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Absolute extreme values are either maximum or
minimum points on a curve.
They are sometimes called global extremes.
They are also sometimes called absolute extrema.
(Extrema is the plural of the Latin extremum.)
4.1 Extreme Values of
Functions
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4.1 Extreme Values of
FunctionsDefinition Absolute Extreme Values
Let f be a function with domain D. Thenf(c) is the
a. absolute minimum value on D if and only
iff(x) f(c) for allx in D.
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Extreme values can be in the interior or the end
points of a function.
0
1
2
3
4
-2 -1 1 2
2
y x
,D AbsoluteMinimum
No Absolute
Maximum
4.1 Extreme Values of
Functions
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0
1
2
3
4
-2 -1 1 2
2
y x 0,2D
Absolute Minimum
Absolute
Maximum
4.1 Extreme Values of Functions
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0
1
2
3
4
-2 -1 1 2
2y x
0,2D
No Minimum
AbsoluteMaximum
4.1 Extreme Values of
Functions
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0
1
2
3
4
-2 -1 1 2
2y x
0,2D No Minimum
NoMaximum
4.1 Extreme Values of
Functions
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Extreme Value Theorem:
Iff is continuous over a closed interval, [a,b] then f has a
maximum and minimum value over that interval.
Maximum &
minimum
at interior points
Maximum &
minimum
at endpoints
Maximum at
interior point,
minimum atendpoint
4.1 Extreme Values of Functions
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Local Extreme Values:
A local maximum is the maximum value within some
open interval.
A local minimum is the minimum value within some
open interval.
4.1 Extreme Values of Functions
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Absolute minimum
(also local minimum)
Local maximum
Local minimum
Absolute maximum
(also local maximum)
Local minimum
Local extremes
are also called
relative extremes.
4.1 Extreme Values of
Functions
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Local maximum
Local minimum
Notice that local extremes in the interior of the function
occur where is zero or is undefined.f f
Absolute maximum
(also local maximum)
4.1 Extreme Values of
Functions
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Local Extreme Values:
If a function f has a local maximum value or a
local minimum value at an interior point c of its
domain, and if exists at c, then
0f c
f
4.1 Extreme Values of
Functions
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Critical Point:
A point in the domain of a function f at which
or does not exist is a critical point off.
0f f
Note:Maximum and minimum points in the interior of a function
always occur at critical points, but critical points are not
always maximum or minimum values.
4.1 Extreme Values of
Functions
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EXAMPLE 3 FINDING ABSOLUTE EXTREMA
Find the absolute maximum and minimum values of
on the interval . 2/3f x x 2,3
2/3f x x
1
323
f x x
3
2
3
f x
x
There are no values of x that will make
the first derivative equal to zero.
The first derivative is undefined at x=0,
so (0,0) is a critical point.
Because the function is defined over a
closed interval, we also must check theendpoints.
4.1 Extreme Values of
Functions
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0 0
f
To determine if this critical point is
actually a maximum or minimum, we
try points on either side, without
passing other critical points.
2/3f x x
1 1f 1 1f
Since 0
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0 0f
2/3f x x 2,3D
At: 2x 232 2 1.5874f
At: 3x
Absolute
minimum:
Absolute
maximum:
0,0
3,2.08
2
33 3 2.08008f
4.1 Extreme Values of
Functions
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4.1 Extreme Values of
Functionsy =x2/3
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Finding Maximums and Minimums Analytically:
1 Find the derivative of the function, and determine
where the derivative is zero or undefined. Theseare the critical points.
2 Find the value of the function at each critical point.
3 Find values or slopes for points between thecritical points to determine if the critical points are
maximums or minimums.
4 For closed intervals, check the end points as
well.
