the above inner product and norm is called n
dimensional Euclidean space.
We then have following theorem:
5
Euclidean Spaces
Theorem: Suppose x, y, zRn, and  is a real. Then
(a) |x|≥0;
(b) |x|=0 iff x=0;
(c) |x|=|||x|;
(d) |xy|≤|x||y|;
(e) |x+y|≤|x|+|y|;
(f) |x-z|≤|x-y|+|y-z|.
Based on property (a), (b), (c) and (e) of the above theorem
we define norm on any vector space over real or complex field.
6
Normed Linear Spaces
Definition: Let X be a vector space over the real or complex field K. Then a
function
║║: X→ℝ is called a norm in X if it satisfies the following properties:
(a) ║x║≥0;
(b) ║x║=0 iff x=0;
(c) ║x║=||║x║;
(d) ║x+y║≤║x║+║y║.
A vector space over the real or complex field is called a linear
space and a linear space with a norm is called a normed linear
space.
7
Normed Linear Spaces
Example: Let X =Rn and let ║║: X→ℝ be defined by
║x║= |x1|+|x2|+  +|xn|, where x=(x1,x2,,xn)X. Then X is a normed linear
space.
Example: Let X =ℝn and let ║║: X→ℝ be defined by
║x║= max{|x1|,|x2|,  ,|xn|}, where x=(x1,x2,,xn)X. Then X is a normed linear
space.
Thus there are many ways to define a norm on the same
linear space.
8
Metric Induced from Norms
For any normed linear space we have the following theorem:
Theorem: Suppose X is a normed linear space . Then for any x, y, zX , we
have ║x-y║≤ ║x-z║+ ║y-z║.
Thus in a normed linear space X if we define
d(x,y)= ║x-y║ then d satisfies all conditions of a metric.
This metric is called metric induced from norm.
It is clear that different norm on the same linear space
induced different metric.
9
Open Sets in Rn
We now consider only the Euclidean norm and the
metric induced from Euclidean norm.
Definition: Let x0ℝn and r>0 be given. We define the set
S(x0,r)={xℝn : |x-x0|<r} and call it a neighborhood of x0 of
radius r.
From the definition it is clear that neighborhood changes with
x0 and r.
Example: In ℝ1 neighborhoods are open intervals and in ℝ2
neighborhoods are open discs.
10
Open Sets in ℝn
r
.
x0
S(x0,r)
O
11
Open Sets in ℝn
Definition: Let E be a subset of a Euclidean space ℝn . A point
x0E is said to be an interior point of E if there exists a
neighborhood S(x0,r) of x0 such that S(x0,r)E.
Definition: A subset E of a Euclidean space ℝn is called an
open set if every point of E is an interior point of E.
Example: For any x0ℝn and for any r>0 the neighborhood
S(x0, r) is an open set.
Example: Union of any number of neighborhoods is an open
set.
Example: A finite subset of ℝn cannot be an open set.
12
Continuous functions
Definition: Let E be a subset of a Euclidean space ℝn . A point
x0 ℝn is said to be an limit point of E if every neighborhood
contains infinitely many points of E.
Definition: Let E be a subset of ℝn and let :E→ ℝm be a
function. Let x0 be a point as well as a limit point of E. Then  is
said to be continuous at x0 if given  there exists a  such
that
|x- x0|
whenever
|x-x0| and xE.
If  is continuous at every point of E then  is called continuous
on E.
13
Continuous functions
Example: The function
(x,y)=1-x2-y2
is continuous on ℝ2 .
Example: The function
 xy
, ( x, y )  (0,0)
 2
2
f ( x, y )   x  y

