continuity stage (2)

7/29/2019 Continuity stage (2)

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Calculus 3rdsecondary Continuitystage (2)-33-

x 1lim f ( x )

x 2 x 2

x 2

At x 2 :

2 1 1 x 1 1f ( 2 ) and lim f ( x ) lim

2 1 3 x 1 3

As lim f ( x ) f ( 2 ) ,Thus f is Continuous at x 2

At x -1 :

2f ( 1 ) , So , The function f ( x ) is undefined at x -1 .

0

Thus f ( x ) is discountinuos at x - 1

2x : x 2let f(x) Then Discuss the Continuity at x 2

4 : x 2

x 2 x 2

x 2

At x 2 : f ( 2 ) 2 2 0 and lim f ( x ) lim x 2 2 2 0

So , As lim f ( x ) f ( 2 ) ,Thus f is Continuous at x 2

f ( 2 ) is not defined f ( x ) is not Continous at x 2

2nd

stage : Continuity of a function at a point

The function Fis said to be continuous at a pointx = a, if three conditions are satisfied :

(1) F(a) is well defined (2)x alim f(x) exists

(3)

x alim f(x) f(a)

Note that :

If any one of these three conditions is not satisfied , then the function is said to be

discontinuous at x a .

----------------------------------------------------------------------------------------- --------------Simple Example

Discuss the continuity of the function : f ( x ) x 3 at x 1

Answerf(1) = 1+ 3 = 4 And

x 1 x 1lim f ( x ) lim x 3 1 3 4

So , f(1) = Then f(x) is Continous at x = 1 .

----------------------------------------------------------------------------------------- --------------Example (1)

x 1Discuss the continuity of the function : f ( x ) at x -1 and x 2

x 1

Answer

----------------------------------------------------------------------------------------- --------------Example (2)

Answer

----------------------------------------------------------------------------------------- --------------Example (3)

Discuss the continuity of the function : f ( x ) x 2 at x 2

Answer


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x 3x

f ( x ) f ( 3 ) x 1 f ( 3 ) 2x 2

x 3

x 3

x 3

x 3

f ( 3 ) 3 1 4

for lim f(x):

left limit : f( 3 ) lim 2x 2 2( 3 ) 2 4

Right limit : f( 3 ) lim x 1 3 1 4

f( 3 ) f( 3 ) f( 3 ) f ( 3 ) lim f(x) 4

f ( x ) is Continous at x 3

4

x 1x

f ( x ) f (1 ) x 3 f (1 ) x 1 2

Example (4)

x 1 : x 3let f(x) Then Discuss the Continuity at x 3

2x 2 : x 3

Answer

----------------------------------------------------------------------------------------- --------------Example (5)

Discuss the continuity of the function f ( x ) x 1 2 at x 1. Answer

x 1 : x 1 x 1 2 : x 1x 1 f ( x )

-( x 1) : x 1 - x 1 2 : x 1

x 1 : x 1f ( x )

- x 3 : x 1

x 1

x 1

x 1

x 1

f (1) 1 1 2

for lim f(x):

left limit : f( 1 ) lim x 3 1 3 2

Right limit : f( 1 ) lim x 1 1 1 2

f( 1 ) f( 1 ) f( 1) f (1) lim f(x) 2



3/28





x2

x

f ( x ) f Sinx Cos x2

x2

x2

x 2

x2

x2

for lim f(x):

left limit : f( ) lim Sin x Cos x 12

Right limit : f( ) lim 2 Cos2x 1

2

f( ) f( ) lim f(x) 12 2

f ( ) lim f(x) f ( x ) is continous at x2 2

1 f 2 Cos2x2

Example (6)

sin x cos x : x2

If f(x) Then Discuss the Continuity at x22 cos 2x : x

2

Answer

f ( ) Sin Cos 12 2 2

----------------------------------------------------------------------------------------- --------------


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x 2x

f ( x )

