Implicit Differentiation ® Local Extrema ® Def 1: Suppose

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Implicit Differentiation
Local Extrema
Def 1: Suppose that
is differentiable
and the equation
defines y as
an implicit function of x, y=f(x), then
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Eg 1: Find
equation
if
defines the
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Def 2: Suppose that
defines
an implicit function f with two variables,
, then
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Eg 2: Find
and
defines the equation
if
.
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Def 3: A function
defines in a
domain D and (a,b) in D.
(i) A function of two variables has a local
maximum at (a,b) if
for all
points (x,y) in a region R where
. The
number is called a local maximum value
(ii) A function of two variables has a local
minimum at (a,b) if
for all
points (x,y) in a region R where
. The
number is called a local minimum value
Local M axima
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Local M inima
Local extrema point value(critical point
or stationary point) can be determine by
using First Derivative Test.
Def 4: If f has a local maximum or
minimum at (a,b), and the first order
partial derivatives of f exists there, then
and
.
[ First Derivative Test for Local Extrema]
But, it happen to has a wired case.....
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(a,b,f(a,b))
[Planar Graph]
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Although graph above have
and
but the point (a,b,f(a,b)) is
neither a local maximum or minimum. Its
actually a saddle point.
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In order to know which properties is the
critical point; either local maximum, local
minimum or saddle, continuation of First
Derivative Test is reveal by a name of
Second Derivative Test.
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Def 5: For a function
with ,
and
all exist on a region, R and (a,b) is
a critical point in R. Define
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Then
if
and
,
is a
local minimum.
(ii) if
and
,
is a
local minimum.
(iii) if
and,
is a saddle point.
(iv) if
and, no conclusion can be
made.
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(i)
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Eg 3: Find the critical points of
. Consequently,
determine whether the critical points
have a local minimum value, local
maximum value or a saddle point.
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Step 1: Find all the partial derivatives.
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Step 2: Find critical points.
Take
and
.
Set them equal,
Therefore, the critical point at
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Step 3: Compute
Therefore,
produce a saddle
point with the value of
.
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