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Use Newton’s method to approximate the indicated root of the equation correct to six decimal places. The positive root of 3 sin x = x

Sagot :

Answer:

[tex]x \approx 2.278863[/tex]

Step-by-step explanation:

Required

The positive root of [tex]3\sin(x) = x[/tex]

Equate to 0

[tex]0 = x -3\sin(x)[/tex]

So, we have our function to be:

[tex]h(x) = x -3\sin(x)[/tex]

Differentiate the above function:

[tex]h'(x) = 1 -3\cos(x)[/tex]

Using Newton's method of approximation, we have:

[tex]x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}[/tex]

Plot the graph of [tex]h(x) = x -3\sin(x)[/tex] to get [tex]x_1[/tex] --- see attachment for graph

From the attached graph, the first value of x is at 2.2; so:

[tex]x_1 = 2.2[/tex]

So, we have:

[tex]x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}[/tex]

[tex]x_{1+1} = x_1 - \frac{h(x_1)}{h'(x_1)}[/tex]

[tex]x_{2} = 2.2 - \frac{2.2 -3\sin(2.2)}{1 -3\cos(2.2)} = 2.28153641[/tex]

The process will be repeated until the digit in the 6th decimal place remains unchanged

[tex]x_{3} = 2.28153641 - \frac{2.28153641 -3\sin(2.28153641)}{1 -3\cos(2.28153641)} = 2.2788654[/tex]

[tex]x_{4} = 2.2788654 - \frac{2.2788654 -3\sin(2.2788654)}{1 -3\cos(2.2788654)} = 2.2788627[/tex]

[tex]x_{5} = 2.2788627 - \frac{2.2788627-3\sin(2.2788627)}{1 -3\cos(2.2788627)} = 2.2788627[/tex]

Hence:

[tex]x \approx 2.278863[/tex]

View image MrRoyal
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