# Use Newton's Method to solve # 2sinx = 2 - x #?

# x = 0.704577 #

We want to solve:

# 2sinx = 2 - x #

Let:

# f(x) = 2sinx + x -2 #

It is important to understand that we cannot use Newton's Method to determine the existence of a root, or establish the number of roots of

graph{2sinx + x -2 [-10, 10, -10, 10]}

From which we can be confident that there is a single root,

# { (x_1,=x_0,), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)),n gt 1 ) :} #

Therefore we need the derivative:

# \ \ \ \ \ \ f(x) = 2sinx+x-2 #

# :. f'(x) = 2cosx+1 #

So our iterative formula is:

# { (x_1,=x_0,), ( x_(n+1), = x_n - (2sinx+x-2)/(2cosx+1), x gt 1 ) :} #

Then using excel working to 8dp with

We get convergence to

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 (to 6dp):

# x = 0.704577 #

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When evaluating a one-sided 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 one-sided 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 one-sided 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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