( x, y )  (0,0) .
 0,
is not continuous at (0,0) .
14
Directional and Partial Derivatives
Definition: Let  be real valued function defined on an open
subset E of ℝn and let x0E be given. We define the
directional derivative of  at x0 in the direction of unit vector u
to be
f ( x 0  tu )  f ( x 0 )
lim
t 0
t
provided the limit exists. It will be denoted by Du(x0).
15
Directional and Partial Derivatives
Example: Let
f ( x, y ) | x 2  y 2 | , x 0  (0,0) .
Then directional derivatives of  exist only in the four
directions (1/2, 1/2).
Definition: The partial derivatives of  are defined as the
directional derivatives in the directions of standard basis
vectors e1, e2,, en. The i th partial derivative of  at x is
f
(x).
usually denoted by Di(x), f xi (x ) or
xi
Thus
f (x  te )  f (x)
Di (x)  lim
t 0
provided the limit exists.
i
t
16
Directional and Partial Derivatives
Problem: Let
f ( x, y)  ( x  1)  y .
2
2
Find the directional derivative at e2 in any direction u.
Solution: Let u=(cos , sin ). Then

f (e 2  tu)  f (e 2 )
Du (e 2 )  lim
t 0
t
f (t cos  ,1  t sin  )  f (0,1)
 lim
t 0
t
(t cos   1)  (1  t sin  ) 2
 lim
t 0
t
 2(cos   sin  ).
2
17
Directional and Partial Derivatives
The following example shows that existence of partial
derivatives at a point does not imply that directional
derivatives exist in all directions.
Example: Let (x,y)=(xy) 1/3 and u=(cos , sin). Then
f (tu)  f (0)
Du (0)  lim
t 0
t
(cos  sin  )1/ 3
 lim
t 0
t 1/ 3
which exists and equal to 0 if and only if sin cos=0. That is if and
only if u=e1, e2, -e1 or –e2.
18
Derivatives
In order to define derivaties of a function whose domain is an
open subset of ℝn, let us take another look at the familiar
case n=1, and let us see how to interpret the derivative in
that case in a way which will naturally extend to n1.
If  is a real function with domain (a,b)ℝ1 and if x(a,b),
then (x) is usually defined to be the real number
f ( x  h)  f ( x )
lim
,
h 0
h
provided the limit exists. Thus
(1)
x+hxxh  rh
where the remainder r(h) is small in the sense that
r ( h)
lim
 0.
h 0 h
19
Derivatives
Thus (1) expresses the difference x+hx as the sum of
the linear function that takes h to xh, plus a small
remainder.
We can therefore regard the derivative of  at x, not as a real
number but as the linear operator on ℝ1 that takes h to
xh.
Let us next consider a function f that maps (a,b)ℝ1 into ℝm .
In that case, fx was defined to be that vector y Rm for
which
 f ( x  h)  f ( x )

lim 
h 0

h
 y   0.

