How do you use Newton's method to find the approximate solution to the equation #2x^3+x+4=0#?
Using NR Iteration we get the solution is
Let
First let us look at the graphs:
graph{2x^3+x^2+4 [4, 4, 10, 10]}
We can see there is one solution in the interval
We can find the solution numerically, using NewtonRhapson method
# \ \ \ \ \ \ \f(x) = 2x^3+x^2+4 #
# :. f'(x) = 6x^2+2x #
The NewtonRhapson method uses the following iterative sequence
# { (x_1,=1), ( x_(n+1), = x_n  f(x_n)/(f'(x_n)) ) :} #
Then using excel working to 8dp we can tabulate the iterations as follows:
We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.
And we conclude that the solution is
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To use Newton's method to find the approximate solution to the equation (2x^3 + x + 4 = 0), follow these steps:
 Start with an initial guess for the root of the equation, denoted as (x_0).
 Use the formula for Newton's method: [x_{n+1} = x_n  \frac{f(x_n)}{f'(x_n)}] where (x_{n+1}) is the next approximation, (x_n) is the current approximation, (f(x_n)) is the function evaluated at (x_n), and (f'(x_n)) is the derivative of the function evaluated at (x_n).
 Iterate using this formula until the desired level of accuracy is achieved.
For the given equation (2x^3 + x + 4 = 0):
 Choose an initial guess, (x_0).
 Calculate (f(x_0)) and (f'(x_0)).
 Use the formula (x_{n+1} = x_n  \frac{f(x_n)}{f'(x_n)}) to find (x_1).
 Continue iterating until convergence.
Let's demonstrate the process:

Choose an initial guess, let's say (x_0 = 1).

Calculate (f(x_0)) and (f'(x_0)):
(f(x_0) = 2(1)^3 + (1) + 4 = 2  1 + 4 = 5)
(f'(x_0) = 6(1)^2 + 1 = 6 + 1 = 7)

Use the formula:
(x_1 = x_0  \frac{f(x_0)}{f'(x_0)} = 1  \frac{5}{7} \approx 1.286)

Now, repeat the process with (x_1) as the new approximation and continue iterating until the desired level of accuracy is reached.
Continue iterating until convergence, or until the desired level of accuracy is achieved. This process provides an approximate solution to the equation (2x^3 + x + 4 = 0).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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