Continuity (Section 1.8)

1
Continuity (Section 1.8)

• Alex Karassev

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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

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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"

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Continuous functions and limits

• TheoremSuppose that f is continuous at
  band Then
• Example

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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.

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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

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Example

• Determine, where is the following function
  continuous

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Solution

• According to the previous theorem, we need to
  find domain of f
• Conditions on x x 1 0 and 2 x gt0
• Therefore x 1 and 2 gt x
• So 1 x lt 2
• Thus f is continuous on 1,2)

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Intermediate Value Theorem
10
River and Road
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River and Road
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Definitions

• A solution of equation is also calleda root of
  equation
• A number c such that f(c)0 is calleda root of
  function f

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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
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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
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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

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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)
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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

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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!

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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

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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 (-8, 8)
• 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 a0 f(0) 05 0 3 -3 lt 0
• Now we need to find b such that f(b) gt0
• Try b1 f(1) 15 1 3 -1 lt 0 does not work
• Try b2 f(2) 25 2 3 31 gt0 works!
• Use IVT to show that root exists in a,b
• So a 0, b 2, f(0) lt0, f(2) gt0 and therefore
  there exists c in 0,2 such that f(c)0, which
  means that the equation has a solution

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x 3 x5 ? x5 x 3 0
y
31
x
0
2
N 0
c (root)
-3
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Example 2

• Find approximate solution of the equationx 3
  x5

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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 (ab)/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

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f(x) x5 x 3 0
We already know that 0,2 contains root
f(x)
gt 0
lt 0
31
-3
-1
Midpoint (02)/2 1
0
2
x
25
f(x) x5 x 3 0
f(x)
31
-3
6.1
-1
1.5
0
2
1
x
Midpoint (12)/2 1.5
26
f(x) x5 x 3 0
f(x)
31
-3
6.1
1.3
-1
0
2
1.5
1
1.25
x
Midpoint (11.5)/2 1.25
27
f(x) x5 x 3 0
f(x)
31
-3
6.1
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 /
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• 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...

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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.