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