Find all the roots of the given function. Use preliminary analysis and graphing to find good initial approximations.

f(x) = cos (4x) - 4x^2 + 8x

The function has a root when x≈

Step-by-step explanation:

The quadratic portion of f(x) factors as (-4x)(x -2), so has zeros at x=0 and x=2. The cosine portion of f(x) will have a value of 1 at x=0 and a value of about -0.15 at x=2. Thus, we might expect roots to be slightly negative and near x=2. (It turns out that a not-unreasonable approximation of the cosine function as cos(x) ≈ 1-x²/2 can be usefully used to get better approximations of the roots in each case.)

A graphing calculator makes it easy to find initial approximations of the two roots. The graph shows them to be -0.108 and 1.988.

__

Many graphing calculators also include the ability to determine a numerical value of the derivative of a function. This makes it possible to write an iteration function after the fashion of Newton's Method iteration:

  g(x) = x -f(x)/f'(x)

where g(x) gives a new value for old guess x, and f'(x) is the derivative of f(x).

The calculator we used is interactive, so the value of g(x) is found even as you type the argument value x. That makes it possible to achieve full calculator accuracy for the root estimate simply by copying the approximate value into the expression g(x).

  x ∈ {−0.107760269824, 1.98779489573}

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.