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Transcript
Continuity
(Section 1.8)
Alex Karassev
Definition

A function f is continuous at a number a if
lim f ( x)  f (a)
xa

Thus, we can use direct substitution to
compute the limit of function that is
continuous at a
Some remarks

Definition of continuity requires three things:




f(a) is defined (i.e. a is in the domain of f)
lim f ( x ) exists
x a
Limit is equal to the value of the function
The graph of a continuous functions does not
have any "gaps" or "jumps"
Continuous functions and limits

Theorem
Suppose that f is continuous at b
g ( x)  b
and lim
Then
xa
lim f ( g ( x))  f (lim g ( x))
x a

x a
Example
lim
x 2
x2  4
x2  4
( x  2)( x  2)
 lim
 lim
x 2 x  2
x 2
x2
x2
 lim ( x  2)  2  2  4  2
x 2
Properties of continuous functions

Suppose f and g are both continuous at a
Then f + g, f – g, fg are continuous at a
 If, in addition, g(a) ≠ 0 then f/g is also continuous
at a


Suppose that g is continuous at a and f is
continuous at g(a). Then f(g(x)) is continuous
at a.
Which functions are continuous?

Theorem

Polynomials, rational functions, root functions,
power functions, trigonometric functions,
exponential functions, logarithmic functions are
continuous on their domains

All functions that can be obtained from the
functions listed above using addition, subtraction,
multiplication, division, and composition, are also
continuous on their domains
Example

Determine, where is the following function
continuous:
 1 
f ( x)  2 x  1  cos

 2 x 
Solution





 1 
f ( x)  2 x  1  cos

 2 x 
According to the previous theorem, we need to
find domain of f
Conditions on x: x – 1 ≥ 0 and 2 – x >0
Therefore x ≥ 1 and 2 > x
So 1 ≤ x < 2
Thus f is continuous on [1,2)
Intermediate Value Theorem
River and Road
River and Road
Definitions


A solution of equation is also called
a root of equation
A number c such that f(c)=0 is called
a root of function f
Intermediate Value Theorem (IVT)


f is continuous on [a,b]
N is a number between f(a) and f(b)


i.e f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a)
then there exists at least one c in [a,b] s.t. f(c) = N
y
y = f(x)
f(b)
N
f(a)
x
a
cb
Intermediate Value Theorem (IVT)


f is continuous on [a,b]
N is a number between f(a) and f(b)


i.e f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a)
then there exists at least one c in [a,b] s.t. f(c) = N
y
y = f(x)
f(b)
N
f(a)
a c1 c2
c3
x
b
Equivalent statement of IVT







f is continuous on [a,b]
N is a number between f(a) and f(b), i.e
f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a)
then f(a) – N ≤ N – N ≤ f(b) – N
or f(b) – N ≤ N – N ≤ f(a) – N
so f(a) – N ≤ 0
≤ f(b) – N
or f(b) – N ≤ 0
≤ f(a) – N
Instead of f(x) we can consider g(x) = f(x) – N
so g(a) ≤ 0 ≤ g(b)
or g(b) ≤ 0 ≤ g(a)
There exists at least one c in [a,b] such that g(c) = 0
Equivalent statement of IVT


f is continuous on [a,b]
f(a) and f(b) have opposite signs


i.e f(a) ≤ 0 ≤ f(b) or f(b) ≤ 0 ≤ f(a)
then there exists at least one c in [a,b] s.t. f(c) = 0
y
y = f(x)
f(b)
a
N=0
c
x
b
f(a)
Continuity is important!
y




Let f(x) = 1/x
Let a = -1 and b = 1
f(-1) = -1, f(1) = 1
However, there is no c
such that f(c) = 1/c =0
1
x
-1
0 1
-1
Important remarks


IVT can be used to prove existence of a root
of equation
It cannot be used to find exact value of the
root!
Example 1


Prove that equation x = 3 – x5 has a solution
(root)
Remarks
Do not try to solve the equation! (it is impossible
to find exact solution)
 Use IVT to prove that solution exists

Steps to prove that x = 3 – x5 has a solution

Write equation in the form f(x) = 0


Check that the condition of IVT is satisfied, i.e. that f(x) is
continuous



x5 + x – 3 = 0 so f(x) = x5 + x – 3
f(x) = x5 + x – 3 is a polynomial, so it is continuous on (-∞, ∞)
Find a and b such that f(a) and f(b) are of opposite signs, i.e.
show that f(x) changes sign (hint: try some integers or some
numbers at which it is easy to compute f)

Try a=0: f(0) = 05 + 0 – 3 = -3 < 0

Now we need to find b such that f(b) >0

Try b=1: f(1) = 15 + 1 – 3 = -1 < 0 does not work

Try b=2: f(2) = 25 + 2 – 3 =31 >0 works!
Use IVT to show that root exists in [a,b]

So a = 0, b = 2, f(0) <0, f(2) >0 and therefore there exists c in [0,2]
such that f(c)=0, which means that the equation has a solution
x = 3 – x5 ⇔ x5 + x – 3 = 0
y
31
N=0
0
-3
2 x
c (root)
Example 2

Find approximate solution of the equation
x = 3 – x5
Idea: method of bisections

Use the IVT to find an interval [a,b] that contains a root

Find the midpoint of an interval that contains root:
midpoint = m = (a+b)/2

Compute the value of the function in the midpoint

If f(a) and f (m) are of opposite signs, switch to [a,m]
(since it contains root by the IVT),
otherwise switch to [m,b]

Repeat the procedure until the length of interval is
sufficiently small
f(x) = x5 + x – 3 = 0
We already know that [0,2] contains root
f(x)≈
-3
0
x
<0
-1
Midpoint = (0+2)/2 = 1
>0
31
2
f(x) = x5 + x – 3 = 0
f(x)≈
-3
0
x
-1
6.1
31
1
1.5
2
Midpoint = (1+2)/2 = 1.5
f(x) = x5 + x – 3 = 0
f(x)≈
-3
0
x
-1
1.3
6.1
31
1
1.25
1.5
2
Midpoint = (1+1.5)/2 = 1.25
f(x) = x5 + x – 3 = 0
f(x)≈
-3
-1 -.07 1.3
1 1.125
0
x
1.25
6.1
31
1.5
2
Midpoint = (1 + 1.25)/2 = 1.125

By the IVT, interval [1.125, 1.25] contains root

Length of the interval: 1.25 – 1.125 = 0.125 = 2 / 16 = =
the length of the original interval / 24

24 appears since we divided 4 times

Both 1.25 and 1.125 are within 0.125 from the root!

Since f(1.125) ≈ -.07, choose c ≈ 1.125

Computer gives c ≈ 1.13299617282...
Exercise

Prove that the equation
sin x = 1 – x2
has at least two solutions
Hint:
Write the equation in the form f(x) = 0 and find three numbers x1, x2, x3,
such that f(x1) and f(x2) have opposite signs AND f(x2) and f(x3) have
opposite signs. Then by the IVT the interval [ x1, x2 ] contains a root AND
the interval [ x2, x3 ] contains a root.