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1)

Since ( ) exists, then

( )

( )

( )

( ( ) ) ( )

( ) ( )

So we get: in case ( ) exists, it equals to 0.

Inductively continuing this statement, we get that the same holds for

Since ( )( ) exists, we get that

( )( )

Inductively getting back to lower derivatives we get:

( )( )

for

So the statement is proved.

2)

Consider a function

( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

|

( ) is a linear combination of functions ( ) ( ) ( ).

Since are continuous on [ ], ( ) is continuous on [ ].

Since are differentiable on ( ), ( ) is differentiable on ( ).

D O M Y A S S I G N M E N T SUBMIT

W W W . A S S I G N M E N T E X P E R T. C O M

Sample: Real Analysis - Continuity and Differentiability

1

( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

|

( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

|

Last 2 equalities hold because matrices the determinant is taken has 2 identical rows.

By Rolle’s theorem,

( ) ( )

( ) | ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

|

The statement is proved.

D O M Y A S S I G N M E N T SUBMIT

W W W . A S S I G N M E N T E X P E R T. C O M

2

3)

Consider a function

( )

( (

) ( ))

It is defined at *

+, continuous at *

+ (because f is continuous) and twice-

differentiable on (

). By mean value theorem,

(

) ( ) (

) ( )

( ) ( )

( ( ) (

) ( ) ( ))

( ) (

) ( )( )

( )

( (

) ( ))

Consider a function on *

+. It is continuous on this interval and

differentiable on (

) (because is twice differentiable). Thus, by mean value theorem,

(

) ( ) (

) ( )

Substituting this into expression we got earlier we get:

D O M Y A S S I G N M E N T SUBMIT

W W W . A S S I G N M E N T E X P E R T. C O M

3

( ) ( ) (

) ( )( )

( ) ( )

( ) (

) ( )

Since (

) ( ) we proved the needed statement.

1)

is continuous at [ ].

∫ ( )

Suppose

( )

for some [ ].

If we take | ( )|

in the definition of continuity of at point we get:

( ) ( ) | ( ) ( )| | ( )|

| ( )|

( ) ( )

| ( )|

So

| ( )| | ( )|

at in the interval ( ) ( ).

D O M Y A S S I G N M E N T SUBMIT

W W W . A S S I G N M E N T E X P E R T. C O M

4

Then

∫ ( )

∫ ( )

( ) ( ) ∫

| ( )|

( ) ( )

| ( )|

( ( ) ( ))

So we get contradiction with

∫ ( )

So ( ) at [ ].

D O M Y A S S I G N M E N T SUBMIT

W W W . A S S I G N M E N T E X P E R T. C O M

5