Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
Calculus Volume 1

4.9 Newton’s Method

Calculus Volume 14.9 Newton’s Method

Learning Objectives

  • 4.9.1 Describe the steps of Newton’s method.
  • 4.9.2 Explain what an iterative process means.
  • 4.9.3 Recognize when Newton’s method does not work.
  • 4.9.4 Apply iterative processes to various situations.

In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form f(x)=0.f(x)=0. For most functions, however, it is difficult—if not impossible—to calculate their zeroes explicitly. In this section, we take a look at a technique that provides a very efficient way of approximating the zeroes of functions. This technique makes use of tangent line approximations and is behind the method used often by calculators and computers to find zeroes.

Describing Newton’s Method

Consider the task of finding the solutions of f(x)=0.f(x)=0. If ff is the first-degree polynomial f(x)=ax+b,f(x)=ax+b, then the solution of f(x)=0f(x)=0 is given by the formula x=ba.x=ba. If ff is the second-degree polynomial f(x)=ax2+bx+c,f(x)=ax2+bx+c, the solutions of f(x)=0f(x)=0 can be found by using the quadratic formula. However, for polynomials of degree 33 or more, finding roots of ff becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. Also, if ff is a polynomial of degree 55 or greater, it is known that no such formulas exist. For example, consider the function


No formula exists that allows us to find the solutions of f(x)=0.f(x)=0. Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of tan(x)x=0.tan(x)x=0. No simple formula exists for the solutions of this equation. In cases such as these, we can use Newton’s method to approximate the roots.

Newton’s method makes use of the following idea to approximate the solutions of f(x)=0.f(x)=0. By sketching a graph of f,f, we can estimate a root of f(x)=0.f(x)=0. Let’s call this estimate x0.x0. We then draw the tangent line to ff at x0.x0. If f(x0)0,f(x0)0, this tangent line intersects the xx-axis at some point (x1,0).(x1,0). Now let x1x1 be the next approximation to the actual root. Typically, x1x1 is closer than x0x0 to an actual root. Next we draw the tangent line to ff at x1.x1. If f(x1)0,f(x1)0, this tangent line also intersects the xx-axis, producing another approximation, x2.x2. We continue in this way, deriving a list of approximations: x0,x1,x2,….x0,x1,x2,…. Typically, the numbers x0,x1,x2,…x0,x1,x2,… quickly approach an actual root x*,x*, as shown in the following figure.

This function f(x) is drawn with points (x0, f(x0)), (x1, f(x1)), and (x2, f(x2)) marked on the function. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x1. From (x0, f(x0)), a tangent line is drawn, and it strikes the x axis at x2. If a tangent line were drawn from (x2, f(x2)), it appears that it would come very close to x*, which is the actual root. Each tangent line drawn in this order appears to get closer and closer to x*.
Figure 4.77 The approximations x0,x1,x2,…x0,x1,x2,… approach the actual root x*.x*. The approximations are derived by looking at tangent lines to the graph of f.f.

Now let’s look at how to calculate the approximations x0,x1,x2,….x0,x1,x2,…. If x0x0 is our first approximation, the approximation x1x1 is defined by letting (x1,0)(x1,0) be the xx-intercept of the tangent line to ff at x0.x0. The equation of this tangent line is given by


Therefore, x1x1 must satisfy


Solving this equation for x1,x1, we conclude that


Similarly, the point (x2,0)(x2,0) is the xx-intercept of the tangent line to ff at x1.x1. Therefore, x2x2 satisfies the equation


In general, for n>0,xnn>0,xn satisfies


Next we see how to make use of this technique to approximate the root of the polynomial f(x)=x33x+1.f(x)=x33x+1.

Example 4.46

Finding a Root of a Polynomial

Use Newton’s method to approximate a root of f(x)=x33x+1f(x)=x33x+1 in the interval [1,2].[1,2]. Let x0=2x0=2 and find x1,x2,x3,x4,x1,x2,x3,x4, and x5.x5.

