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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)
x2
x2
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
x2
x2
lim f ( x)  lim (b  1) x  (b  1)(2)  2b  2
x2
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
b3
What will be the formula for this
function?

f ( x)  

2
 2x x ;; x12< x < 2
  2
; x  1

( 2 1) x 2 ; x  2
( 31) x ; 1< x < 2
13
; 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
xa
; 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
xa
Therefore f would be continuous at x  1, if :
1 a  6
a5
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
; x0
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 11


2
4
 x sin( x ) ; x > 0

  2 x ; x0


; x>0
; x0