We can again rewrite this in the form
f ( x  h)  f ( x)  hy  r (h)
(2)
where r(h)/h→0 as h→0. The main term on the RHS of (2) is
again a linear function of h.
20
Derivatives
Each yℝm induces a linear transformation of ℝ1 into ℝm, by
associating to each h ℝ1 the vector hy ℝm. This
identification of ℝm with L(ℝ1, ℝm) allow us to regard f(x) as a
member of L(ℝ1, ℝm).
Thus, if f is a differential mapping of (a,b)ℝ1 into ℝm, and if
x(a,b), then f(x) is the linear transformation of ℝ1 into ℝm
that satisfies
(3)
f ( x  h)  f ( x )  f ' ( x ) h
lim
 0,
h 0
h
or, equivalently,
(4)
lim
h 0
f ( x  h)  f ( x )  f ' ( x ) h
h
 0.
21
Derivatives
Motivating from the above fact, we define derivative of a
function from ℝn into ℝm.
Definition: Suppose E is an open set in ℝn, f maps E into ℝm,
and xE. If there exists a linear transformation A of Rn into
ℝm such that
(5)
lim
h 0
f (x  h)  f (x)  Ah
h
 0,
then we say that f is differentiable at x, and we write
(5)
f(x)=A.
If f is differentiable at every xE , we say that f is differentiable in
E
22
Derivatives
The following theorem proves that if a function f is
differentiable at x then f(x) is unique.
Theorem: Suppose E is an open set in ℝn, f maps E into ℝm,
and xE. If there exists a linear transformation A1 and A2 of
ℝn into ℝm such that
lim
h 0
f (x  h)  f (x)  Ah
h
 0,
holds with A=A1 and with A=A2. Then A1=A2.
23
Derivatives
We see that if a function f:E→ ℝm is differentiable at a point
xE then f(x)L(ℝn, ℝm). Since every mn matrix with real
entries are in L(ℝn, ℝm) and every member of L(ℝn, ℝm) has
a matrix representation with respect to given bases of ℝn and
ℝm. In order to make everything simple we will always
consider the standard bases.
Thus when we see f(x), we shall think of f(x) as an mn
matrix which is the matrix of f(x) with respect to standard
bases.
24
Derivatives
The following theorem extends chain rule of derivatives of
single variable.
Theorem: Suppose E is an open set in ℝn, f maps E into ℝm,
f is differentiable at x0E, g maps an open set containing f(E)
into ℝk, and g is differentiable at f(x0). Then the mapping F of
E into Rk defined by
F(x)=g(f(x))
is differentiable at x0, and
F(x0)=g(f(x0))f(x0).
x0
f
f(x0)
F
g
g(f(x0))
25
Derivatives
Suppose E is an open set in ℝn, f maps E into ℝm. Let {e1,
e2,…., en} and {u1, u2,…., um} be the standard bases of ℝn
and ℝm respectively. Let the component functions of f be 1,
2,…., m. Then for xE,
f(x)=(1(x), 2(x), … , m(x))
and each i is a real valued function on E. We denote partial
derivative of i with respect to xj by the notation Dji. These
partial derivatives are called partial derivatives of f.
If we consider i as row index and j as column index then
partial derivatives of f form an mn matrix.
26
Derivatives
The following theorem shows that if f is differentiable at a
point x then its partial derivatives exist at x, and they
determine f(x) completely.
Theorem: Suppose f maps an open set E ℝn into ℝm, and f
is differentiable at point xE. Then the partial derivatives
(Dji)(x) exist, and the matrix of f(x) is
 ( D1 f1 )( x)  ( Dn f1 )( x) 
 .......................................... .


( D1 f m )( x)  ( Dn f m )( x)
27
Derivatives
The following theorem shows that differentiability at a point x
is sufficient for a function to be continuous at the point.
Theorem: Suppose f maps an open set E ℝn into ℝm, and f
is differentiable at point x0E. Then f is continuous at x0.
28
Derivatives
The following example shows that existence of partial
derivatives at a point is not enough for a function to be
differentiable or even continuous.
Example: Let
 2 xy2

, ( x, y )  (0,0)
f ( x, y )   x 2  y 4
 0 ,
( x, y )  (0,0),
then both partial derivatives x and y exist at (0,0) but the
function  is not continuous at (0,0).
29
Derivatives
Definition: A differentiable mapping f of an open set Eℝn
into Rm is said to be continuously differentiable in E if f is a
continuous mapping of E into L(ℝn , ℝm ).
More explicitly, it is required that to every 
corresponds a  such that
f ' ( y)  f ' ( x)  
If yE and |x-y|.
If this is so, we also say that f is a 𝒞- mapping, or that f𝒞(E).
30
Derivatives
The following theorem shows that existence of continuous
partial derivatives at a point for a function is a sufficient
condition to be differentiable at the point.
Theorem: Suppose f maps an open set E ℝn into ℝn. Then
f𝒞(E) if and only if the partial derivatives Dji exists and are
continuous on E for 1≤i ≤m, 1 ≤j ≤n.
31
Derivatives
The following example shows that continuity of partial
derivatives at a point is not a necessary condition for a
function to be differentiable there.
Example: Let
( x  y) 2 sin ( x1 y ) , x  y
f ( x, y)  
0,
x  y,

