How do you use Newton's approximation method with #f(x) = x^2  2# to iteratively solve for the positive zero of #f(x)# accurately to within 4 decimal places using at least #6# iterations?
It's basically a recursion problem to repeatedly guess until you converge onto your answer. You take the result you got previously and utilize it in the next iteration, getting closer each time.
The fractions get ugly, but that's what Wolfram Alpha is for.
which reads as...
In summary, we went through the following iterations, with approximate decimals:
Again, doing this on a TI83, TI84, some calculator of that sort, would make visualizing this a bit easier.
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To use Newton's approximation method to iteratively solve for the positive zero of ( f(x) = x^2  2 ) accurately to within 4 decimal places using at least 6 iterations, follow these steps:

Choose an initial guess ( x_0 ) close to the root.

Use the formula for Newton's iteration:
[ x_{n+1} = x_n  \frac{f(x_n)}{f'(x_n)} ]

Compute the derivative of ( f(x) ), which is ( f'(x) = 2x ).

Iteratively apply the formula to find ( x_{n+1} ) from ( x_n ) until the desired accuracy is achieved.

Continue iterating until the difference between successive approximations is less than ( 0.0001 ) (4 decimal places).
Here is an example of the iterative process:
Initial guess: ( x_0 = 2 ) (since ( f(2) = 2^2  2 = 2 > 0 ) and we're looking for the positive zero).
Iteration 1: [ x_1 = x_0  \frac{f(x_0)}{f'(x_0)} = 2  \frac{2^2  2}{2 \cdot 2} = 2  \frac{2}{4} = 2  0.5 = 1.5 ]
Iteration 2: [ x_2 = x_1  \frac{f(x_1)}{f'(x_1)} = 1.5  \frac{1.5^2  2}{2 \cdot 1.5} = 1.5  \frac{0.25}{3} \approx 1.41667 ]
Iteration 3: [ x_3 = x_2  \frac{f(x_2)}{f'(x_2)} = 1.41667  \frac{(1.41667)^2  2}{2 \cdot 1.41667} \approx 1.41422 ]
Continue this process until you reach the desired accuracy or until you have completed at least 6 iterations.
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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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