2x 4f ( 2 )

x 2

2x 4f ( 2 )

x 2

x 2

2

x 2 x 2 x 2

x 2

f ( 2 ) 2 " Given"

for lim f(x):

x 2 x 2x 4 0left limit : f( 2 ) f( 2 ) lim lim lim x 2 4

x 2 0 x 2

f ( 2 ) lim f(x) f ( x ) is discontinous at x 2

Note : we can redefine f(x) to make it continous by puttin

x 2

2

g f ( 2 ) 4 , so that we will

make f ( 2 ) lim f(x) 4

x 4

: x 2So , the Correct f(x) to be Continous is f(x) x 2

4 : x = 2

2

Example (7)2x 4

: x 2let f(x) Then Discuss the Continuity at x 2x 2

2 : x = 2

Answer

----------------------------------------------------------------------------------------- --------------Example (8)

3

2

x 8Redefine the following function f(x) to be Continous at x 2

2x x 6

Answer

Note :To redefine any function to be Continous , you have to find its limit and equalize it to

its value of f(x) at x = 2

2 2 23

2x 2 x 2 x 2

x 2

3

2

x 2 x 2x 4 x 2x 4 2 2 2 4x 8 0 12lim lim lim

2x x 6 0 2x 3 x 2 2x 3 2 2 3 7

12And f(x) is Continous at x 2 f ( 2 ) lim f ( x )

7

x 8x 2

2x x 6 So f(x)= 12

x 27


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Example (9)

9 x 3Redefine the following function f (x) to be Continous at x 0

x

Answer

x 0 x 0 x 0

x 0

x 0

9 x 3 0 9 x 3 9 x 3 9 x 9lim lim lim

x 0 x 9 x 3 x 9 x 3

1 1 1lim

3 3 69 x 3

1And f(x) is Continous at x 0 f (0 ) lim f ( x )

6

9 x 3 x 0xSo f(x)=1

x 06

---------------------------------------------------------------------------------------------------------------------

Example (10)

4

x 3Redefine the following function f (x) to be Continous at x 3

x 81

Answer

2 2 2

2 2 2

2

2

x 3 x 3, x 3 , x 3

x 9 x 9 x 3 x 3 x 9f ( x ) f ( x )

x 3 x 3, x 3 , x 3

x 9 x 9 x 3 x 3 x 9

1, x 3

x 3 x 9

f ( x ) 1, x 3

x 3 x 9

2 2x 3 x 3

x 3

-1 -1 1 1f ( 3 ) lim And f ( 3 ) lim

108 108x 3 x 9 x 3 x 9

f ( 3 ) f ( 3 ) lim f ( x ) doesn't exist

So we cannot define f at x 3 to be continuous.


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Example (11)

Tan 5 xRedefine the following function f (x) to be Continous at x 0

Sin 8x

Answer

x 0 x 0

Tan 5 xx 0

Tan 5 x 5 Sin 8 xlim f ( x ) lim So f(x)=

Sin 8 x 8 5x 0

8

---------------------------------------------------------------------------------------------------------------------

Example (12)2Cos x

Redefine the following function f (x) to be Continous at x1 Sin x 2

Answer

2 2

x x x x x2 2 2 2 2

x2

2

1 sin x 1 sin xcos x 0 1 sin xlim f ( x ) lim So lim lim lim 1 sin x 2

1 sin x 0 1 sin x 1 sin x

f(x) is Continous at x f ( ) lim f ( x ) 22 2

cos xx

1 sin x 2So f(x)=

2 x2

---------------------------------------------------------------------------------------------------------------------


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2

x 2

2

x 2

x 2

x 3

At x 2 : f ( 2 ) 2 2 2 2 2

for lim f(x):

left limit : f( 2 ) lim x 2x 2 2

Right limit : f( 2 ) lim - x 3 1

f( 2 ) f( 2 ) limit does't exist at x 2 f ( x ) is discontinous at x 2

At x 3 : f ( 3 ) 3 3 0

for lim f(x

x 3

x 3

x 3 x 3

):

left limit : f( 3 ) lim - x 3 0

Right limit : f( 3 ) lim x 3 0

f( 3 ) f( 3 ) lim f(x) 3 f ( 3 ) lim f(x)

