Extremum and inflection of point

Extremum and inflection of point

Presented by:-:: Bothaina AL-sobai ::

:: Meji Le :::: Hind Nader ::

:: Hanan ::::Aisha alsulati ::

Definition (Hanan)

Extreme Value Theorem (Hanan)

Critical Number (Hanan)

The First Derivative Test

(Bothaina)

The Second Derivative Test (Hind)

Inflection Point (MiJi)

Outline…Outline…

Definition of ExtremumDefinition of Extremum

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.

Maximum

Minimum

Extreme Value Theorem

Extreme Value Theorem

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.

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

Keep in mind….Keep in mind….

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 )

Example (2)Example (2)• Locate all relative extrema for

the function

143632 23 xxx

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

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)

+ _ +

Remember…Remember…

• f '(X) changes sign from + to -).

• f '(x) changes sign from – to +).

MAX(-2)

MIN(3)

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

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)

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)

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

Inflection PointInflection Point

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

Inflection PointInflection PointA point on a curve at which the slope

of graph does not changes

• 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

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

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

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

THANK YOU!

THANK YOU!

For Your AttentionFor Your Attention