MTH132 - SS20 - Section 24 Chapter 1 - Functions and Limits · 01/01/2020 · Chapter 1 -...
Transcript of MTH132 - SS20 - Section 24 Chapter 1 - Functions and Limits · 01/01/2020 · Chapter 1 -...
MTH132 - SS20 - Section 24
Chapter 1 - Functions and Limits
1.8 - Continuity
Name:
We will be learning about:
• Continuity
• Discontinuity
• Intermediate Value Theorem
1
1 Continuous Functions
• A function f is said to be continuous at a number a if and only if
• This requires
1. f(a) to be defined.
2. limx!a
f(x) exists.
3. limx!a
f(x) = f(a).
• A function f is continuous on an interval if it is continuous at each point in that interval.
• Polynomial functions, rational functions, root functions, trigonometric functions are con-
tinuous everywhere in their domain.
Theorem: If f and g are continuous at a and c is a constant, then the following are
also continuous at a.
Theorem: If g is continuous at a and f is continuous at g(a), then the composite
function f � g given by (f � g)(x) = f(g(x)) is continuous at a.
Example 1.8.1.1: Determine where g(x) =p1�x2
x is continuous.
2
him fcx ,= fca )
x - a
CD f -18 3 f - g Ciii , c. f
CD f. g a ) Fg Tf 81=0
G is continuous oh its domain -
Domain : X # O and I - XZ 30 ⇒ ( I - x)LltX ) 30
-0 ⑦ -0I I
- I I
⇒ - I
EXEIF) e ,
C→ ⑦
So,
8 is continuous on C- 1,0 ) UCO,
I ].
2 Discontinuity
• We say that a function f is discontinuous at a if f is not continuous at a.
Example 1.8.2.1: Consider the function f(x) =
8><
>:
x+ 1 , x < 0
2 cos (⇡x) , 0 x 2
6� x2 , x > 2
Determine the values of x at which f(x) is discontinuous.
Types of Discontinuities:
1. Removable Discontinuity: Discontinuity can be removed by redefining f at just a single
number.
2. Infinite Discontinuity: The function has a vertical asymptote.
3. Jump Discontinuity: The function jumps from one value to another.
3
#
f is continuous in each interval XLO,
OLXCZ
and X > 2 .
Now,
we Will Check the end Points .
At fcoj-2coscoj-zltjmgfcxs-l.li#so+fcxs= ZX
Since l×i→Mo fax ) D .N -
E.
, f is discontinuousat O .
¥2 : f- (2) = 2 Cos (2-4)=2,
lim f CX ) = zoos (2-4)=2,
him fcx )= 6-4=2
x→z- x→zt r
A
Eg : % case
•
>
EG : % case a
onI s
Eg : Piecewise defined a
Functions •
O
>
Example 1.8.2.2: Determine and classify the discontinuities of f(x) = x2+5x+6x2�3x�10 .
Example 1.8.2.3: f(x) =3x2�3xx2�1 has a removable discontinuity at x = 1. Find a function
g(x) that agrees with f(x) for x 6= 1 and is continuous at 1.
4
o NO jump discontinuities Since this is not piecewise defined .
• For infinite discontinuities : fix )=X2t5Xt6= Lxt 3) Cxtz )
- -
XZ - 3×-10 ( x - 5) C x -12 )
When X=5 ,f (5) = (8)C ← I
O 0
•For removable discontinuities '
.
When X=- 2
,fc -2 ) = 8-
fcxj= 3×2-3 X= 3X
#= 3X
--
-
XZ - I C xtl ) Xtl
Let gcx ) = 3€Xt ,
-
Then fcx ) = gcx , for Xfl and G Cx >is
continuous at x=I .
3 Intermediate Value Theorem (IVT)
Theorem: (IVT)Suppose that f is continuous on the closed interval [a, b] and let N be any number between
f(a) and f(b), where f(a) 6= f(b).Then, there exists a number c 2 (a, b) such that f(c) = N .
• The IVT states that a continuous function takes on every intermediate value between the
function values f(a) and f(b).
Example 1.8.3.1: Use the IVT to show that there is a root of the function f(x) = x4+ x� 3
on the interval (1, 2).
5
d
fcb ) --
- -
-
yN a -- -
-
y I÷fca )- -
- fI
l l. I 7
a c b
To find roots ,Itt X - 3=0 .
f- C D= IF c- 3 = - I ← negative - I < 0<15
f- Cz ) = 24T 2-3 = 16 - I = 15 ← positive
So,
there is some CE Cl ,2) such that
Fcc ) = O .
Example 1.8.3.2: Suppose f(x) is a continuous function with values given by the table below.
x 0 1 2 3 4 5
f(x) 10.1 3.4 2.9 -1.5 0 0.8
On which interval must there be a c for which f(c) = 4?
A. (0,1) B. (1,2) C. (2,3) D. (3,4) E. (4,5)
Example 1.8.3.3: Prove that the equation cosx = x3has atleast one solution. What in-
terval is it in?
Exercises:
1. Use interval notation to indicate the interval where the following functions are continuous.
(a) f(x) = 8x7 � 5x3+ 2
(b) f(x) = x2�1x2�7x+6
(c) f(x) =p3x� 5
(d) f(x) = 3p2x� 4
2. f(x) =
(2x
x2�5x , x < 0
cos x+ a , x � 0
Find the value of a so that f is continuous at x = 0.
6
Let fcx ) = Cos x -x3
We need to find a CE Cobb ) Such that f- CC )=o
for some a and b .
f- Co ) =Cos o - O = I ,
f CTI ) =Cos Tyz - z}=0-
z } negative
positive .
Since - ⇐zP co < I, there is some CE (9%2)
such that fcc )=0 .
C- x,
co )
= CX-1) Cxtl ) Discontinuousat X=l
,6
-
⇐- 6) Cx - I )
continuous on C- As 13 U ( I,
6) UCG ,
3×-5 70 as X I I3
continuous on (5/3,0)
Continuous on C- Ex )
=¥*sf¥s
him fcxs -_him ¥g=
-
¥ ,
l-yygotfcxs-lxirygcosxta-itax-so-x-s.ci
For continuity,
-
z -
- ita ⇒ a =-
¥ - t.EE I