So f ( x ) is Continous at x 3

x 2x

f ( x ) 2x 2x 2 - x 3

x 3

x 32 0

Example (13)2x 2x 2 : x 2

let f(x) Then Discuss the Continuity at x 2 and x 3x 3 : x > 2

Answer

2

x 3 , x 3x 3

- x 3 , x 3

x 2x 2 , x 2x 3 , x 3

So for x > 2 : x 3 Then f(x) - x 3 , 2 x 3- x 3 , 2 x 3

x 3 , x 3

----------------------------------------------------------------------------------------- --------------


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x 1

x 1

22

x 1 x 1 x 1

x 1

At x 1: f (1) 0 " given"

for lim f(x):

left limit : f( 1 ) lim 0 0

2 x 1 2 x 1 x 12x 2 0Right limit : f( 1 ) lim lim lim

x 1 0 x 1 x 1

lim 2 x 1 4

f( 1 ) f( 1 ) limit does't exist at x 1 f ( x ) is di

2

x 2

2

x 2

x 2

x 2 x 2

scontinous at x 2

And we can't redefine f(x)

2 2 2At x 2 : f ( 2 ) 6

2 1

for lim f(x)

2 x 2left limit : f( 2 ) lim 6

x 1

Right limit : f( 2 ) lim 10 2x 10 2(2 ) 6

f( 2 ) f( 2 ) lim f(x) 6 f ( 2 ) lim f(x)

So


x 0x

f ( x ) Zero

22x 2

x 1

10 2x0

x 1

0

x 2

6

x 3

0

Example (14)

2

0 : 0 x 1

2x 2let f(x) : 1 x 2x 1

10 2x : 2 x 3

Then Discuss the Continuity at x 1 and x 2 , And if f(x) is discontinous , redefine f(x) if

possible

Answer

----------------------------------------------------------------------------------------- --------------


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Calculus 3rdsecondary Continuitystage (2)--

Example (15)

a x + 3 : x 1find the value of "a" to make f ( x ) Continous at x 1

3 a x : x < 1

Answer

x 1

x 1

f (1) a 3 , f ( 1 ) lim ( 3 a x ) 3 a

f (1 ) lim ( ax 3 ) a 3

f ( x ) is Continous at x 1 f ( 1 ) f (1 )

3 a a 3 a 0

----------------------------------------------------------------------------- --------------------------

Example (16)

find the value of a in the following function to make f(x) Continous

2x a 1 x a, x 1

f ( x ) at x 1x 1

2 , x = 1

Answer

x 1

2

x 1 x 1 x 1

x 1

f (1) 2 " Given"

for lim f(x):

x a 1 x a x a x 10f( 1 ) f( 1 ) lim lim lim x a 1 a

x 1 0 x 1

f ( x ) is Continous at x 1 f (1) lim f(x) 1 a 2 a 1

----------------------------------------------------------------------------------------- --------------


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x 0

x 0

x 0 x 0

x 0

f (0 ) a " Given"

for lim f(x):

5x Tan3x 0f( 0 ) f( 0 ) lim Divide both numerator and denominator by x

Sin4x 05x Tan3x Tan3x

55 3x x xlim lim 2

Sin4x Sin4x 4

x x

f ( x ) is Continous at x 0 f (0 ) lim f(x) a 2

Example (17)

5x Tan 3x: x 0

Sin4xIf f(x) is Continous at x 0 , Then find the value of "a"

a : x 0

Answer

----------------------------------------------------------------------------------------- --------------Example (18)

2

Find a and b, such that

x ax 2 : x 2

f ( x ) 4 : x 2 becomes continuous at x 2

5a bx : x 2

Answer

x 2

2

x 2

x 2

x 2

f ( 2 ) 4 " Given"

for lim f(x):

f( 2 ) lim x a x 2 4 2a 2 2a 2

f( 2 ) lim 5a bx 5a 2b

f ( x ) is Continous at x 2 f ( 2 ) lim f(x)

f ( 2 ) f ( 2 ) f ( 2 ) f ( 2 ) f ( 2 ) 4

2a 2 4 2a 2 a 1 Then substitu

te in the second equation

1And 5a 2b 4 5 2b 4 b -

2


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x 0x

f ( x ) Sin2x

f 22x

x

f 2Tanx

x 0

x 0

Sin 2xRemember that : Sin 2x 2 Sin x Cos x Sin x Cos x

2

Sin 2x: x 0

2x

So f(x) a : x 0

x: x 0

Tan x

f (0 ) a " Given"

for lim f(x):

