Transcript Document
A continuous member of a family
of functions
Example (1)
Let f represents the general member of a
family of functions, with each member
corresponding to a real number c, where:
f ( x)
2 cx ; x 3
2
x
; x <3
Find for which value of the constant c, we will
have a continuous function ( on R)
Solution
For f to be continues on R , it has to be continuous at x =3, as well.
For f to be continuous at x =3, we must have:
lim f ( x) f (3) lim f ( x)
x 3
x 3
we have :
f (3) 2c(3) 6c and
lim f ( x) lim (2cx) 2c(3) 6c and
x 3
x 3
lim f ( x) lim x 2 (3) 2 9
x 3
x 3
Therefore f would be continuous at x 3, if :
9 6c
9 3
c
6 2
What is the formula of this function?
The formula of this continuous funcion is :
f ( x)
3 x ; x 3
x 2 ; x <3
3
2 ( ) x ; x 3
2
x 2 ; x <3
Example (2)
Let f represents the general member of a family
of function, with each member corresponding to
a pair real numbers a, b, where:
f ( x)
( 2 a ) x 2 ; x 2
( b 1) x ; 1< x < 2
a b
; x 1
Find for which value of the pair of constants a
and b,we will have a continuous function ( on R)
Solution
For f to be continues , it has to be continuous at x =2 and x=-1, as well.
For f to be continuous at x =2, we must have:
lim f ( x) f (2) lim f ( x)
x2
x2
we have :
f (2) (2 a ) (2) 2 8 4a and
lim f ( x) lim (2 a ) x 2 (2 a ) (2) 2 8 4a and
x2
x2
lim f ( x) lim (b 1) x (b 1)(2) 2b 2
x2
x 3
Therefore f would be continuous at x 2, if :
2b 2 8 4a
b 2a 5
Continue
For f to be continuous at x =-1, we must have:
lim f ( x) f (1) lim f ( x)
x ( 1)
x ( 1)
we have :
f (1) a b and
lim f ( x) lim (a b) a b and
x ( 1)
x ( 1)
lim f ( x) lim (b 1) x (b 1)(1) b 1
x ( 1)
x ( 1)
Therefore f would be continuous at x 1, if :
b 1 a b
a 1
Substituting that in b 2a 5, we get : b 2 5
b3
What will be the formula for this
function?
f ( x)
2
2x x ;; x12< x < 2
2
; x 1
( 2 1) x 2 ; x 2
( 31) x ; 1< x < 2
13
; x 1
Example (3)
Let f represents the general member of a family of functions,
with each member corresponding to a real number a, where:
f ( x) 6
x2 a 2
xa
; x >1
; x 1
Find for which value of the constant a, we will have a
continuous function at x=1. At which point this function will
be discontinuous?
Solution
For f to be continuous at x =1, we must have:
lim f ( x) f (1) lim f ( x)
x 1
x 1
we have :
f (1) 6 and
lim f ( x) lim 6 6 and
x 1
x 1
x2 a2
lim f ( x) lim
lim ( x a ) 1 a
x 1
x 1
x 1
xa
Therefore f would be continuous at x 1, if :
1 a 6
a5
Continue
The formula of this funcion is :
f ( x) 6
x 2 (5) 2
x 5
; x >1
; x 1
6
x 2 25
x 5
; x >1
; x 1
This function is not defined at x=5, and hence it is discontinues at x=5
Notice that is continuous on the intervals (-∞, 5), (5, ∞).
Example (4)
Let f represents the general member of a family of functions,
with each member corresponding to a real number a, where:
f ( x) 2 x 1 a
2
x sin( )
x
4
; x>0
; x0
Find for which value of the constant a, we will have a
continuous function (on R)
Solution
For f to be continues on R , it has to be continuous at x =0, as well.
For f to be continuous at x =0, we must have:
lim f ( x) f (0) lim f ( x)
x 0
x 0
we have :
f (0) 6 and
lim f ( x) lim (2 x 1 a ) 2(0) 1 a 1 a and
x 1
x 1
1
lim f ( x) lim x 4 sin( ) 0 Why ?
x 0
x 0
x
[ we can prove, using the sandwitch theorem, that
1
lim x 4 sin( ) 0 ]
x 0
x
Therefore f would be continuous at x 0, if :
1 a 0
a 1
The formula of this function
x 4 sin( 2x )
f ( x) 2 x 11
2x
2
x sin( )
x
4
; x>0
; x0
; x>0
; x0