Checkpoint 4.45

Letting x0=0,x0=0, let’s use Newton’s method to approximate the root of f(x)=x33x+1f(x)=x33x+1 over the interval [0,1][0,1] by calculating x1x1 and x2.x2.

Newton’s method can also be used to approximate square roots. Here we show how to approximate 2.2. This method can be modified to approximate the square root of any positive number.

Example 4.47

Finding a Square Root

Use Newton’s method to approximate 22 (Figure 4.79). Let f(x)=x22,f(x)=x22, let x0=2,x0=2, and calculate x1,x2,x3,x4,x5.x1,x2,x3,x4,x5. (We note that since f(x)=x22f(x)=x22 has a zero at 2,2, the initial value x0=2x0=2 is a reasonable choice to approximate 2.)2.)

Checkpoint 4.46

Use Newton’s method to approximate 33 by letting f(x)=x23f(x)=x23 and x0=3.x0=3. Find x1x1 and x2.x2.

When using Newton’s method, each approximation after the initial guess is defined in terms of the previous approximation by using the same formula. In particular, by defining the function F(x)=x[f(x)f(x)],F(x)=x[f(x)f(x)], we can rewrite Equation 4.8 as xn=F(xn1).xn=F(xn1). This type of process, where each xnxn is defined in terms of xn1xn1 by repeating the same function, is an example of an iterative process. Shortly, we examine other iterative processes. First, let’s look at the reasons why Newton’s method could fail to find a root.

Failures of Newton’s Method

Typically, Newton’s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton’s method might fail include the following:

  1. At one of the approximations xn,xn, the derivative ff is zero at xn,xn, but f(xn)0.f(xn)0. As a result, the tangent line of ff at xnxn does not intersect the xx-axis. Therefore, we cannot continue the iterative process.
  2. The approximations x0,x1,x2,…x0,x1,x2,… may approach a different root. If the function ff has more than one root, it is possible that our approximations do not approach the one for which we are looking, but approach a different root (see Figure 4.80). This event most often occurs when we do not choose the approximation x0x0 close enough to the desired root.
  3. The approximations may fail to approach a root entirely. In Example 4.48, we provide an example of a function and an initial guess x0x0 such that the successive approximations never approach a root because the successive approximations continue to alternate back and forth between two values.
A function is drawn with two roots, labeled root sought and root found. A point x0 is chosen such that when the tangent of x0 is taken, even though it is nearer to the root sought, the tangent points to the root found.
Figure 4.80 If the initial guess x0x0 is too far from the root sought, it may lead to approximations that approach a different root.

Example 4.48

When Newton’s Method Fails

Consider the function f(x)=x32x+2.f(x)=x32x+2. Let x0=0.x0=0. Show that the sequence x1,x2,…x1,x2,… fails to approach a root of f.f.

Checkpoint 4.47

For f(x)=x32x+2,f(x)=x32x+2, let x0=−1.5x0=−1.5 and find x1x1 and x2.x2.

From Example 4.48, we see that Newton’s method does not always work. However, when it does work, the sequence of approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a root found using Newton’s method are included in texts on numerical analysis.

Other Iterative Processes

As mentioned earlier, Newton’s method is a type of iterative process. We now look at an example of a different type of iterative process.

Consider a function FF and an initial number x0.x0. Define the subsequent numbers xnxn by the formula xn=F(xn1).xn=F(xn1). This process is an iterative process that creates a list of numbers x0,x1,x2,…,xn,….x0,x1,x2,…,xn,…. This list of numbers may approach a finite number x*x* as nn gets larger, or it may not. In Example 4.49, we see an example of a function FF and an initial guess x0x0 such that the resulting list of numbers approaches a finite value.