Sin 2x 0left limit : f( 0 ) lim

2x

x 0

x 0 x 0

x 0

x 0

Sin 2x 2lim 1

0 2x 2

x 0 xRight limit : f( 0 ) lim lim 1

Tan x 0 Tan x

f( 0 ) f( 0 ) lim f(x) 1

f(x) is Continous at x 0 f (0 ) lim f(x) a 1

a

Example (19)

If the function f ( x ) is continuous at x 0 , Then find a , where :

1

sin xcos x : x < 0x

f(x) a : x = 0

xcos x: x > 0

sin x

Answer


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2

x a

x a

22 3

x a

3

x a

3

a x x a a x 4 x aa x f ( x )

- a x x a 2ax 3b x a

f ( a ) a a 4 4

for lim f(x):

f( a ) lim a x 4 4

f( a ) lim 2ax 3b 2a a 3b 2a 3b

f ( x ) is Continous at x 0 f ( a ) lim f(x) 2a 3b 4 (1)

And f ( a 4 ) b

3 3

3 3 3

a a 4 4 b b 8 b 2

Substitute in (1) to get a , 2a 3( 2 ) 4 2a 2 a 1 a 1

Example (20)

3

2

a x 4 : x aIf f(x) is Continous at x a , and f ( a 4 ) b

2ax 3b : x a

Then find the value of a and b

Answer

----------------------------------------------------------------------------------------- --------------


13/28





x 1

x 1

x 1

x 1

x 2

At x 1: f (1) a b

for lim f(x):

left limit : f( 1 ) lim x 2 1

Right limit : f( 1 ) lim ax b a b

f ( x ) is Continous at x 1 f (1) lim f(x) a b -1 (1)

At x 2 : f ( 2 ) 2a b

for lim f(x):

left limit : f( 2 ) l

x 2

x 2

x 2

im ax b 2a b


f ( x ) is Continous at x 2 f ( 2 ) lim f(x) 2a b 2 ( 2 )

from (1) and (2) by using subtraction : a 3 and by substituting b -4

x 1x

f ( x ) x 2 ax b

x 2

x 2

Example (21)

let f ( x ) be a continuous function at x 1 and at x 2 , then find a and b ,where :

x 2 : x 1

f ( x ) a x b : 1 x 2

x 2 : x 2

Answer


14/28





Continui ty of a function Over an interval

Any function is said to be Continuous over an intervalif the function is Continuous at every

point in the interval .

For example: f(x) = 7 is a constant function , its domain is R and the limit for eachpoint in Ris 7 , So the function is Continuous on R .

SO , Any Constant function is Continuous over R

Another example: f(x) = 4 23x 8x 7x 3 is a polynomial function , its domain is R and the

limit for any point in R equals the value of the function at the same point .let me Convince you

4 2

4 2 4 2

x a x a

x a

let a R where a is any point you have

So f ( a ) 3a 8a 7a 3

for lim f(x): lim 3x 8x 7x 3 3a 8a 7a 3

So as you see , at x a f ( a ) lim f ( x ) and that's true for each a R

f(x) is Continuous on R

SO , Any Polynomial function is Continuous over R

Another example: f(x) =1

x 1is Continuous on R {1}

But what if we ask you about the Continuity of f(x) =1

x 1on the interval ] 0 , 2 [

Then the answer would be f(x) is discontinuous over ]0 ,2 [ because f(x) is not defined at

x=1 which belongs to ]0 , 2 [ .

Remember that: each point belongs to the interval must be defined in order to make f(x) a

Continuous function over this interval .

