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Continuity of a Function
Over an interval a function is continuous if its graph can be drawn
without lifting the pen from the paper, otherwise the function is discontinuous.
We now state formally the definition of a continuous function as follows:
A function f is continuous at “a” if
lim f ( x )  f ( a ) .
xa
Three conditions implied by the definitions
1) f(a) is defined
2) lim f ( x )  L exists
xa
3) L=f(a)
The function f( x ) =
x2  x  2
is discontinuous at x = 2 as shown in
x2
Figure (a) where there is a hole or break at x = 2. Notice that the limit of
x2  x  2
as x  2 exists but f( 2 ) does not. This kind of discontinuity is
x2
removable because the discontinuity can be removed by redefining the
function into g( x ) = x + 1.
Another kind of discontinuity is2called jump discontinuity because
the function value jumps from one value to another. The function is
discontinuous at x = - 2. Here f( - 2 ) is defined but the left and right limits are
not the same. Same thing for x = 2.
There is also an infinite discontinuity. The function is discontinuous
at x = 0 because f(0) is not defined. The discontinuity of the function is called
an infinite discontinuity because y approaches to infinity as x  0. This
resulted to the presence of an asymptote as shown below,
3numbers at the continuity of the
Domain is a set of real
function. The elements of which if substituted to the independent variable, x in
y=f(x) can produce a real number value of the function, y.
Range is a set of real numbers values of the dependent variable, y for
which the relation is defined. In curve tracing it is important to know them
specially the domain. As when curve are traced by point plotting method it is
done by assigning values to independent variable and correspondingly the
values of y will then follow.Points on curves are identified and curves are traced.
But nowadays curve tracing is not so difficult to do as computer apps
like Mathway can help you do the task easily.
Show the graph of the following, give the domain and range:
1) y  ( x  1)
3
2) y  e
2x
3) y 
4  x2
Solution:
y  ( x  1) 3
As polynomial functions x can be all real numbers thus domain, D 
x
y=(x+1)
Graph:
3
x / x  R
-3
-2
-1
0
1
2
3
-8
-1
0
1
8
27
64
4
2) y  e 2 x
As polynomial functions x can be all real numbers thus domain, D 
x / x  R
x
-3
-2
-1
0
1
2
3
y=e2x
0.00248
0.0183
0.135
1
7.389
54.598
403.429
Graph:
y=e2x
domain, D  {x / xR}
range, R  { y / y  0} from the graph
or by solving for x in y  e 2 x
ln y  ln e 2 x
(recall : log b x  y iff b y  x)
ln y  2 x, x  (1 / 2) ln y
If inverse function is needed, after solving for x let y=x
y=(1/2)lnx
(inverse function)
If the given is f(x)=e2x
considering y=f(x)
Inverse function is f 1(x)=(1/2)lnx
-
(in inverse function notation)
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3) y  4  x 2
For irrational function, take note of the following:
if y  n u if n is odd u can be any real number
but if n is even u≥0.
So when,
4  x 2  0 and solving for zero of 4  x 2
4  x 2  0,
x 2  4 and x  2
Intervals: x≤-2 or (, 2]
test No : x=-3,
: -2≤x≤2 or [-2, 2]
x≥2 or [2,+∞)
: x=0
: x=3
Graph :
Domain : D={x/-2 ≤ x ≤ 2}
Range
: R={y/0 ≤ y ≥ 2
4-(-3)2= -5 not ok
4-02= 4 ok (continuous)
4-32= -5 not ok
Rational Function
f ( x) 
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N ( x) ao x m  a1 x m1  a 2 x m2  . . .  a m1 x  a m

D( x)
bo x n  b1 x n 1  b2 x n 2  . . .  bn 1 x  bn
Rational functions, f(x) are continuous for all values of x except for the
value/s at which the denominator is zero. The asymptotes of this function is
very useful in sketching its graph. Asymptote is a line which can be vertical,
horizontal or slant for which the curve approaches more closely without
intersecting through that line.
VERTICAL ASYMPTOTES
The line x=a is the vertical asymptote of f(x) if,
N ( x)
lim D( x)  
xa
To find the vertical asymptotes, let D(x)=0 and solve for x, when D(a)=0, the
x=a is the vertical asymptote. This is where a discontinuity of the rational
function exists.