4.1 Extreme Values of
Functions
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4.1 Extreme Values of
Functions
Find the absolute maximum and minimum of the function
]2,1[,2452)(23 onxxxxf
4106)('2 xxxf
410602 xx
Find the critical numbers2530
2 xx
)1)(23(0 xx 13
2 xx or
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4.1 Extreme Values of
Functions
Find the absolute maximum and minimum of the function
]2,1[,2452)( 23 onxxxxf
Check endpoints and critical numbers
The absolute maximum is 2 whenx = -2The absolute minimum is -13 whenx = -1
22
11
27
26
3
2
131
xfx
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4.1 Extreme Values of
Functions
Find the absolute maximum and minimum of the function
]3,0[,1
3)(
2
on
x
xxf 2
2
)1(
)1)(3()2)(1()('
x
xxxxf
3202 xx
Find the critical numbers
)1)(3(0 xx 13 xx or
2
2
)1(
32)('
x
xxxf
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4.1 Extreme Values of
Functions
Find the absolute maximum and minimum of the function
]3,0[,1
3)(
2
on
x
xxf
3321
30
xfx
Check endpoints and critical numbers
The absolute maximum is 3 whenx = 0, 3
The absolute minimum is 2 whenx = 1
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4.1 Extreme Values of Functions
Find the absolute maximum and minimum of the function
2,0,sinsin)( 2 onxxxf
xxxxf cossin2cos)('
Find the critical numbers
xxx cossin2cos0
)sin21(cos0 xx
0cos x 0sin21 x
2
3,
2
x
6
5,
6
x
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4.1 Extreme Values of
Functions
Find the absolute maximum and
minimum of the function
2,0,sinsin)( 2 onxxxf
02
22
3
4
1
6
5
02
4
1
6
00
xfx
The absolute maximum is 1/4 whenx =
/6, 5
/6The absolute minimum is2 whenx =3/2
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Critical points are not always extremes!
-2
-1
0
1
2
-2 -1 1 2
3y x
0f (not an extreme)
4.1 Extreme Values of
Functions
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-2
-1
0
1
2
-2 -1 1 2
1/3y x
is undefined.f
(not an extreme)
4.1 Extreme Values of
Functions
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Iff(x) is a differentiable function over [a,b],
then at some point between a and b:
f b f a
f cb a
Mean Value Theorem for Derivatives
4.2 Mean Value Theorem
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Iff(x) is a differentiable function over [a,b],then at some point between a and b:
f b f af c
b a
Mean Value Theorem for Derivatives
Differentiable implies that the function is also continuous.
4.2 Mean Value Theorem
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Iff(x) is a differentiable function over [a,b],then at some point between a and b:
f b f af c
b a
Mean Value Theorem for Derivatives
Differentiable implies that the function is also continuous.
The Mean Value Theorem only applies over a closed interval.
4.2 Mean Value Theorem
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Iff(x) is a differentiable function over [a,b],then at some point between a and b:
f b f af c
b a
Mean Value Theorem for Derivatives
The Mean Value Theorem says that at some point
in the closed interval, the actual slope equals
the average slope.
4.2 Mean Value Theorem
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y
x0
A
B
a b
Slope of chord:
f b f a
b a
Slope of tangent:
f c
y f x
Tangent parallel
to chord.
c
4.2 Mean Value Theorem
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Iff(x) is a differentiable function over [a,b],and iff(a) =f(b) = 0, then there is at least one
point c between a and b such thatf (c)=0:
Rolles Theorem
4.2 Mean Value Theorem
(a,0) (b,0)
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4.2 Mean Value TheoremShow the function
satisfies the hypothesis of
the Mean Value Theorem
3,0oncos)(
xxf
The function is continuous on [0,/3] and differentiable on
(0,/3). Sincef(0) = 1 andf(/3) = 1/2, the Mean Value
Theorem guarantees a point c in the interval (0,/3) forwhich
f b f af c
b a
csin
03/
12/1
c = .498
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4.2 Mean Value Theorem(0,1)
(/3,1/2)
atx = .498, the slope
of the tangent line is
equal to the slope of
the chord.
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4.2 Mean Value Theorem
Definitions Increasing Functions, Decreasing FunctionsLet f be a function defined on an interval I and letx1 andx2be any two points in I.
1. f increases on I ifx1 f(x2).
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A function is increasing over an interval if thederivative is always positive.
A function is decreasing over an interval if the
derivative is always negative.
A couple of somewhat obvious definitions:
4.2 Mean Value TheoremCorollary Increasing Functions, Decreasing Functions
Let f be continuous on [a,b] and differentiable on (a,b).1. If f > 0 at each point of (a,b), then f increases on [a,b].