Example 4.49

Finding a Limit for an Iterative Process

Let F(x)=12x+4F(x)=12x+4 and let x0=0.x0=0. For all n1,n1, let xn=F(xn1).xn=F(xn1). Find the values x1,x2,x3,x4,x5.x1,x2,x3,x4,x5. Make a conjecture about what happens to this list of numbers x1,x2,x3,xn,…x1,x2,x3,xn,… as n.n. If the list of numbers x1,x2,x3,…x1,x2,x3,… approaches a finite number x*,x*, then x*x* satisfies x*=F(x*),x*=F(x*), and x*x* is called a fixed point of F.F.

Checkpoint 4.48

Consider the function F(x)=13x+6.F(x)=13x+6. Let x0=0x0=0 and let xn=F(xn1)xn=F(xn1) for n2.n2. Find x1,x2,x3,x4,x5.x1,x2,x3,x4,x5. Make a conjecture about what happens to the list of numbers x1,x2,x3,…xn,…x1,x2,x3,…xn,… as n.n.

Student Project

Iterative Processes and Chaos

Iterative processes can yield some very interesting behavior. In this section, we have seen several examples of iterative processes that converge to a fixed point. We also saw in Example 4.48 that the iterative process bounced back and forth between two values. We call this kind of behavior a 22-cycle. Iterative processes can converge to cycles with various periodicities, such as 2cycles,4cycles2cycles,4cycles (where the iterative process repeats a sequence of four values), 8-cycles, and so on.

Some iterative processes yield what mathematicians call chaos. In this case, the iterative process jumps from value to value in a seemingly random fashion and never converges or settles into a cycle. Although a complete exploration of chaos is beyond the scope of this text, in this project we look at one of the key properties of a chaotic iterative process: sensitive dependence on initial conditions. This property refers to the concept that small changes in initial conditions can generate drastically different behavior in the iterative process.

Probably the best-known example of chaos is the Mandelbrot set (see Figure 4.83), named after Benoit Mandelbrot (1924–2010), who investigated its properties and helped popularize the field of chaos theory. The Mandelbrot set is usually generated by computer and shows fascinating details on enlargement, including self-replication of the set. Several colorized versions of the set have been shown in museums and can be found online and in popular books on the subject.

A very complicated and organic looking fractal.
Figure 4.83 The Mandelbrot set is a well-known example of a set of points generated by the iterative chaotic behavior of a relatively simple function.

In this project we use the logistic map


as the function in our iterative process. The logistic map is a deceptively simple function; but, depending on the value of r,r, the resulting iterative process displays some very interesting behavior. It can lead to fixed points, cycles, and even chaos.

To visualize the long-term behavior of the iterative process associated with the logistic map, we will use a tool called a cobweb diagram. As we did with the iterative process we examined earlier in this section, we first draw a vertical line from the point (x0,0)(x0,0) to the point (x0,f(x0))=(x0,x1).(x0,f(x0))=(x0,x1). We then draw a horizontal line from that point to the point (x1,x1),(x1,x1), then draw a vertical line to (x1,f(x1))=(x1,x2),(x1,f(x1))=(x1,x2), and continue the process until the long-term behavior of the system becomes apparent. Figure 4.84 shows the long-term behavior of the logistic map when r=3.55r=3.55 and x0=0.2.x0=0.2. (The first 100100 iterations are not plotted.) The long-term behavior of this iterative process is an 88-cycle.

