Curve Sketching(3)

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Curve Sketching



Relative Extrema



Definition. A function f has a relative

maximum value at a number c if there

exists an open interval I containing c such that ( ) ( ) f c f x≥ ∀ x in I .

a bc

f (c) f (c)

a bc



Definition. A function f has a relative

minimum value at a number c if there

exists an open interval I containing c such that ( ) ( ) f c f x≤ ∀ x in I .

a b

f (c) f (c

)a bc c

If a function has either a relative maximum or relative minimum

value at c, then the function has a relative extremum at c.



Illustration: Consider a function hose graph

is shon belo, at numbers does the function

have relative extrema!

A

C

B"he function has relative minima at

x # $ and x # % and a relative

maximum at x # &.

A, C ' relative minimum pts.

B ' relative maximum pt.



Definition. If c is a number in the domain of

the function f , and if either f ()c* # + or f ()c*

does not exist, then c is called a critical

number of f .

Illustration $ Determine the critical numbers ofthe function defined b- ( ) 2 8 2. f x x x= − +

Solution

( )' 2 8 f x x= −

'( ) 0 f x =

ote that f () x* is + hen x # & and there is no value of x that

ill make f () x* undefined.

"hus, the onl- critical number of f is &.

4 x

⇒ =2 8 0 x

⇒ − =



Illustration / Determine the critical numbers of

the function defined b- ( )

6 15 5

12 . f x x x= −

Solution

( ) 1 4

5 56 12'5 5

f x x x−

= − ( ) ( )45

45

6 26 25 5

x x x x

− −= − =

ote that f () x* is + hen x # / and it is undefined hen x # +.

"hus, the critical numbers of f are + and /.



0onotonicit-



( ) ( )1 2 f x f x<

1 2 1 2, . x x I x x∀ ∈ <such that

Definition. A function f is increasing on an

interval I if and onl- if

x1 x2

f ( x2)

f ( x1)

x1 x2

f ( x2)

f ( x1)



( ) ( )1 2 f x f x>

1 2 1 2, . x x I x x∀ ∈ <such that

Definition. A function f is decreasing on an

interval I if and onl- if

x1 x2

f ( x1)

f ( x2)

x1 x2

f ( x1)

f ( x2)

Definition. A function f is said to be monotonic

on an interval I if it is either increasing or

decreasing on I .



Illustration: Determine hich of the folloing graphs

represent an increasing or a decreasing function.



[ ]b,a( )b,a

( )' 0 f x >

( )b,a[ ], .a b

( )' 0 f x < ( )b,a[ ], .a b

Theorem 1 Suppose a function f is

continuous on the closed interval anddifferentiable on the open interval .

a. If for all x in then f

isincreasing on

b. If for all x in then f isdecreasing on



Theorem 2 The First-Derivative Test for

Relative Extrema

1et f be a function hich is continuous

at each number in some open interval

containing the number c and suppose thatexists at each number in except possibl-

at c.

( )b,a'f ( )b,a



a. If for all values of x in some openinterval having c as its right endpoint,

and if for all values of x in someopen interval having c as its leftendpoint, then f has a relative

maximum value at c.

( )' 0 f x >

( )' 0 f x <

a bc



b. If for all values of x in some openinterval having c as its right endpoint,

and if for all values of x in someopen interval having c as its leftendpoint, then f has a relative minimum

value at c.

( )' 0 f x <

( )' 0 f x >

a bc



2rocedure in determining the relative

extrema of a function f $. Compute .

/. Determine the critical numbers of f )i.e.,

the value)s* of x for hich or

does not exist*.

3. Appl- the 4irst5Derivative "est.

( ) x 'f

( )' 0 f x = ( ) x 'f



Example 1. 1et .

Determine the intervals on hich f is

increasing and the intervals here f isdecreasing. Also, determine the relative

extrema of f , if an-.

( ) 162 3

+−= x x x f

( ) 66' 2

−= x x f

( )' 0 f x = 26 6 0 1 x x⇒ − = ⇒ = ±

Solution

:Step 1. Find f ’( x).

Step2. Determine the critical numbers of f .

critical numbers: 1 and 1



( )1!−∞− ( )1!1− ( )+∞!1

ep 3. "onsider the inter#als determined b$ 1 and

%hen determine the si&n of f () x* in each

interval and appl- "heorem &.3.$ and the

4irst5Derivative "est.

( ) x f ( ) x f '

( )1!−∞−

1−= x

( )1!1−

1= x

( )+∞!1

Conclusion

6 f is increasing

7 + f has a relativemaximum value

5 f is decreasing

53 + f has a relativeminimum value

6 f is increasing



Interval here f is increasing ( ) ( ), 1 1,−∞ − ∪ +∞

( )1, 3−

( )1!1−Interval here f is decreasing

Relative minimum point :

( )1,5−

Relative maximum point :

( ) x f ( ) x f '

( )1!−∞−

1−= x

( )1!1−

1= x

( )+∞!1

Conclusion

6 f is increasing

7 + f has a relativemaximum value

5 f is decreasing

53 + f has a relativeminimum value

6 f is increasing



Concavit-



Definition. "he graph of a function f is said to be

concave upward at a point( )( )cf ,c if

( )c'f a. exists and

b. there is an open interval I containing

c such that for all values of x in I such

that , the point is above the tangent line to the graph at .

x ≠ ( )( ) x f , x ( )( )cf ,c

c c



Definition. "he graph of a function f is said to be

concave downward at a point( )( )cf ,c if

( )c'f a. exists and

b. there is an open interval I containing

c such that for all values of x in I such

that , the point is belo the tangent line to the graph at .

x ≠ ( )( ) x f , x ( )( )cf ,c

c c



Illustration: Determine hich of the folloing graphs

are concave upard or concave donard.