2. If f < 0 at each point of (a,b), then f decreases on [a,b].
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4.2 Mean Value TheoremFind where the function
is increasing and decreasing and find the local
extrema.
xxxxf 249)( 23
xxxxf 249)(23
24183)('2 xxxf
)86(302
xx
)86(02 xx
)2)(4(0 xx
2 4
0 0f(x)
+-+
),4()2,( inc)4,2(dec
x = 2, local maximum
x = 4, local minimum
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y
x0
y f x
y g x
These two functions have the
same slope at any value ofx.
Functions with the same
derivative differ by a constant.
C4.2 Mean Value Theorem
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Find the function whose derivative is and whose
graph passes through f x sin x
0,2
cos sind
x xdx
cos sind
x xdx
so:
cosf x x C
2 cos 0 C
4.2 Mean Value Theorem
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Find the functionf(x) whose derivative is sin(x) and
whose graph passes through (0,2).
cos sind
x xdx
cos sind
x xdx so:
cosf x x C
2 cos 0 C
2 1 C 3 C
cos 3f x x Notice that we had to have
initial values to determine
the value ofC.
4.2 Mean Value Theorem
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The process of finding the original function from the
derivative is so important that it has a name:
Antiderivative
A function is an antiderivative of a function
if for all x in the domain off. The process
of finding an antiderivative is antidifferentiation.
F x f x
F x f x
You will hear much more about antiderivatives in the future.
This section is just an introduction.
4.2 Mean Value Theorem
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Since acceleration is thederivative of velocity,
velocity must be the
antiderivative of
acceleration.
Example 7b: Find the velocity and position equations
for a downward acceleration of 9.8 m/sec2 and an
initial velocity of 1 m/sec downward.
9.8a t
9.8 1v t t
1 9.8 0 C 1 C
9.8v t t C (We let down be positive.)
4.2 Mean Value Theorem
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Since velocity is the derivative of position,
position must be the antiderivative of velocity.
9.8a t
9.8 1v t t
1 9.8 0 C
1 C
9.8v t t C
29.8
2s t t t C
The power rule in reverse:
Increase the exponent by one and
multiply by the reciprocal of the
new exponent.
4.2 Mean Value Theorem
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9.8a t
9.8 1v t t
1 9.8 0 C
1 C
9.8v t t C
29.8
2
s t t t C
24.9s t t t C The initial position is zero at time zero.
20 4.9 0 0 C 0 C
2
4.9s t t t
4.2 Mean Value Theorem
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In the past, one of the important uses of derivatives was
as an aid in curve sketching. We usually use a calculator
of computer to draw complicated graphs, it is still
important to understand the relationships between
derivatives and graphs.
4.3 Connectingf andf with the
Graph off
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First Derivative Test for Local Extrema at a critical point c
4.3 Connectingf andf with the
Graph off
1. If f changes sign from positive to
negative at c, then f has a localmaximum at c.
local max
f>0 f0 f>0
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First derivative:
y is positive Curve is rising.
y is negative Curve is falling.
y is zero Possible local maximum orminimum.
4.3 Connectingf andf with
the Graph off
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4.3 Connectingf andf with the
Graph offDefinition Concavity
The graph of a differentiablefunctiony =f(x) is
a. concave up on an open interval
I ify is increasing on I. (y>0)b. concave down on an open interval
I ify is decreasing on I. (y
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4.3 Connectingf andf with the
Graph off
Second Derivative Test for Local Extrema at a critical point c
1. If f(c) = 0 andf(c) < 0, then f has a local maximum atx = c.2. If f(c) = 0 andf(c) > 0, then f has a local minimum atx = c.
+ +
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Second derivative:
y is positive Curve is concave up.
y is negative Curve is concave down.
y is zero Possible inflection point(where concavity changes).
4.3 Connectingf andf with the
Graph off
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4.3 Connectingf andf with the
Graph off
Definition Point of Inflection
A point where the graph of a function has a tangent line and
where the concavity changes is called a point of inflection.