In the first quadrant, f(x) = 3.55x(1 – x) is graphed as is y = x. From some point on the x axis, a line is drawn up to the line y = x, at which point it turns to be horizontal and continues until it touches the outside edge of f(x), at which point it turns again to be vertical until it each the line y = x. This process continues a number of times and creates an interesting series of boxes.
Figure 4.84 A cobweb diagram for f(x)=3.55x(1x)f(x)=3.55x(1x) is presented here. The sequence of values results in an 88-cycle.
  1. Let r=0.5r=0.5 and choose x0=0.2.x0=0.2. Either by hand or by using a computer, calculate the first 1010 values in the sequence. Does the sequence appear to converge? If so, to what value? Does it result in a cycle? If so, what kind of cycle (for example, 2cycle,4cycle.)?2cycle,4cycle.)?
  2. What happens when r=2?r=2?
  3. For r=3.2r=3.2 and r=3.5,r=3.5, calculate the first 100100 sequence values. Generate a cobweb diagram for each iterative process. (Several free applets are available online that generate cobweb diagrams for the logistic map.) What is the long-term behavior in each of these cases?
  4. Now let r=4.r=4. Calculate the first 100100 sequence values and generate a cobweb diagram. What is the long-term behavior in this case?
  5. Repeat the process for r=4,r=4, but let x0=0.201.x0=0.201. How does this behavior compare with the behavior for x0=0.2?x0=0.2?

Section 4.9 Exercises

For the following exercises, write Newton’s formula as xn+1=F(xn)xn+1=F(xn) for solving f(x)=0.f(x)=0.


f ( x ) = x 2 + 1 f ( x ) = x 2 + 1


f ( x ) = x 3 + 2 x + 1 f ( x ) = x 3 + 2 x + 1


f ( x ) = sin x f ( x ) = sin x


f ( x ) = e x f ( x ) = e x


f ( x ) = x 3 + 3 x e x f ( x ) = x 3 + 3 x e x

For the following exercises, solve f(x)=0f(x)=0 using the iteration xn+1=xncf(xn),xn+1=xncf(xn), which differs slightly from Newton’s method. Find a cc that works and a cc that fails to converge, with the exception of c=0.c=0.


f(x)=x24,f(x)=x24, with x0=0x0=0


f(x)=x24x+3,f(x)=x24x+3, with x0=2x0=2


What is the value of cc for Newton’s method?

For the following exercises, start at

a. x0=0.6x0=0.6 and

b. x0=2.x0=2.

Compute x1x1 and x2x2 using the specified iterative method.


x n + 1 = x n 2 1 2 x n + 1 = x n 2 1 2


x n + 1 = 2 x n ( 1 x n ) x n + 1 = 2 x n ( 1 x n )


x n + 1 = x n x n + 1 = x n


x n + 1 = 1 x n x n + 1 = 1 x n


x n + 1 = 3 x n ( 1 x n ) x n + 1 = 3 x n ( 1 x n )


x n + 1 = x n 2 + x n 2 x n + 1 = x n 2 + x n 2


x n + 1 = 1 2 x n 1 x n + 1 = 1 2 x n 1


x n + 1 = | x n | x n + 1 = | x n |

For the following exercises, solve to four decimal places using Newton’s method and a computer or calculator. Choose any initial guess x0x0 that is not the exact root.


x 2 10 = 0 x 2 10 = 0


x 4 100 = 0 x 4 100 = 0


x 2 x = 0 x 2 x = 0


x 3 x = 0 x 3 x = 0


x + 5 cos ( x ) = 0 x + 5 cos ( x ) = 0


x+tan(x)=0,x+tan(x)=0, choose x0(π2,π2)x0(π2,π2)


1 1 x = 2 1 1 x = 2


1 + x + x 2 + x 3 + x 4 = 2 1 + x + x 2 + x 3 + x 4 = 2


x 3 + ( x + 1 ) 3 = 10 3 x 3 + ( x + 1 ) 3 = 10 3


x = sin 2 ( x ) x = sin 2 ( x )

For the following exercises, use Newton’s method to find the fixed points of the function where f(x)=x;f(x)=x; round to three decimals.


sin x sin x


tan(x)tan(x) on x=(π2,3π2)x=(π2,3π2)


e x 2 e x 2


ln ( x ) + 2 ln ( x ) + 2

Newton’s method can be used to find maxima and minima of functions in addition to the roots. In this case apply Newton’s method to the derivative function f(x)f(x) to find its roots, instead of the original function. For the following exercises, consider the formulation of the method.


