`4(e^(-x^2))sin(x) = x^2 - x + 1` Use Newton's method to find all roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.

Textbook Question

Chapter 4, 4.8 - Problem 27 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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To solve equation using Newton's method apply the formula,




Plug in f(x) and f'(x) in the formula,


See the attached graph for getting the initial values of x . The curve intersects the x axis at `~~`  0.20 and 1.1.

Let's solve for the first zero x_1=0.2, carry out iteration until we have same approximations at decimal places.






Now we have two approximations that have same decimal places,

Now let's solve for second zero x_1=1.1,






So the roots of the equation to the eight decimal places are 0.21916367 , 1.08422461

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