inflection point
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23 2
3 4 1 2y x x x x
23 6y x x
0y Set2
0 3 6x x 2
0 2x x
0 2x x
0, 2x
First derivative test:
y
0 2
0 0
2
1 3 1 6 1 3y
negative
2
1 3 1 6 1 9y positive
23 3 3 6 3 9
y positive
Possible extreme at .0, 2x
4.3 Connectingf andf with the
Graph offSketch the graph
zeros atx = -1,x = 2
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23 6y x x
0y Set
20 3 6x x
20 2x x
0 2x x
0, 2x
First derivative test:
y
0 2
0 0
maximum at 0x
minimum at 2x
Possible extreme at .0, 2x
4.3 Connectingf andf with the
Graph off
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23 6y x x
0y Set
20 3 6x x
20 2x x
0 2x x
0, 2x
Possible extreme at .0, 2x
Or you could use the second derivative test:
maximum at 0x minimum at 2x
6 6y x
0 6 0 6 6y negativeconcave down
local maximum
2 6 2 6 6y positiveconcave up
local minimum
4.3 Connectingf andf with the
Graph off
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6 6y x
We then look for inflection points by setting the second
derivative equal to zero.
0 6 6x
6 6x
1 x
Possible inflection point at .1x
y
1
0
0 6 0 6 6y negative
2 6 2 6 6y positive
inflection point at 1x
4.3 Connectingf andf with the
Graph off
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43210-1-2
5
4
3
2
1
0
-1
43210-1-2
5
4
3
2
1
0
-1
Make a summary table: x y y y
1 0 9 12 rising, concave down
0 4 0 6 local max
1 2 3 0 falling, inflection point
2 0 0 6 local min
3 4 9 12 rising, concave up
4.3 Connecting f and f with
the Graph of f
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A Classic Problem
You have 40 feet of fence to enclose a rectangular garden along
the side of a barn. What is the maximum area that you can
enclose?
4.4 Modeling and Optimization
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x x
40 2x
40 2A x x
240 2A x x
40 4A x
0 40 4x
4 40x
10x
10 40 2 10A
10 20A
2200 ftA
40 2l x
w x 10 ftw
20 ftl
4.4 Modeling and Optimization
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To find the maximum (or minimum) value of a function:
4.4 Modeling and Optimization
1. Understand the Problem.2. Develop a Mathematical Model.
3. Graph the Function.
4. Identify Critical Points and Endpoints.
5. Solve the Mathematical Model.
6. Interpret the Solution.
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What dimensions for a one liter cylindrical can will use
the least amount of material?
We can minimize the material by minimizing the area.
22 2A r rh
area of
ends
lateral
area
We need another
equation that relates
rand h:
2V r h
31 L 1000 cm
21000 r h
2
1000h
r
2
2
10 02
02A r r
r
2 20002A rr
2
20004A r
r
4.4 Modeling and Optimization
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22 2A r rh
area of
ends
lateral
area
2V r h
31 L 1000 cm2
1000 r h
2
1000h
r
22
10 02
02A r r
r
2 20002A r
r
2
20004A r
r
2
2000
0 4 r r
2
20004 r
r
32000 4 r
3500r
3 500r
5.42 cmr
2
1000
5.42h
10.83 cmh
4.4 Modeling and Optimization
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4.4 Modeling and Optimization
Find the radius and height of
the right-circular cylinder of
largest volume that can beinscribed in a right-circular
cone with radius 6 in. and
height 10 in. h
r
10 in
6 in
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4.4 Modeling and Optimization
h
r
10 in
6 in
The formula for the volume of
the cylinder is hrV 2
To eliminate one variable, we
need to find a relationship
between rand h.
61010 r
h
rh3
510
6
h
10-h
r10
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4.4 Modeling and Optimization
h
r
10 in
6 in
hrV2
322
3
5103
510 rrrrV
2520 rr
dr
dV
)4(50 rr
4,0 rr
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4.4 Modeling and Optimization
h
r
10 in
6 in
Check critical points and endpoints.
r= 0, V= 0
r= 4 V= 160/3
r= 6 V= 0
The cylinder will have amaximum volume when
r= 4 in. and h = 10/3 in.
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Determine the point on the
curvey =x2 that is closest to
the point (18, 0).
4.4 Modeling and Optimization
22)18( yxd
42)18( xxd
Substitute forx
42)32436( xxxd
)3624()32436(2
1 321
24
xxxxx
dx
ds
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Determine the point on the
curvey =x2 that is closest to
the point (18, 0).
4.4 Modeling and Optimization
)3624()32436(
2
1 321
24
xxxxx
dx
ds
0dx
dsset 36240
3 xx 1820 3 xx
2x 4y
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Determine the point on the
curvey =x2 that is closest to
the point (18, 0).
