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Section 4-13 : Newton’s Method
1. Use Newton’s Method to determine x 2 for ( )
3 2 f x = x − 7 x + 8 x − 3 if x (^) 0 = 5
Step 1
There really isn’t that much to do with this problem. We know that the basic formula for Newton’s
Method is,
1
n n n n
f x x x f x
+ =^ −^
so all we need to do is run through this twice.
Here is the derivative of the function since we’ll need that.
2 f ′^ x = 3 x − 14 x + 8
We just now need to run through the formula above twice.
Step 2
The first iteration through the formula for x 1 is,
0 1 0 0
f x (^) f x x f x f
Step 3
The second iteration through the formula for (^) x (^) 2 is,
1 2 1 1
f x (^) f x x f x f
So, the answer for this problem is x (^) 2 = 5.71875.
Although it was not asked for in the problem statement the actual root is 5.68577952608963. Note as
well that this did require some computational aid to get and it not something that you can, in general,
get by hand.
2. Use Newton’s Method to determine x 2 for ( ) ( )
2 f x = x cos x − x if (^) x (^) 0 = 1
Step 1
There really isn’t that much to do with this problem. We know that the basic formula for Newton’s
Method is,
1
n n n n
f x x x f x
+ =^ −^
so all we need to do is run through this twice.
Here is the derivative of the function since we’ll need that.
f ′ ( x ) = cos ( x ) − x sin ( x )− 2 x
We just now need to run through the formula above twice.
Step 2
The first iteration through the formula for x 1 is,
0 1 0 0
f x (^) f x x f x f
′ ′^ −
Don’t forget that for us angles are always in radians so make sure your calculator is set to compute in
radians.
Step 3
The second iteration through the formula for x (^) 2 is,
1 2 1 1
f x (^) f x x f x f
So, the answer for this problem is x (^) 2 = 0.7440943985.
Although it was not asked for in the problem statement the actual root is 0.739085133215161. Note as
well that this did require some computational aid to get and it not something that you can, in general,
get by hand.
- Use Newton’s Method to find the root of 4 3 x − 5 x + 9 x + 3 = 0 accurate to six decimal places in the
interval [ 4, 6].
The fourth iteration through the formula for x (^) 4 is,
3 4 3 3
f x x x f x
At this point we are accurate to 4 decimal places so we need to continue.
Step 6
The fifth iteration through the formula for x (^) 5 is,
4 5 4 4
7 1.714694911* 4.52891796 4.
f x x x f x
− = − = − = ′
At this point we are accurate to 8 decimal places which is actually better than we asked and so we can
officially stop and we can estimate that the root in the interval is,
x ≈ 4.
Using computational aids we found that the actual root in this interval is 4.52891795729. Note that this
wasn’t actually asked for in the problem and is only given for comparison purposes.
- Use Newton’s Method to find the root of
2 2 5
x x + = e accurate to six decimal places in the interval
[ 3, 4^ ].
Step 1
First, recall that Newton’s Method solves equation in the form f ( x ) = 0 and so we’ll need move
everything to one side. Doing this gives,
2 2 5
x f x = x + − e
Note that we could have just as easily gone the other direction. All that would have done was change
the signs on the function and derivative evaluations in the work below. The final answers however
would not be changed.
Next, we are not given a starting value, x (^) 0 , but we were given an interval in which the root exists so we
may as well use the midpoint of this interval as our starting point or, (^) x (^) 0 = 3.5. Note that this is not the
only value we could use and if you use a different one (which is perfectly acceptable) then your values
will be different that those here.
At this point all we need to do is run through Newton’s Method,
1
n n n n
f x x x f x
+ =^ −^
until the answers agree to six decimal places.
Step 2
The first iteration through the formula for x 1 is,
0 1 0 0
f x x x f x
Step 3
The second iteration through the formula for x (^) 2 is,
1 2 1 1
f x x x f x
We’ll need to keep going because even the first decimal is not correct yet.
Step 4
The third iteration through the formula for x (^) 3 is,
2 3 2 2
f x x x f x
At this point we are accurate to two decimal places so we need to continue.
Step 5
The fourth iteration through the formula for x (^) 4 is,
3 4 3 3
f x x x f x
At this point we are accurate to 6 decimal places which is what we were asked to do and so we can
officially stop and we can estimate that the root in the interval is,
x ≈ 3.