SO , Any rational function is Continuous on R

except the set of zeroes of its denominator


15/28





The above can be summerized in the following :

2 n

0 1 2 n

2 n

0 1 2 n

2 m

0 1 2 m

( 1) The polynomial function f ( x ) a a x a x .... a x , is continous on R

or any subset of R .

p( x ) a a x a x .... a x( 2 ) The rational function f ( x )

Q( x ) b b x b x .... b x

is continuous on

R or any subset of R except at the zero s of Q( x )

( 3 ) The Trigonometric function.

a ) f ( x ) Sin x is continunous on R or any subset or R

b ) f ( x ) Cos x is continunous on R or any subset or R

c ) f ( x

) Tan x is continunous on R or any subset or R . exceptfor

2n 1 5 3The point s x : n Z " , , , ,........"

2 2 2 2 2

----------------------------------------------------------------------------------------- --------------

Simple examplesDiscuss the Continuity of f(x) on R of each of the following :

(1) f(x)=5

Answer

f(x) is Constant function , then it is Continuous on R

----------------------------------------------------------------------------------------- --------------

(2) f(x) = 23x 4x 3 Answer

f(x) is a polynomial function , then it is Continuous on R

----------------------------------------------------------------------------------------- --------------

(3) f(x) =3

x 2

Answer

The domain of f(x) is R {2} , Then f(x) is Continuous on R {2}

----------------------------------------------------------------------------------------- --------------

(4) f(x) =2

2

2x 3

x 9

Answer

The domain of f(x) is : 2 2x 9 0 x 9 x 3

Then f(x) is Continuous on R { 3 }

----------------------------------------------------------------------------------------- --------------


16/28





2

2 2

The domain of x 1 :

x 1 0 x 1

Continuous on R

The domain of 2x 3 :

3

2x 3 0 2x 3 x 2

3Continuous on R

2

3R

2

2

2

cos x sin x is continuous on R and x 3 is continuous on R .

f ( x ) is continuous on R except at the zeros of denominator

f ( x ) is continuous on R {-3}

3

x 1f ( x )

cos x

3

nf ( x ) is continuous on R x : x , n Z

2

(5) f(x) =2

x 3

x 4x 3

Answer

The domain of f(x) is 2x 4x 3 0 x 3 x 1 0 x 3 and x 1 Then f(x) is Continuous on R {3 , 1}

----------------------------------------------------------------------------------------- --------------

(6) f(x) =2x 1

2x 3

Answer

Then f(x) is Continuous on

----------------------------------------------------------------------------------------- --------------

(7)2

2xf ( x )

x 5x 8

Answer

2b 4ac for the denominator 25 4 1 8 25 32 -7 0Therefore there are no real zeroes for the denominator f is continuous on R .

----------------------------------------------------------------------------------------- --------------

1( 8 ) f ( x ) x 2 sin x Answer

1 2Each of the two functions g ( x ) x 2 "Polynomial fn" and g ( x ) sin x "Sin fn"

is continuous on R .

----------------------------------------------------------------------------------------- --------------

2

cos x sin x( 9 ) f ( x )

x 3

Answer

----------------------------------------------------------------------------------------- --------------

(10)

Answer

which makes Cos x = 0


17/28





The domain of x 3 :

x 3 0 x 3 Domain is 3,

x 3x

f ( x ) f ( 3 ) x 3

Examples(1) Discuss the Continuity of : f ( x ) x 3 over its domain .

Answer

At x = 3 : f(3)= 3 3 0

x 3for lim f(x):

x 3 x 3lim x 3= lim x 3 0

So as f(3) =x 3lim f ( x ) Then f(x) is Continuous at x 3

On the interval 3, :

x a x a

x a

let a R

So f ( a ) a 3

for lim f(x): lim a 3 a 3

f ( a ) lim f ( x ) f(x) is Continuous on R

----------------------------------------------------------------------------------------- --------------


18/28





x -3

x -3

x 3

2

x 2

x 2

At x -3 : f ( -3 ) 3 -3 2 -7

for lim f(x):

