We want to use Newton's method
to approximate a root or a zero of the equation
three x to the seventh,
plus eight x to the fourth,
plus two equals zero.
We want to let x sub one equal three
be the initial approximation.
The formula for Newton's method for
approximating zeros of a function is given here below.
We can derive this by determining the
x intercept or zero of the equation of a tangent line.
The first thing we need to recognize is
that this formula is used for finding zeros of functions,
and we're given an equation.
But, notice how this polynomial equation is equal to zero,
which means for our formula f of x will be equal to
three x to the seventh,
plus eight x to the fourth, plus two.
Notice to find the zero of this function
we would set f of x equal to zero and solve for x,
which would give us the initial equation.
Notice how Newton's method also calls for
the derivative, or f prime, of x.
Let's also find that now.
F prime of x is equal to the derivative of
three x to the seventh,
that would be 21x to the sixth,
plus the derivative of eight x to the fourth,
that would be 32x to the third,
and the derivative of two would be zero.
Let's take these two functions to the next slide
with our initial approximation and find
the second and third approximations.
Again, we're given that x sub one equals three,
so now we want to find the
second approximation, or x sub two.
Notice in our formula here,
if we want x sub two, n is equal to one,
since one plus one is two.
So completing the formula we would have
x sub two equals x sub one,
minus f of x sub one,
divided by f prime of x sub one.
We know x sub one is three,
which means this is equal to
three minus f of three,
divided by f prime of three.
If we were required to show this work by hand,
we would have three minus the quotient
where the numerator is f of three,
which would be three times three to the seventh,
plus eight times three to the fourth,
plus two, divided by f prime of three,
which would be 21 times three to the sixth,
plus 32, times three to the third.
But I'll also show how we can use
a graphing calculator to determine this value.
The first thing we'll do is enter f of x and y sub one,
and f prime of x and y sub two,
so we'll press y equals,
type in f of x three x to the seventh,
plus eight x to the fourth,
plus two, enter.
Y sub two will be f prime of x,
21x to the sixth,
plus 32x to the third.
Now, we'll go back to the home screen.
To determine our second approximation,
we'll just enter three minus y sub one of three,
divided by y sub two of three.
We'll have three minus,
to access y sub one we press vars,
right arrow, enter, enter,
and in parenthesis we want three as our input,
divided by f prime of three,
which will be y sub two of three,
so we press vars, right arrow,
enter, arrow down to two, enter,
and in parenthesis enter three.
And then enter one more time,
so our second approximation to
four decimal places would be 2.5541.
If we did this by hand, of course,
we'd get the same approximation.
Now, we'll find x sub three, the third approximation.
Notice for x sub three, n would be two,
since two plus one equals three.
So, x sub three equals x sub two,
minus f of x sub two,
divided by f prime of x sub two,
which, in our case, x sub two is 2.5541,
so we have 2.5541,
minus f of 2.5541,
divided by f prime of 2.5541.
For x sub three, you can see why it would be helpful
to use a calculator to find f of 2.5541.
Performing substitution into function f
would be pretty messy as well as for f prime of 2.5541.
So, for this example, I'm going to
go straight to the calculator.
We'll have 2.5541,
minus f of 2.5541,
that's y sub one of the input,
divided by f prime of 2.5541,
which would be y sub two,
so vars, right arrow, enter, enter,
the input is 2.5541 for f of x,
divided by y sub two of the input,
so vars, right arrow, enter,
select y sub two by pressing two,
then the input is 2.5541.
Close parenthesis, and enter.
Our third approximation to four decimal places,
notice how we have an eight in the fifth decimal place,
so we round up, and since we have a nine here,
it's going to be approximately 2.1660.
So now we have our second and third approximations.
I hope you found this helpful.