So, it looks like we are going to have three roots here ( i.e. the graph crosses the x -axis three times and
so three roots…).
For each root we’ll use the graph to pick a value of x (^) 0 that is close to the root we are after (we’ll go
from left to right for the problem) and then run through Newton’s Method,
1
n n n n
f x x x f x
+ =^ −^
until the answers agree to six decimal places.
Note as well that unlike Problems 3 & 4 we are not going to put in all the function evaluations for this
problem. We’ll leave that to you to check and verify our final answers for each iteration.
Step 2
For the left most root let’s start with x (^) 0 = −3.5. Here are the results of iterating through Newton’s
Method for this root.
0 1 0 0
1 2 1 1
2 3 2 2
3.443478261 No decimal places agree
3.442146902 Accurate to two decimal places
3.44214617 Accurate to six decimal places
f x x x f x
f x x x f x
f x x x f x
So, it looks like the estimate of the left most root is : x ≈ −3..
Step 3
For the middle root let’s start with x (^) 0 = 0. Be careful with this root. From the graph we may be
tempted to just say the root is zero. However, as we’ll see the root is not zero. It is close to zero, but is
not exactly zero!
Here are the results of iterating through Newton’s Method for this root.
0 1 0 0
1 2 1 1
2 3 2 2
0.06666666667 No decimal places agree
0.06639231824 Accurate to three decimal places
0.06639231426 Accurate to eight decimal places
f x x x f x
f x x x f x
f x x x f x
So, it looks like the estimate of the middle root is : (^) x ≈ 0.06639231426.
Step 4
For the right most root let’s start with x (^) 0 = 4.5. Here are the results of iterating through Newton’s
Method for this root.
0 1 0 0
1 2 1 1
2 3 2 2
3 4 3 3
4.380952381 No decimal places agree
4.375763556 Accurate to one decimal place
4.375753856 Accurate to four decimal places
4.375753856 Accurate to nine
f x x x f x
f x x x f x
f x x x f x
f x x x f x
decimal places
So, it looks like the estimate of the right most root is : x ≈ 4.375753856.
Step 5
Using computational aids we found that the actual roots of this equation to be,
x = − 3.44214616993 x = 0.0663923142603 x =4.
So, let’s graph both the quadratic and sine function to see if our intuition on this is correct. Doing this
gives,
So, it looks like we guessed correctly and should have two roots here.
For each root we’ll use the graph to pick a value of (^) x (^) 0 that is close to the root we are after (we’ll go
from left to right for the problem) and then run through Newton’s Method,
1
n n n n
f x x x f x
+ =^ −^
until the answers agree to six decimal places.
Note as well that unlike Problems 3 & 4 we are not going to put in all the function evaluations for this
problem. We’ll leave that to you to check and verify our final answers for each iteration.
Also note that the analysis that we had to do to estimate the number of roots is something that does
need to be done for these kinds of problems and it will differ for each equation. However, if you do
have a basic knowledge of how most of the basic functions behave you can do this for most equations
you’ll be asked to deal with.
Step 2
For the left most root let’s start with x (^) 0 = −1.5. Here are the results of iterating through Newton’s
Method for this root.
0 1 0 0
1 2 1 1
2 3 2 2
3 4 3 3
1.755181948 No decimal places agree
1.728754674 Accurate to one decimal place
1.728466353 Accurate to three decimal places
1.728466319 Accurate to
f x x x f x
f x x x f x
f x x x f x
f x x x f x
seven decimal places
So, it looks like the estimate of the left most root is : x ≈ −1..
Step 3
For the right most root let’s start with x (^) 0 = 1. Here are the results of iterating through Newton’s
Method for this root.
0 1 0 0
1 2 1 1
2 3 2 2
1.062405571 No decimal places agree
1.061549933 Accurate to two decimal places
1.061549775 Accurate to six decimal places
f x x x f x
f x x x f x
f x x x f x
So, it looks like the estimate of the right most root is : (^) x ≈ 1.061549775.
Step 4
Using computational aids we found that the actual roots of this equation to be,
x = − 1.72846631899718 x =1.
Note that these weren’t actually asked for in the problem and are only given for comparison purposes.
As a final warning about Newton’s Method, be careful to not assume that you’ll get six (or better in
some cases) decimal places of accuracy with just a few iterations.
These problems were chosen with the understanding that it would only take a few iterations of the
method. There are problems and/or choices of x (^) 0 for which it will take significantly more iterations to