Right limit : f( -3 ) lim 3x 2 3 -3 2 -7

f ( 3 ) lim f(x) f ( x ) is Continous at x -3

At x 2 : f ( 2 ) 2 4 8

for lim f(x):

left limit : f( 2 ) lim 3x 2 8

Right limit

2

x 2

x 2

2

x 5

22

x 5

x 5

: f( 2 ) lim x 4 8

f ( x ) is Continous at x 2 f ( 2 ) lim f(x)

At x 5 : f ( 5 ) 5 4 29

for lim f(x):

left limit : f( 5 ) lim x 4 5 4 29

f ( 5 ) lim f(x) f ( x ) is Continous at x 5

x -3x

f ( x ) 3x 2

x 5

2x 4

x 2

7 8 29

2) Discuss the continuity of the function :2

3x 2 : -3 x 2f ( x )

x 4 : 2 x 5

Answer

For intervals :

(1) for all x -3,2 :

f ( x ) 3x 2 is a polynomial f ( x ) is continuous on - 3,2

for all x 2 , 5 : 2f ( x ) x +4 is a polynomial f ( x ) is continuous on x 2 , 5

(2) For points :

From (1) and ( 2 ) , we see that f ( x ) is continuous on -3 , 5


19/28





x 0x

f ( x ) Sin x Cos x

x

22Cos x 12 1

x 0

x 0

x 0

2

x

x

At x 0 : f (0 ) Sin 0 Cos 0 1

for lim f(x):

Right limit : f( 0 ) lim Sin x Cos x Sin 0 Cos 0 1

f (0 ) lim f(x) f ( x ) is Continous at x 0

At x : f ( ) 2Cos 1 1

for lim f(x):

left limit : f( ) lim Sin x Cos

2

x

x

x 1

Right limit : f( ) lim 2Cos x 1 1

f ( ) lim f(x) f ( x ) is Continuous at x

3) Discuss the continuity of the function :2

sin x cos x : 0 xf ( x )

2cos x 1 : x

Answer

For intervals :

(1) for all x 0 , : f ( x ) Sinx Cosx is continuous "Theorem"

for all x , 2f ( x ) 2Cos x 1 is continuous "Theorem"

(2) For points :

From (1) and ( 2 ) , we see that f ( x ) is continuous on 0 , ----------------------------------------------------------------------------------------- --------------


20/28





1x

2

1x

2

1x

2

1x

2

1 1 7 1At x : f ( ) 3

2 2 2 2

for lim f(x):

1 7 7 1left limit: f( ) lim x 3

2 2 2 2

1 3Right limit : f( ) lim 1

2 2x 4

1 1f( ) f( ) lim f(x) doesn' t exist

2 2

1f ( x ) is discontinous at x2

From (1) and ( 2 ) , we

1see that f ( x ) is continuous on R ,2

2

x

f ( x )7

x2

3

2x 4

1x

2

32x 4 0 2x 4 x 2

2x 4

3So the domain of is R- 22x 4

4) Discuss the continuity of the function on R where :

7 1x x

2 2f ( x )

3 1

x2x 4 2

Answer

For intervals :

(1)

1for all x - , :2

7 1f ( x ) x is a polynomial f ( x ) is continuous on - ,

2 2

1for all x , :

2

Domain of

1

f ( x ) is continuous on , 22

(2) For points :


21/28





x 0x

f ( x ) 23x 2

9 2x

x 5

5

x 1 x 2 x 3

2

x 0

2

x 0

x 0

x 1

2

x 1

x 1

At x 0 : f (0 ) 3 0 2 2

for lim f(x):

Right limit : f( 0 ) lim 3x 2 2


At x 1: f (1) 5

for lim f(x):

left limit : f(1 ) lim 3x 2 5

Right limit : f( 1 ) lim 5 5


23x +2 : 0 x 1

let f(x) 5 : 1 x 2

9 2x : 2 x 3x 5

(5) Then Discuss the Continuity over its domain

Answer

For intervals :

(1)

for all x 0,1 :

2f ( x ) 3x 2 is a polynomial f ( x ) is continuous on 0 ,1

for all x 1 , 2 :

f ( x ) 5 is a Constant function f ( x ) is continuous on 1 , 2

for all x 2 , 3 :