4.4 Modeling and Optimization
18203 xx
2x 4y
)942)(2(02 xxx
2
- 0 +
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If the end points could be the maximum or
minimum, you have to check.
Notes:
If the function that you want to optimize has morethan one variable, use substitution to rewrite the
function.
If you are not sure that the extreme youve found is a
maximum or a minimum, you have to check.
4.4 Modeling and Optimization
4 5 Li i i d
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For any functionf(x), the tangent is aclose approximation of the function for
some small distance from the tangent
point.
y
x
0 x a
f x f aWe call the equation of the
tangent the linearization of
the function.
4.5 Linearization and
Newtons Method
4 5 Li i ti d
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The linearization is the equation of the tangent line, and you
can use the old formulas if you like.
Start with the point/slope equation:
1 1y y m x x 1x a 1y f a m f a
y f a f a x a y f a f a x a
L x f a f a x a linearization offat a
f x L x is the standard linear approximation offat a.
4.5 Linearization and
Newtons Method
4 5 Li i ti d
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Find the linearization off(x) =x4 + 2x atx = 2
L x f a f a x a
4.5 Linearization and
Newtons Method
f (x) = 4x3 + 2
L (x) =f(3) +f(3)(x - 3)
L (x) = 87 + 110(x - 3)
L (x) = 110x - 243
4 5 Li i ti d
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Important linearizations forx near zero:
1k
x 1 kx
sinx
cosx
tanx
x
1
x
1
21
1 1 1
2
x x x
13 4 4 3
4 4
1 5 1 5
1 51 5 1
3 3
x x
x x
f x L x
This formula also leads to
non-linear approximations:
4.5 Linearization and
Newtons Method
4 5 Li i ti d
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4.5 Linearization and
Newtons Method
Estimate using local linearization.37
2
1
2
1)('
)(
xxf
xxf L x f a f a x a
)3637)(36(')36()37( ffL
)1(1216)37( L
0833.6)37( L
4 5 Li i ti d
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4.5 Linearization and
Newtons Method
Estimate sin 31 using local linearization.
xxf
xxf
cos)('
sin)(
L x f a f a x a
180)30(')30()31(
ffL
1802
321)31( L
360
3180)31(
L
Need to
be in radians
4 5 Li i ti d
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Differentials:
When we first started to talk about derivatives, we said
that becomes when the change in x and
change in y become very small.
y
x
dy
dx
dy can be considered a very small change iny.
dx can be considered a very small change inx.
4.5 Linearization and
Newtons Method
4 5 Li i ti d
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Let y =f(x) be a differentiable function.The differential dx is an independent
variable.
The differential dy is: dy =f(x)dx
4.5 Linearization and
Newtons Method
4 5 Li i ti d
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Example: Consider a circle of radius 10. If the radius increases by
0.1, approximately how much will the area change?
2
A r2dA r dr
2dA dr rdx dx
very small change in A
very small change in r
2 10 0.1dA
2dA
(approximate change in area)
4.5 Linearization and
Newtons Method
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Compare to actual change:
New area:
Old area:
2
10.1 102.01
2
10 100.00
4.5 Linearization and
Newtons Method
01.2A
2dAAbsoluteerror
%2
100
2
A
dA
%01.2100
01.2
A
Apercenterror
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4.5 Linearization and
Newtons Method
True Estimated
Absolute Change
Relative Change
Percent Change
)()( afdxaff dxafdf )('
)(af
f
)(af
df
%100)(
xaf
df%100)(
xaf
f
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4.5 Linearization and
Newtons Method
Newtons Method
0 x
y
y =f(x)
Root
sought
x1
First
(x1,f(x1))
x2
Second
x3
Third
(x2,f(x2))
(x3,f(x3))
)( 1212 xxmyy ))((')(0 121 xxxfxf
))((')(0 121 xxxfxf
)(')(')( 1121 xfxxfxxf
)('
)(
1
112
xf
xfxx
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This isNewtons Method of finding roots. It is an
example of an algorithm (a specific set of
computational steps.)
Newtons Method:
1n
n n
n
f xx x
f x
This is a recursive algorithm because a set of steps are
repeated with the previous answer put in the next
repetition. Each repetition is called an iteration.
