extremum and inflection of point
DESCRIPTION
Extremum and inflection of point. Presented by:- :: Bothaina AL-sobai :: :: Meji Le :: :: Hind Nader :: :: Hanan :: ::Aisha alsulati ::. Outline…. Definition (Hanan) Extreme Value Theorem (Hanan) Critical Number (Hanan) The First Derivative Test (Bothaina) - PowerPoint PPT PresentationTRANSCRIPT
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Extremum and inflection of point
Extremum and inflection of point
Presented by:-:: Bothaina AL-sobai ::
:: Meji Le :::: Hind Nader ::
:: Hanan ::::Aisha alsulati ::
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Definition (Hanan)
Extreme Value Theorem (Hanan)
Critical Number (Hanan)
The First Derivative Test
(Bothaina)
The Second Derivative Test (Hind)
Inflection Point (MiJi)
Outline…Outline…
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Definition of ExtremumDefinition of Extremum
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Extreme Value Theorem
Extreme Value Theorem
States that if a function f(x) is
continuous in the closed interval [a,
b] then f(x) must attain its maximum
and minimum value, each at least
once.
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Maximum
Minimum
Extreme Value Theorem
Extreme Value Theorem
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Critical NumberCritical Number
Critical number means:
If a number c is in the domain of a function f,
then c is called a critical number of
provided f ' (c) = 0 or f ‘ (c) doesn't exist.
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Example (1)Example (1)
• Find the critical number for the function f(x) =
• F '(x) =
• Then f '(x) is zero at x=3, and x=-1 and f ' is
undefined at x=1, however x=1 is not in the
domain of f and hence the only critical
numbers of f are x=3 and x=-1
• SO the critical point (3,6) and(-1,-2)
22
2
2
2
)1(
)1)(3(
)1(
32
)1(
)1)(3()2)(1(
x
xx
x
xx
x
xxx
)1(
)3( 2
x
x
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Keep in mind….Keep in mind….
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The First Derivative Test
The First Derivative Test
To find extrema we should follow 3 steps:
1)Find the derivative of f and set the equation to zero.
2) Find the critical points 3)Check the signs (by taking test
points )
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Example (2)Example (2)• Locate all relative extrema for
the function
143632 23 xxx
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Example (2)Example (2)• Step number 1,2
03666 2 xx
0)6(6 2 xx0)2)(3(6 xx
F '(x)=
=
=
The only critical numbers of f are -2 and 3.
Step number 1
Step number 2
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Third step…Third step…
• Using theorem we conclude that the critical number -2 yields a relative maximum (f ' changes sign from + to -) and 3 yields a relative minimum
Interval (-∞ ,-2) (-2,3) (3, ∞)
Point in
interval
-3 0 4
Sign of f
'(x)
+ _ +
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Remember…Remember…
• f '(X) changes sign from + to -).
• f '(x) changes sign from – to +).
MAX(-2)
MIN(3)
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The Second Derivative Test
The Second Derivative Test
Suppose f'(x)= 0• If f"(x) < 0, then f has a relative maximum
at x• If f"(x) > 0, then f has a relative minimum
at x
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1) f(x)= 18x – (2/3)x³
f'(x)= 18 – 2x² = 0 2 (9 – x2) = 0
2 (3 – x) (3 + x) = 0
x = 3, -3 f'' (x)= -4x
when x = 3, y'' = -4(3) = -12 < 0 so there is
a relative maximum when x = -3, y'' = -4(-3) = 12 > 0 so there is
a relative minimum
Example (3)Example (3)
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2) f(x)= 6x4 – 8x³ + 1
f' (x) = 24x³ – 24x² = 0 24x² (x – 1) = 0
x = 0, 1
f" (x) = 72x² – 48x
Example (4)Example (4)
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when x = 1, f'' (x)= 72(1) – 48(1) = 24 > 0
so there is a relative minimum
when x = 0
we can't apply the second derivative test when x
= 0
if x < 0 then f' (x)< 0
if 0 < x < 1, then f' (x)< 0
thus, no maximum or minimum exist when x=0
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Inflection PointInflection Point
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A point on a curve at which the tangent crosses the curve
itself.
A point on a curve at which the tangent crosses the curve
itself.
f
D2f
Tangent at a point of
inflexion.
P
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Inflection PointInflection PointA point on a curve at which the slope
of graph does not changes
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• An important point on a graph is one which marks a transition between a region where the graph is concave up and one where it is concave down. We call such a point an inflection point.
• If (c,f(c)) is a point of a inflection of f, then either f’’(c) =0 or does not exist.
Property of point of inflection
Property of point of inflection
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Example(6)Example(6)
F(x) = 6x4 – 8x3 + 1
• Step 1. f' (x) = 24x³ – 24x²• Step 2. f'' (x) = 72x² – 48x
= 24x (3x – 2)• Step 3. To find where f''(x)= 0
24x (3x - 2) = 024x = 0 OR 3x – 2 = 0X = 0 x = 2/3
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Step 4. Check on ConcavitySign of second derivative
Step 4. Check on ConcavitySign of second derivative
f'' (x) = 24x (3x – 2)f’’ (-1) = 24x (3x – 2)
= 24(-1) (3(-1) – 2)= -24 (-3 -2)= -24 (-5)= 120
f’’ (0.1) = 24(0.1) (3(0.1) – 2)
= 2.4 (0.3 – 2)= 2.4 (-1.7)= -4.08
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Step 5. Check on slopeStep 5. Check on slope
f' (x) = 24x³ – 24x²f‘(-1) = 24(-1)³ – 24(-1)²
= -24 – 24= -48
f‘(0.1) = 24x³ – 24x²= 24(0.1)³ – 24(0.1)²= 0.024 – 0.24= -0.216
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