9 2xf ( x ) has a domain : x 5 But 5 2 , 3

x 5

f ( x ) is continuous on 2 , 3

(2) For points :


22/28





x 2

x 2

x 2

x 2

x 3

9 2 2 5At x 2 : f ( 2 ) -

2 5 3

for lim f(x):


9 2x 5Right limit : f( 2 ) lim -

x 5 3

f( 2 ) f( 2 ) lim f(x) doesn' t exist

f ( x ) is discontinous at x 2

9 2 3 3At x 3 : f ( 3 )

3 5 2

for lim

x 3

x 3

f(x):

9 2x 3left limit: f( 3 ) lim

x 5 2

f ( 3 ) lim f(x) f ( x ) is Continous at x 3

From (1) and ( 2 ) , we see that f ( x ) is continuous on 0 , 3 2

----------------------------------------------------------------------------------------- --------------


23/28





x 0

x 0

x 0

x 3

x 3

x 3

At x 0 : f (0 ) 5 0 5

for lim f(x):

Right limit : f( 0 ) lim 5 x 5


At x 3 : f ( 3 ) 3 1 2

for lim f(x):

left limit : f( 3 ) lim 5 x 2


x 3f ( 3 ) lim f(x) f ( x ) is Continous at x 3

At x 4 : f ( 4 ) Undefined

f ( x ) is discontinous at x 4From (1) and ( 2 ) , we see that f ( x ) is continuous on 0 , 4

x

f ( x ) 5 x

x 4

x 1

x 3x 0

Discuss the continuity of the function : f ( x ) x 3 2 on 0 , 4 (6)

Answer

x 3 : x 3 x 3 : 0 x 3

x 3 x 3x 3 : x 3 x 3 : 3 x 4

3 x 2 : 0 x 3Thus f ( x ) x 3 2

x 3 2 : 3 x 4

5 x : 0 x 3f ( x )

x 1

: 3 x 4

For intervals :

(1) for all x 0,3 :

f ( x ) 5 x is a polynomial f ( x ) is continuous on 0 ,3

for all x 3 , 4 :

f ( x ) x 1 is a polynomial f ( x ) is continuous on x 3 , 4

(2) For points :


24/28





x

f ( x ) 5 2x 11 2x

3x

2

Discuss the continuity of the function : f ( x ) 8 3 2x in R

3x2

3x

2

3x

2

3 3 3At x : f ( ) 5 2 8

2 2 2

for lim f(x):

3left limit : f( ) lim 5 2x 8

2

3Right limit : f( ) lim 11 2x 8

2

3f ( x ) is Continous at x

2

From (1) and ( 2 ) , we see that f ( x ) is continuous on R

(7)

Answer

3 33 2x : x 8 3 2x : x2 2

3 2x f ( x )3 3

3 2x : x 8 3 2x : x2 2

35 2x : x

2Thus f ( x ) f ( x )

311 2x : x

2

For intervals :

(1)3

for all x - , :2

3f ( x ) 5 2x is a polynomial f ( x ) is continuous on - ,

2

3for all x , :2

3f ( x ) 11 2x is a polynomial f ( x ) is continuous on ,

2

(2) For points :


25/28





x

f ( x ) -2x 3 7

x 5x -2

Discuss the continuity of the function : f ( x ) x 2 x 5 in R

x 5 : x 5x 5

x 5 : x 5

x 2 : x 2x 2

x 2 : x 2

2x 3

x 2

x 2

x 2

x 5

x 5

At x 2 : f ( 2 ) 7

for lim f(x):

left limit : f( 2 ) lim 2x 3 7

Right limit : f( 2 ) lim 7 7

f ( x ) is Continuous at x 2

At x 5 : f ( 5 ) 2 5 3 7

for lim f(x):


Right limit : f( 5

x 5) lim 2x 3 7


From (1) and ( 2 ) , we see that f ( x ) is continuous on R

(8)

Answer

Forx 5 : f(x)= (x 5) + (x+2)= 2 x 3Forx 2 : f(x)= -( x + 2) ( x 5 )= -2 x + 3

For 2 x 5 : f(x)= ( x + 2) ( x 5 )= 7

For intervals :

