<meta http-equiv="refresh" content="1; url=/nojavascript/"> Approximating Real Zeros of Polynomial Functions | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Math Analysis Go to the latest version.

2.7: Approximating Real Zeros of Polynomial Functions

Created by: CK-12
 0  0  0

Learning Objectives

  • Understand the statement of the Intermediate Value Theorem
  • Apply the intermediate value theorem to find bounds on the zeros of a function
  • Use numerical methods to find roots of a polynomial

Intermediate Value Theorem & Bounds on Zeros

The intermediate value theorem offers one way to find roots of a continuous function. Recall that our informal definition of continuous is that a function is continuous over a certain interval if it has no breaks, jumps asymptotes, or holes in that interval. Polynomial functions are continuous for all real numbers x. Rational functions are often not continuous over the set of real numbers because of asymptotes or holes in the graph. But for intervals without holes, rational functions are continuous.

If we know a function is continuous over some interval [a,b], then we can use the intermediate value theorem:

Intermediate Value Theorem

If f(x) is continuous on some interval [a,b] and n is between f(a) and f(b), then there is some c\in[a,b] such that f(c)=n.

The following graphs highlight how the intermediate value theorem works. Consider the graph of the function f(x)=\frac{1}{4} \left ( x^{3}-\frac{5x^{2}}{2}-9x \right ) below on the interval [-3, -1].

f(-3)=-5.625 and f(-1)=1.375. If we draw bounds on [-3, -1] and [f(-3),f(-1)], then we see that for any y-value between y=-5.625 and y=1.375, there must be an x value in [-3, -1] such that f(x)=y.

So, for example, if we choose c=-2, we know that for some x\in[-3,-1], f(x)=-2, even though solving this by hand would be a chore!

The Bounds on Zeros Theorem

The Bounds on Zeros Theorem is a corollary to the Intermediate Value Theorem:

Bounds on Zeros Theorem

If f is continuous on [a,b] and there is a sign change between f(a) and f(b) (that is, f(a) is positive and f(b) is negative, or vice versa), then there is a c\in(a,b) such that f(c)=0.

The bounds on zeros theorem is a corollary to the intermediate value theorem because it is not fundamentally different from the general statement of the IVT, just a special case where n=0.

Looking back at f(x)=\frac{1}{4} \left (x^{3}-\frac{5x^{2}}{2}-9x \right ) above, because f(-3)<0 and f(-1)>0, we know that for some x\in[-3,-1], f(x) has a root. In fact, that root is at x=2. and we can test that using synthetic division or by evaluating f(2) directly.

Example 1

Show that f(x)=-3x^{3}+5x has at least one root in the interval [1, 2]

Solution

Since f(x) is a polynomial we know that it is continuous. f(1)=2 and f(2)=-14. Let n=0\in[-14,2]. Applying the Intermediate Value Theorem, there must exist some point c\in[1,2] such that f(c)=0. This proves that f(x) has a root in [1, 2].

Example 2

The table below shows several sample values of a polynomial p(x).

& x && -4 && -2 && \qquad 0 && \qquad 1 && \quad 4 && 6 && \qquad 8 && \qquad 10 && \qquad 15 && 18\\& p(x) && 44.15 && 6.62 && -4.12 && -4.09 && 1.16 && 0 && -8.74 && -24.07 && -49.89 && 3.41

Based on the information in the table

(a) What is the minimum number of roots of p(x)?

(b) What are bounds on the roots of p(x) that you identified in (a)?

Solution

Since p(x) is a polynomial we already know that it is continuous. We can use the Intermediate Value Theorem to identify roots by looking at when p(x) changes from negative to positive, or from positive to negative.

(a) There are four sign changes of p(x) in the table, so at minimum, p(x) has four roots.

(b) The roots are in the following intervals x\in[-2,0], x\in[1,4], x\in[15,18], and the table also tells us that one root is at x=6.

Approximate Zeros of Polynomials Functions

In calculus you will learn several methods for numerically approximating the roots of functions. In this section we show one elementary numerical method for finding the zeros of a polynomial which takes advantage of the Intermediate Value Theorem.

Given a continuous function g(x),

  1. Find two points such that g(a)>0 and g(b)<0. Once you have found these two points, you can iteratively use the steps below to find the root of g(x) on the interval [a,b]. (Note, we will assume a<b, the same algorithm works with minor adjustments if b>a)
  2. Evaluate g \left ( \frac{a+b}{2} \right ).
    1. If g \left ( \frac{a+b}{2} \right )=0, then the root is x=\frac{a+b}{2}.
    2. If g \left ( \frac{a+b}{2} \right )>0, replace a with \frac{a+b}{2}. and repeat steps 1-2 using \left [ \frac{a+b}{2},b \right ]
    3. If g\left ( \frac{a+b}{2} \right )<0, replace b with \frac{a+b}{2}. and repeat steps 1-2 using \left [ a,\frac{a+b}{2} \right ]

This algorithm will not usually find the exact root of g(x), but it will allow you to find a reasonably small interval for the root. For example, you could repeat this process enough times so that you find an interval with |a-b|<0.01, and you will know the root of g(x) within a reasonably good approximation. The quality of the approximation you use (and the number of steps you use) will depend on why you are looking for the root. For most applications coming within 0.01 of the root is a reasonable approximation, but for some applications (such as building a bridge or launching a rocket) you need much more accuracy.