To find candidates for maxima and minima, we need to find the critical points f(x)=0.f(x)=0. Show that to solve for the critical points of a function f(x),f(x), Newton’s method is given by xn+1=xnf(xn)f(xn).xn+1=xnf(xn)f(xn).


What additional restrictions are necessary on the function f?f?

For the following exercises, use Newton’s method to find the location of the local minima and/or maxima of the following functions; round to three decimals.


Minimum of f(x)=x2+2x+4f(x)=x2+2x+4


Minimum of f(x)=3x3+2x216f(x)=3x3+2x216


Minimum of f(x)=x2exf(x)=x2ex


Maximum of f(x)=x+1xf(x)=x+1x


Maximum of f(x)=x3+10x2+15x2f(x)=x3+10x2+15x2


Maximum of f(x)=xx3xf(x)=xx3x


Minimum of f(x)=x2sinx,f(x)=x2sinx, closest non-zero minimum to x=0x=0


Minimum of f(x)=x4+x3+3x2+12x+6f(x)=x4+x3+3x2+12x+6

For the following exercises, use the specified method to solve the equation. If it does not work, explain why it does not work.


Newton’s method, x2+2=0x2+2=0


Newton’s method, 0=ex0=ex


Newton’s method, 0=1+x20=1+x2 starting at x0=0x0=0


Solving xn+1=xn3xn+1=xn3 starting at x0=−1x0=−1

For the following exercises, use the secant method, an alternative iterative method to Newton’s method. The formula is given by


Find a root to 0=x2x30=x2x3 accurate to three decimal places.


Find a root to 0=sinx+3x0=sinx+3x accurate to four decimal places.


Find a root to 0=ex20=ex2 accurate to four decimal places.


Find a root to ln(x+2)=12ln(x+2)=12 accurate to four decimal places.


Why would you use the secant method over Newton’s method? What are the necessary restrictions on f?f?

For the following exercises, use both Newton’s method and the secant method to calculate a root for the following equations. Use a calculator or computer to calculate how many iterations of each are needed to reach within three decimal places of the exact answer. For the secant method, use the first guess from Newton’s method.


f ( x ) = x 2 + 2 x + 1 , x 0 = 1 f ( x ) = x 2 + 2 x + 1 , x 0 = 1


f ( x ) = x 2 , x 0 = 1 f ( x ) = x 2 , x 0 = 1


f ( x ) = sin x , x 0 = 1 f ( x ) = sin x , x 0 = 1


f ( x ) = e x 1 , x 0 = 2 f ( x ) = e x 1 , x 0 = 2


f ( x ) = x 3 + 2 x + 4 , x 0 = 0 f ( x ) = x 3 + 2 x + 4 , x 0 = 0

In the following exercises, consider Kepler’s equation regarding planetary orbits, M=Eεsin(E),M=Eεsin(E), where MM is the mean anomaly, EE is eccentric anomaly, and εε measures eccentricity.


Use Newton’s method to solve for the eccentric anomaly EE when the mean anomaly M=π3M=π3 and the eccentricity of the orbit ε=0.25;ε=0.25; round to three decimals.


Use Newton’s method to solve for the eccentric anomaly EE when the mean anomaly M=3π2M=3π2 and the eccentricity of the orbit ε=0.8;ε=0.8; round to three decimals.

The following two exercises consider a bank investment. The initial investment is $10,000.$10,000. After 2525 years, the investment has tripled to $30,000.$30,000.


Use Newton’s method to determine the interest rate if the interest was compounded annually.


Use Newton’s method to determine the interest rate if the interest was compounded continuously.


The total cost for printing xx books can be given by the equation C(x)=1000+12x+(12)x2/3.C(x)=1000+12x+(12)x2/3. Use Newton’s method to find the break-even point if the printer sells each book for $20.$20.

Order a print copy

As an Amazon Associate we earn from qualifying purchases.


This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at
Citation information

© Feb 5, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.