then (x and y exist for all (x,y))  is differentiable at every
point on the line x=y but the functions x and y are not
continuous on the line x=y.
32
Derivatives
The existence of a derivative for a real valued function of one
variable is a fact of considerable interest. Geometrically it
says that a tangent line exists. However, the fact that a real
valued function of several variables has partial derivatives is
not in much interest. For, one thing, the existence of
directional derivatives in the directions of standard basis
vectors e1, e2,…., en does not imply that directional
derivatives in all direction exist. Moreover, the function need
not have a tangent hyper plane even if there is a derivative in
every direction.
It can be shown that most of the basic properties of differentiable
functions of one variable remain true for differentiable functions of
several variables.
33
Derivatives
Now we will see the geometrical similarities:
Theorem: Suppose a real function  defined on (a,b)ℝ1 is
differentiable at a point x0(a,b). Then the tangent line to the
curve y=(x) has the equation y=(x0)+(x0)(x-x0).
y
(x0,f(x0))
o
a
x0
y=f(x)
b
x
34
Derivatives
Theorem: Suppose a real function  defined on an open set
ERn is differentiable in E and x0E is a point on the hyper
surface (x)=c . Then the tangent hyper plane to the hyper
surface z=(x) at the point (x0,(x0))
n
has the equation
z  f ( x 0 )   Di f ( x 0 )( xi  x0 i ).
i 1
z  z0  f x ( x0 , y0 )( x  x0 )  f y ( x0 , y0 )( y  y0 ).
E
35
Derivatives
Theorem: Suppose a real function  defined on an open set
Eℝ2 is differentiable in E and (x0,y0)E be a point on the
curve (x,y)=c. Then the gradient vector (x, y) of  at (x0,y0)
represents the direction of the normal to the curve (x,y)=c at
(x0,y0).
y
f(x,y)=c
(x0,y0)
o
x
36
Derivatives
Theorem: Suppose a real function  defined on an open set
Eℝn is differentiable at a point x0E. Then the gradient vector
(D1, D2, … , Dn) of  at x0 represents the direction of the
normal to the hyper surface (x)=c at x0.
x0
37
Jacobian
Suppose f maps an open set E ℝn into itself, and x0E.
Then f is continuous at x0. Let f=(1, …. ,n),
so that 1, …. ,n are real valued functions with domain E.
We suppose all first order partial derivatives (Dji)(x0) exist
for 1≤i,j≤n. In this case we say that f possesses first order
partial derivatives at x0. The determinant
( D1 f1 )( x 0 )  ( Dn f1 )( x 0 )
..........................................
( D1 f n )( x 0 )  ( Dn f n )( x 0 )
is called the Jacobian of the function f at the point x0 and is
denoted by Jf(x0). Another common notation of Jacobian is
 ( f1 , f 2 ,  , f n )
.
 ( x10 , x20 ,  , xn 0 )
38
Jacobian
Definition: A function f is said be one-to-one, or invertible,
if f(x1) and f(x2) are distinct whenever x1 and x2 are distinct
points of Df.
Example: The linear map f(x,y)=(x-y, x+y) is one-to-one in
the domain Df=ℝ2.
Example: The map f(x,y)=(x2-y2, 2xy) is not one-to-one in
the domain Df=ℝ2, since f(x,y)=f(-x,-y).
Definition: A function f may fail to be one-to-one, but be oneto-one on a subset S of Df. By this we mean that f(x1) and
f(x2) are distinct whenever x1 and x2 are distinct points of S. In
this case, f is not invertible, but if fS is defined on S by
fS(x)=f(x), xS,
and left undefined for xS, then fS is invertible. We say that fS
is the restriction of f to S.
39
Jacobian
Definition: If a function f is one-to-one on a neighborhood
of x0, we say that f is locally invertible at x0.
Example: The map f(x,y)=(ex cos y, ex sin y) is not one-toone in the domain Df=ℝ2, since f(x,y)=f(x,y+2). But this map
is one-to-one on S={(x,y):   x  ,  ≤ y 2 }. This map is
locally invertible on the entire plane.
Remark: The above example proves that locally invertible
at every point of the domain of a function does not imply that
the function is invertible.
40
Jacobian
Definition: A function f : ℝn → ℝn is said to be regular on an
open set S if f is one-to-one and continuously differentiable
on S, and Jf(x)0 if xS.
Example: If f(x,y)=(x-y, x+y), then
1 1
Jf ( x, y ) 
 2, for all (x,y)ℝ2.
1 1
So f is regular on ℝ2.
Example: If f(x,y)=(x+y, 2x+2y), then
1 1
Jf ( x, y ) 
 0, for all (x,y)ℝ2.
2 2
So f is not regular on any open subset of ℝ2.
41
Jacobian
Example: If f(x,y)=(ex cos y, ex sin y), then
Jf ( x, y) 
e x cos y  e x sin y
e x sin y
e x cos y
 e 2 x ( 0),
for all (x,y)ℝ2.
So f is regular on any open set S on which f is one-to-one.
Remark: The above example proves that non zero Jacobian
in the entire plane does not prove that the function is one-toone in the entire plane.
42
Inverse function theorem
The inverse function theorem states, roughly speaking, that a
continuously differentiable mapping f is invertible in a
neighborhood of any point x at which the linear transformation
f(x) is invertible.
43
Inverse function theorem
Theorem (Inverse function theorem): Suppose f is a
continuously differentiable mapping of an open set Eℝn into
ℝn, f(a) is invertible for some aE, and b=f(a). Then
(a) there exists open set U and V in ℝn such that aU,
bV, f is one-to-one on U, and f(U)=V;
(b) if g is the inverse of f [ which exists by (a)], defined in V
by g(f(x)=x
(xU),
then g is continuously differentiable in V and
g(y)={f(g(y))}-1
(yV).
44
Hypothesis: (i) Eℝn is open. (ii) f : E→ ℝn . (iii) f𝒞(E) .
(iv) a E such that Jf(a)0. (v) b=f(a).
Conclusion:
(i)  open sets U and V in ℝn such that aUE, bVf(E) , f is
one-to-one and f(U)=V .
(ii) if g is the inverse of f defined in V then g𝒞(V) and
g(y)={f(g(y))}-1
(yV).
b=f(a)
a
f
.
U
g
.
V=f(U)
f(E)
E
Rn
Rn
45
2