Example 3

Show the first 5 iterations of finding the root of h(x)=x^{2}-x-1 using the starting values a=0 and b=2.

Solution:

  1. First we verify that there is a root between x=0 and x=2. h(0)=-1 and h(2)=1 so we know there is a root in the interval [0, 2]. Check h\left ( \frac{2+0}{2} \right )=h(1)=-1. Since -1<0 we know the root is between x=1 and x=2, and we use the new interval [1, 2].
  2. Now we use the interval [1, 2]. h\left ( \frac{1+2}{2} \right )=h(1.5)=-0.25. Since -0.25<0, we use the interval [1.5, 2].
  3. h\left ( \frac{1.5+2}{2} \right )=h(1.75)=0.31. Since 0.31>0, we know that the zero is in the interval [1.5, 1.75].
  4. h\left ( \frac{1.5+1.75}{2} \right )=h(1.625)=0.02. Since 0.02>0, we know the root is between 1.5 and 1.625.
  5. h\left ( \frac{1.5+1.625}{2} \right )=h(1.5620)=-0.12. Since -0.12<0, we know the root is between 1.5620 and 1.625.

This example shows that after five iterations we have narrowed the possible location of the root to within 0.06 units. Not bad!

Recall that we have already reviewed using the CALC menu on a graphing calculator to find the roots of a function. This algorithm is not the one used by a calculator, but the calculator uses a similar, more efficient, algorithm for approximating the root of a function to 13 decimal places. When the calculator prompts for a GUESS? it is asking for a starting value to run the iterations.

Optional: An Interesting Corollary of the IVT

One surprising result of the Intermediate Value Theorem is that if you draw any great circle around the globe, then there must two antipodal points on that great circle that have exactly the same temperature.

Recall that a great circle is a path around a sphere that gives the shortest distance between any two points on the sphere. The equator is a great circle around the globe. Antipodal points are two points on opposite sides of the sphere. In the diagram below, B and B' are antipodal.

(Source: http://commons.wikimedia.org/wiki/File:Spherical_triangle_3d_opti.png License: GNU FDL)

For an informal proof of this result, look at the the image of a sphere with three great circles above. Suppose that the temperature at B is 75^{\circ} and the temperature B' is 50^{\circ}. The difference between the temperature at B and at B' is 75-50=25. Now imagine rotating the segment \overline{BB'} around the blue great circle. When the segment has rotated 180 degrees (i.e. when B has rotated to where B' is), then The difference between the temperatures at these two points is 50-75=-25. Since temperatures vary continuously, by the intermediate value theorem, there must be some point on that circle when the difference was 0, implying two antipodal points had the same temperature.

Notice that this little demonstration does not tell us which two antipodal points had the same temperature, only that there must be two such points on any great circle.

Exercises

For Exercises 1-5 use the intermediate value theorem to show the bounds on the zeros of each function. Your bounds should be within the whole number

  1. f(x)=2x^{3}-3x+4
  2. g(x)=-5x^{2}+8x+12
  3. h(x)=\frac{1}{2}x^{4}-x^{3}-3x^{2}+1
  4. j(x)=-\frac{2}{x^{2}+1}+\frac{1}{2}
  5. k(x) is a polynomial and selected values of k(x) are given in the following table: & x && -3 && -2 && -1 && \ \ \ 0 && \ 1 && \ \ \ 2 && \ \ \ \ 3\\& k(x) && -23.5 && -1 && 0.5 && -1 && .5 && -1 && -23.5
  6. Stephen argues the function r(x)=\frac{4x+1}{x+3.5} has two zeros based on the following table and an application of the Bounds on Zeros Theorem. What is faulty about Stephen's reasoning? & x && -5 && -4 && -3 && \ -2 && \ -1 && \ 0 && \ 1 &&  \ 2 &&  \ 3 && \ 4\\& r(x) && 12.67 && \ 30 && -22 && -4.67 && -1.20 && 0.29 && 1.11 && 1.64 && 2.0 && 2.27
  7. Apply the numerical algorithm five times to find a bound on the zeros of the following functions given the indicated starting values. What is your final estimate for the zero?
    1. k(x)=x^{4}-3x+1 on [0, 1]
    2. b(x)=-0.1x^{5}+3x^{3}-5x^{2} on [1, 3]
    3. c(x)=\frac{3x^{2}-2}{x^{4}+2} on [0, 2]

Answers

  1. There is a zero in [-2, -1]
  2. There is a zero in [-1, 0] and [2, 3]
  3. There is a zero in [-2, -1], [-1, 0], [0, 1], and [3, 4]
  4. There is a zero in [-2, -1] and [1, 2]
  5. There is a zero in [-2, -1], [-1, 0], [0, 1] and [1, 2]
  6. Unlike the previous question which specified the function was a polynomial (and hence continuous), r(x)=\frac{4x+1}{x+3.5} has a vertical asymptote at x=-3.5, so it is not continuous in the interval [-4, -3]. Therefore we cannot use the Bounds on Zeros theorem to claim there is a zero in that interval.
    1. The zero is in [0.3125, 0.34825]
    2. The zero is in [1.5000, 1.5625]
    3. [0.8125, 0.875]

Image Attributions

Description

Categories:

Concept Nodes:

Grades:

Date Created:

Feb 23, 2012

Last Modified:

Sep 24, 2014
Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.1.Math-Analysis.2.7
ShareThis Copy and Paste

Original text