4.5 Linearization and
Newtons Method
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Newtons Method
21
32f x x
Finding a root for:
-3
-2
-1
0
1
2
3
4
5
-4 -3 -2 -1 1 2 3 4
We will use
Newtons Method to
find the rootbetween 2 and 3.
4.5 Linearization and
Newtons Method
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Newtons Method 2
13
2f x x
-3
-2
-1
0
1
2
3
4
5
-4 -3 -2 -1 1 2 3 4
4.5 Linearization and
Newtons Method
xxf )('
Guessx1 = 2
)('
)(
1
112
xf
xfxx
5.22
122
x
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Newtons Method 2
13
2f x x
-3
-2
-1
0
1
2
3
4
5
-4 -3 -2 -1 1 2 3 4
4.5 Linearization and
Newtons Method
xxf )('
Guessx2 = 2.5
)('
)(
2
223
xf
xfxx
45.25.2
125.5.23 x
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Find where crosses .3
y x x 1y
3
1 x x 3
0 1x x
3 1f x x x
23 1f x x
4.5 Linearization and
Newtons Method
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nx nf xn nf x 1
n
n n
n
f xx x
f x
0 1 1 2 11 1.52
1 1.5 .875 5.75.875
1.5 1.34782615.75
2 1.3478261 .1006822 4.4499055 1.3252004
3
1.3252004 1.3252004 1.0020584 1
4.5 Linearization and
Newtons Method
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There are some limitations to Newtons Method:
Wrong root found
Looking for this root.
Bad guess.
Failure to converge
4.5 Linearization and
Newtons Method
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First, a review problem:
Consider a sphere of radius 10 cm.
If the radius changes 0.1 cm (a very small amount)
how much does the volume change?
343
V r24dV r dr
2
4 10cm 0.1cmdV 3
40 cmdV
The volume would change by approximately 40 cm3
4.6 Related Rates
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Now, suppose that the radius is
changing at an instantaneous rate
of 0.1 cm/sec.
34
3V r 24
dV dr r
dt dt
2 cm
4 10cm 0.1
sec
dV
dt
3cm
40
sec
dV
dt
The sphere is growing at a rate of 40 cm3/sec .
Note: This is an exact answer, not an approximation like
we got with the differential problems.
4.6 Related Rates
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Water is draining from a cylindrical tank
at 3 liters/second. How fast is the surface
dropping?
L3
sec
dV
dt
3cm3000
sec
Finddh
dt
2V r h
2dV dhr
dt dt
(ris a constant.)
32cm
3000
sec
dhr
dt
3
2
cm3000
secdh
dt r
(We need a formula to
relate Vand h. )
4.6 Related Rates
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Steps for Related Rates Problems:
1. Draw a picture (sketch).
2. Write down known information.
3. Write down what you are looking for.
4. Write an equation to relate the variables.
5. Differentiate both sides with respect to t.
6. Evaluate.
4.6 Related Rates
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Hot Air Balloon Problem:
Given:
4
rad0.14
min
d
dt
How fast is the balloon rising?
Finddh
dttan
500
h
2 1sec
500
d dh
dt dt
2
1sec 0.14
4 500
dh
dt
h
500ft
4.6 Related Rates
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Hot Air Balloon Problem:
Given:
4
rad0.14
min
d
dt
How fast is the balloon rising?
Find
dh
dt tan 500
h
2 1sec
500
d dh
dt dt
21
sec 0.144 500
dh
dt
h
500ft
2
2 0.14 500dh
dt
1
1
2
4
sec 24
ft140
min
dh
dt
4.6 Related Rates
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4x
3y
B
A
5z
TruckProblem:
Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the
trucks changing 6 minutes later?
r t d
140 410
130 310
2 2 23 4 z
2
9 16 z
2
25 z 5 z
4.6 Related Rates
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4x
3y
30dy
dt
40dx
dt
B
A
5z
Truck Problem:
How fast is the distance between the
trucks changing 6 minutes later?
r t d
140 410
130 310
2 2 23 4 z
29 16 z
2 2 2x y z
2 2 2dx dy dzx y zdt dt dt
4 40 3 30 5dz
dt
Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
4.6 Related Rates
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250 5dz
dt 50
dz
d
miles50
hour
4.6 Related Rates
TruckProblem:
How fast is the distance between the
trucks changing 6 minutes later?
Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
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