Theorem 1et f be a function hich is

differentiable on some open intervalcontaining c.

( )'' 0 f c >

( )( )cf ,c

( )'' 0 f c <

( )( )cf ,c

a. If , the graph of f is concave

upard at .

b. If , the graph of f is concavedonard at .



2oint of Inflection



Definition. A point is a point of inflection of

the graph of the function f

ifa. the graph has a tangent line at and

b. there exists an open interval I containing

c such that if x is in I , then either

i. henever , and

henever 8 or

ii. henever

, and

henever .

( )( )cf ,c

( )'' 0 f x < c x <

( )'' 0 f x > x >

c x <

x >

)c, f )c**

)c, f )c**( )'' 0 f x >

( )'' 0 f x <



Illustration: In 4igures &.3.9)a* and &.3.9)b*, e see portions of

the graphs of to different functions, each having a point of

inflection at .( )( )cf ,c



Theorem ! The Secon"-Derivative Testfor Relative Extrema1et c be a critical number of a function f athich .Suppose further that f '’

exists for each value of x in some openinterval containing c.

a. If ,then f has a relative

maximum value at c. b. If , then f has a relativeminimum value at c.

( )' 0 f c =

( )'' 0 f c <

( )'' 0 f c >



Example 2. 1et .

Determine the intervals here the graph

of f is concave upard and concave

donard. Also, determine the inflection

points of the graph of f , if an-.

( ) 162 3

+−= x x x f

( )'' 12 0 f x x= = 0 x⇒ =

Solution

:Step 1. Find f ’’( x) and e:uate to ;ero to find the

possible <point)s* of inflection=.

( ) 66'

2−=

x x f



( ), 0−∞ ( )0, +∞

tep 2. "onsider the inter#als determined b$ .

%hen determine the si&n of f (() x* in each

interval and appl- "heorem &.3.3 and the

Second5Derivative "est.

( ) x f ( ) x f '

( ), 0−∞

0 x =

( )0, +∞

Conclusion

5 graph of f is concavedonard

$ 59 + graph of f has point ofinflection

6 graph of f is concaveupard

( )'' 12 0 f x x= =( ) 66'

2

−= x x f

( )'' f x

( ) 162 3

+−= x x x f



( ) x f ( ) x f '

( ), 0−∞

0 x =

( )0, +∞

Conclusion

5 graph of f is concavedonard

$ 59 + graph of f has point ofinflection

6 graph of f is concaveupard

( )'' f x

Interval here graph of

f is concave upard ( )0, +∞

( )0,1

( ), 0−∞

Inflection point :

Interval here graph of f is concave donard

( ) 162 3 +−= x x x f



In order to make the graphing of a given

function f easier, e summari;e the things e

have done as follos.

Step 2. "onsider all the critical numbers of f

and thesolution to the euation f ’’( x) * .

+rran&ethese numbers in their natural order.Step 3. ,btain the inter#als determined b$

these numbers.

Step 1. Sol#e for f ’) x* and f

’’( x).

Step -. +ppl$ theorems 2 and 3 and the rstand

second deri#ati#e tests for relati#e

e/trema.



Example . Sketch the graph of the

function f defined b- ( ) 1623

+−= x x x f

( )'' 12 f x x=

0 x⇒ =

Solution:

Step 1. ( ) 66' 2

−= x x f

Step 2. 12 0 x =2

6 6 0 x − =

1 x⇒ = ±

Arrange these numbers in the real number line.

-1 # 1

( )1!−∞− ( )(!1− ( )1!( ( )+∞!1 Step 3.



Step -.

( ) x f ( ) x f ' ( ) x f 0

( )1!−∞− +

1−= x (

( )(!1−

(= x

( )1!(1= x 3− (

( )+∞!1 +

Conclusion

( ) 162 3

+−= x x x f ( )'' 12 f x x=( ) 66' 2

−= x x f

1 −

−

(

+−

−

+

+

−

−

f has relative maximum value

f is decreasing8 graph is concave donard

f is decreasing8 graph is concave upard

f is increasing8 graph is concave upard

f is increasing8 graph is concave donard

f has relative minimum value

graph has a point of inflection

Step



Step .

f ↓, CD f ↑, CD

f ↑, C> f ↓, C>

+ $5$

( ) x f

( )1!−∞−

1−= x

( )(!1−

(= x

( )1!(

1= x 3−( )+∞!1

Conclusion

1








)5$,7*

)+,$*

)$,53*

( )xf ( )xf ' ( )xf0



( ) x f ( ) x f ' ( ) x f

( )1!−∞− +

1−= x (

( )(!1−

(= x

( )1!(

1= x 3− (

( )+∞!1 +

Conclusion

1 −

−

(

+−

−

+

+

−

−








f ↓

f ↑

CD

C>

(-1,1)

( ) ( ), 1 1,−∞ − ∪ +∞

Rel min pt

( ), 0−∞

( )0, +∞

Rel max pt

(1,-3)

(-1,5)

(0 1)

( ) 32 6 1 f x x x= − +

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