Continuity ( Section 1.8)

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Continuity ( Section 1.8). Alex Karassev. Definition. A function f is continuous at a number a if 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:

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### Continuity (Section 1.8)

Alex Karassev

Definition
• A function f is continuous at a number a if
• 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)
• exists
• 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
• TheoremSuppose that f is continuous at band Then
• Example
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:
Solution
• 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

Definitions
• A solution of equation is also calledarootof equation
• A number c such that f(c)=0 is calledarootof 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

c

a

b

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

c3

c1

c2

a

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)

c

x

a

N = 0

b

f(a)

y

1

x

-1

0

1

-1

Continuity is important!
• 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
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 IVTto prove that solution exists
Steps to prove that x = 3 – x5 has a solution
• Write equation in the form f(x) = 0
• x5 + x – 3 = 0 so f(x) = x5 + x – 3
• Check that the condition of IVT is satisfied, i.e. that f(x) is continuous
• 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
Example 2
• Find approximate solution of the equationx = 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)≈

> 0

< 0

31

-3

-1

Midpoint = (0+2)/2 = 1

0

2

x

f(x) = x5 + x – 3 = 0

f(x)≈

31

6.1

-3

-1

1.5

0

2

1

x

Midpoint = (1+2)/2 = 1.5

f(x) = x5 + x – 3 = 0

f(x)≈

31

6.1

-3

1.3

-1

0

2

1.5

1

1.25

x

Midpoint = (1+1.5)/2 = 1.25

f(x)≈

f(x) = x5 + x – 3 = 0

31

6.1

-3

1.3

-.07

-1

1.25

1.125

1

0

2

1.5

Midpoint = (1 + 1.25)/2 = 1.125

x

• 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 equationsin x = 1 – x2has 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) haveopposite signs. Then by the IVT the interval [ x1, x2 ] contains a root ANDthe interval [ x2, x3 ] contains a root.