HORIZONTAL ASYMPTOTE
The line y=b is the horizontal asymptote if,
N ( x)
lim D( x)  b
x 
Recall the limits at infinity
N ( x)
N ( x)
ao x m  a1 x m1  a 2 x m2  . . .  a m1 x  a m
n
n 1
 b2 x n 2  . . .  bn 1 x  bn
o
1
lim D( x)  lim D( x)  b x  b x
x 
x 
1) If m<n (proper fraction), L=0, thus the horizontal asymptote is y=0.
2) If m=n (improper fraction with equal degree for N and D), L=ao/bo and the
horizontal asymptote is y=ao/bo.
3) If m>n (an improper fraction),
and there is no limit under this
situation and therefore there is also no horizontal asymptote but a slant
asymptote may exist provided m>n by only 1.
SLANTING ASYMPTOTE
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By division of polynomial reduce the function to
y
N ( x)
R( x)
 Q( x) 
D( x)
D( x)
and from this result, the slanting asymptote is y=Q(x).
Identify the asymptotes and show the graph of the following curves:
1) y  4
x2
Solution :
Vertical asymptote : let D(x)=0 and solve for x, x2=0 so x=0.
V.A. is x=0 (the y-axis)
Horizontal asymptote: (as proper fraction in limit @ infinity, L=0)
H.A. is y=0 (the x-axis)
Graph preparation :
f(-1)=4/(-1)2=4
Since, y=4 and H.A. is
y=0, it means that the
curve is above the H.A.
f(1)=4/(1)2=4
Since, y=4 and H.A. is y=0,
it means that the curve is
above the H.A.
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Graph:
2) y 
x2
x2  4
Solution:
Vertical Asymptotes : D(x)=0, x2 - 4=0, x=- 2 and x=2
Horizontal Asymptote: since m=n=2, L=1 and y=1
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Checking if the curve is above or below the HA. y 
Left of x=-2 or x<-2, f (  3 ) 
x2
x2  4
(  3) 2
 9 / 5  1 . 8 above y  1
(  3) 2  4
2
In between -2 and 2: f (0)  (0)  0 pt(0,0) it is belowy  1
2
(0)  4
More information needed, so assign more values of x between -2 and 2.
f (1) 
(1) 2
 0.333
(1) 2  4
f (1) 
(1) 2
 0.333
(1) 2  4
With these two values of y or function values the idea is enough to conclude
that the curve concaves downward from (0,0)
For last interval, right of x=2 or x>2
f (3) 
(3) 2
 9 / 5  1.8 above y  1
(3) 2  4
2x 2  x  1
3) y 
x 1
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Solution :
Vertical Asymptote : let D(x)=0, x + 1=0, x = -1
Horizontal Asymptote : none (improper fraction, degree of N>D)
Slanting Asymptote exists since the degree of N>D by 1
Divide the polynomial :
The Slanting Asymptote is y=Q(x)
SA:
2x 2  x  1
Graph Preparation: f ( x ) 
x 1
checking the curve location
@ the right of VA or x>-1
f(0)=1
@ the left of VA or x<-1
f(-2)=-7
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Graph :
For exercises only not required for submission
I. Sketch the graph of the following; find the domain and range, and tell if
the given is a function or not.
1. y = 2x – 4
3.
x
y
5. y2 = 4x
-2
4
2. y = 4 – x2
0
1
-2
0
1
3
4. y =  x 
6. y =
9  x2
II. Find the value/s of x for which
5
.
2x  1
1) y 
x2
x 2  4x
4x  1
3) y  3
x  25 x
2) f( x ) =
12
the function is discontinuous.
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27  3 x
7x  2
5) y  3
x  2 x 2  3x
4) y 
III. Show the graph of the following curves with their asymptotes if any
x2  9
x3
2
3. f( x ) =
x
1. f ( x ) 
5. f( x ) =
7. y=
x2
x 2  4x
x2
x3
x2  x  6
x3
2
x 1
4. f( x ) = 2
x 1
2. f( x ) =
x2 1
x
3x
8. f ( x )  2
x 4
6. f( x ) =
Prepared by : Engr. Joselito U. Ferraren/CE-MAEd-Math
Faculty,
DEMPC/CIT-U
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