x  y  ( z  1)  1 
Example: Let a=(1,2,1) and


2
f ( x, y, z )   y  z  ( x  1)  1 .
 z  x  ( y  2) 2  3


1
2 z  2
 1
f ' (a)  2 x  2
1
1 ,
 1
2y  4
1  1 1 0
Jf (a)  0 1 1  2.
so
1 0 1
Then
In this case, it is difficult to describe U of find g=fU-1 explicitly;
however we know that f(U) is a nbd. of
b=f(a)=(4,2,5), that
g(b)=a=(1,2,1), and that
46
1
1 1 0
 1 1 1 
1


1
g ' (b)  [f ' (a)]  0 1 1   1 1  1.
2
1 0 1
 1 1 1 
Therefore
1
 1  1 1   y1  4 
1





g(y )  2   1 1  1  y2  2  e(y ),
2
1
 1 1 1   y3  5 
where
lim
y ( 4 , 2 , 5 )
e( y )
( y1  4)  ( y2  2)  ( y3  5)
2
2
2
 0.
Thus we have approximated g near b=(4,2,5).
47
IImplicit function theorem
If  is a continuously differentiable real function in the plane,
then the equation (x,y)=0 can be solved for y in terms of x in
a nbd. of any point (a,b) at which (a,b)=0 and /y0.
Similarly x can be solved in terms of y near (a,b) if /x0. For
a simple example which illustrates the need for assuming
/y0 , consider
y
(x,y)=x2+y2-1.
(x,y)=0
.
O
x
48
IImplicit function theorem
We now consider mappings from ℝn+m to ℝm. It will be
convenient to denote points in ℝn+m by
(x, u)  ( x1 , x2 ,, xn , u1 , u2 ,, um ).
To motivate the problem we are interested in, we first ask
whether the linear system of m equations in m+n variables
a11x1  a12 x2    a1n xn  b11u1  b12u2    b1mum  0
a21x1  a22 x2    a2 n xn  b21u1  b22u2    b2 mum  0

