Intermediate Value Theorem - PowerPoint by rt3463df

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									Continuity




    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
    and lim g ( x)  b
         x a
                         Then

        lim f ( g ( x))  f (lim g ( x))
        xa                            xa

   Example
                   x2  4        x2  4       ( x  2)(x  2)
          lim              lim          lim
          x 2     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 
                                       1 
                f ( x)  2 x  1  cos      
                                       2 x 
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
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)
                                                     x
                            a c1 c2      c3
                                              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    c                     x
        N=0
                                             b
                 f(a)
Continuity is important!

                                        y
   Let f(x) = 1/x
   Let a = -1 and b = 1
   f(-1) = -1, f(1) = 1
                                   1
                                             x
   However, there is no c    -1
    such that f(c) = 1/c =0            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
       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
x = 3 – x5 ⇔ x5 + x – 3 = 0

              y

              31




               0      2 x
        N=0
                   c (root)

              -3
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)≈
                               <0               >0
-3                             -1               31

0           Midpoint = (0+2)/2 = 1               2
 x
         f(x) = x5 + x – 3 = 0
 f(x)≈
-3                        -1                6.1           31

0                          1                1.5            2
 x
                               Midpoint = (1+2)/2 = 1.5
         f(x) = x5 + x – 3 = 0
 f(x)≈
-3                         -1     1.3       6.1    31

0                           1     1.25      1.5     2
 x
                     Midpoint = (1+1.5)/2 = 1.25
                      f(x) = x5 + x – 3 = 0
  f(x)≈
-3                                -1 -.07 1.3       6.1             31
                                   1 1.125   1.25
0                                                   1.5             2
    x          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.

								
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