(1) for all x - , 2 :

f ( x ) -2x 3 is a polynomial f ( x ) is continuous on - , 2

for all x 2 , 5 :

f ( x ) 7 is Constant function So , it is continuous on 2 , 5

for all x 5 , : f ( x ) 2x 3 is a polynomial f ( x ) is continuous on 5 ,

(2) For points :


26/28





2

ax 7 : x -1

If f(x) x +b : -1 x 2

3x a : x 2

x

f ( x ) 3x 2 a x b

x 5x -2

2x 12

x

f ( x ) ax 7 2x b

x 2x -1

3x a

x 2

f ( x ) is continuous on R

f ( x ) is continuous at x -2 f ( 2 ) 3 2 2 -8

And f ( -2 ) f ( -2 ) f ( -2 )

lim ax b -2a b -8 (1)


f ( x ) is continuous

2

x 5

at x 5 f ( 5 ) 5 12 13

f ( 5 ) f ( 5 ) f ( 5 )

lim ax b 13 5a b 13 ( 2 )

From (1) and ( 2 ) and by subtraction : we get 7a 21 a 3 and b -8 6 -2

2

x -1


f ( x ) is continuous at x -1 f ( 1) 1 b 1 b

And f ( -1 ) f ( -1 ) f ( -1)

lim ax 7 1 b a 7 1 b a b 6 (1)

And f ( x ) is continuous on R

f ( x )

2

x 2

is continuous at x 2 f ( 2 ) 3 2 a 6 a

f ( 2 ) f ( 2 ) f ( 2 )

lim x b 6 a 4 b 6 a a b 2 ( 2 )

From ( 1) and ( 2 ) and by addition: we get 2a 4 a 2 and b 4

2

3x 2 : x -2

(9 ) If the function f ( x ) ax b : - 2 x 5 is Continuous on R , then find a and b

x - 12 : x 5

Answer

----------------------------------------------------------------------------------------- --------------

(10) is Continuous on R , then find a and b

Answer


27/28





x 0

x 0 x 0

x 0

x3

At x 0 : f (0 ) 1 Cos 0 2

for lim f(x):

sin2x 0 sin2xleft limit : f( 0 ) lim lim 2

x 0 x

Right limit : f( 0 ) lim 1 Cos x 2


At x : f ( ) 1 Cos3 3 3 2

for lim f(x):

left limit : f(3

x3

3) lim 1 Cos x

2


From (1) and ( 2 ) , we see that f ( x ) is continuous on - ,6 3

x

f ( x )sin2x

x

x3

1 Cos x

x 0x

6

(11) Discuss the Continuity of

sin2x -: x 0

x 6f ( x )

1 cos x : 0 x 3

Answer

For intervals :

(1)for all x - , 0 :6

sin2xf ( x ) is a Sin function f ( x ) is continuous on - , 0

x 6

for all x 0 , :3

f ( x ) 1 Cos x is Cos function So , it is continuous on 0 ,3

(2) For points :


28/28




x

f ( x )

x sin2x

1sin x

2

x a

x x - x 0

x 0

x 0

f ( x ) is continuous on 0 ,

f ( x ) is continuous at x 0 f (0 ) 0 a a

And f (0 ) f (0 ) f (0 )

x sin2x 0lim " Divide both numerator and denominator by x"

1 0sin x

2

sin2x1 1 2xlim 6 a 6 1 1

sin x2 2

x

2

x 5find the value of a and b such that f ( x ) is continuous where f ( x ) on R

x ax 25

(12)

Answer

2 2

2

f ( x ) is continuous on R The denominator has no zeros

Discriminant " b 4ac" is negative a 4 25 0

a 100 a -10 ,10

----------------------------------------------------------------------------------------- --------------(13 ) find the value of a and b such that f ( x ) is continuous where:

x sin2x: - x 0

1sin xf ( x )2

x a : 0 x

Answer

----------------------------------------------------------------------------------------- --------------

continuity stage (2)

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