am1 x1  am 2 x2    amn xn  bm1u1  bm 2u.2    bmmum  0
determines u1 , u2 ,, um uniquely in terms of x1 , x2 ,, xn .
49
By rewriting the system in matrix form as
(1)
where
Ax+Bu=0,
 a11
a
A   21
 

 am1
 x1 
x 
x   2 ,

 
 xn 
a12
a22

am 2




and
a1n 
a2 n 
,
 

amn 
 b11
b
B   21
 

 bm1
b12
b22

am 2




b1m 
b2 m 
,
 

bmm 
 u1 
u 
u   2 .

 
un 
. uniquely for u in
We see that the equation (1) can be solved
terms of x if the square matrix B is nonsingular. In this case the
solution is u=-B-1Ax.
50
For our purpose it is convenient to restate this: If
(2)
f(x,u)= Ax+Bu,
where B is nonsingular, then the system f(x,u)= 0 determines u
as a function of x, for all x in ℝn .
Notice that f in (2) is a linear transformation from ℝn+m to ℝm.
Now the question of whether an arbitrary system
f(x,u)= 0
determines u as a function of x is too general to have a useful
answer.
The implicit function theorem answers this question in an
important special case.
51
Notations: We partition the differential matrix of
f : ℝn+m →ℝm :
 f1
 x
 1
 f 2
f '   x1
 
 f
 m
 x1
f1
x2
f 2
x2

f m
x2
f1
xn
f 2

xn
 
f m

xn

f1
u1
f 2
|
u1
| 
f m
|
u1
|
f1
u2
f 2
u2

f m
u2
f1 
um 

f 2 

um 

 
f m 

um 

or
f=[fx,fu],
where fx denotes the submatrix to the left of the dashed line
and fu is to the right.
52
Theorem(Implicit function theorem): Suppose f : ℝn+m →ℝm is
continuously differentiable on an open set E ℝn+m containing
(x0,u0). Let f (x0,u0)=0, and suppose that fu (x0,u0) is
nonsingular. Then there exists a neighborhood M of (x0,u0),
contained in E, on which fu (x,u) is nonsingular and a
neighborhood N of x0 in ℝn on which a unique continuously
(
differentiable map g : ℝn→ℝm is defined, such that g(x0)=u0
and
(x,g(x))M and f(x,g(x))=0 if xN.
Moreover,
g(x)=-[fu(x,g(x)]-1fx(x,g(x), xN.
53
Eℝn+m
u
f(x,u)=0
ℝm
E
M
u=g(x)
*
(x0,u0)
x0
ℝn
*
N
x
54
Example: Take m=2, n=3, and consider the mapping f=(1,2) of
ℝ5
into
ℝ2
x1
f
(
x
,
x
,
u
,
u
,
u
)

2
e
 x2u1  4u2  3
given by
1 1
2
1
2
3
f 2 ( x1 , x2 , u1 , u2 , u3 )  x2 cos x1  6 x1  2u1  u3 .
If x0=(0,1) and u0=(3,2,7), then f(x0,u0)=0.
With respect to the standard bases, the matrix of f(x0,u0) is
 2 3 1 4 0 
[f ' (x 0 , u 0 )]  
.

 6 1 2 0  1
Hence
 2 3
1  4 0 
f x (x 0 , u 0 )  
,
.
 fu (x 0 , u 0 )  


6
1


2 0  1
Since fx(x0,u0) is nonsingular, by the implicit function theorem,
there exists a continuously differentiable map g defined in a
nbd. of u0, such that g(u0)=x0. and f(g(u),u)=0.
55
We can compute g(u0) using the formula
g(u0)=-[fx(x0,u0)]-1fu(x0,u0).
Since
1 1  3
[f x (x 0 , u 0 )] 
20 6 2 
1
1 1  3 1  4 0   14
g' (u 0 )   
 1



20 6 2  2 0  1  2
1
5
6
5
 203 
.
1 
10 
56