Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 453: 64


$x= 0.21916368, 1.08422462$

Work Step by Step

Draw a graph for the curve $y=4e^{-x^{2}}sinx-x^{2}+x-1$with the help of computer graphics (as depicted below) and move the cursor at the point of intersection of these curves. We observe that x-coordinate of the point is about 0.2. 1. Therefore, we start with initial approximation $x_{1}=0.2$ Using Newton’s approximation formula. $x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$ $f(x) =4e^{-x^{2}}sinx-x^{2}+x-1$ $f'(x)=4e^{-x^{2}}(cosx-sinx)-2x+1$ Thus, $x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{1})}$ $x_{2}=0.2-\frac{f(0.2)}{f'(0.2)}\approx0.2188327306$ $x_{3}=0.2188327306-\frac{f(0.2188327306)}{f'(0.2188327306)}\approx0.2191635713$ $x_{4}=0.2191635713-\frac{f(0.2191635713)}{f'(0.2191635713)}\approx0.2191636772$ $x_{5}=0.2191636772-\frac{f(0.2191636772)}{f'(0.2191636772)}\approx0.2191636772$ Hence, root $x= 0.21916368$ 2. Second approximation $x_{1}=1.1$ $x_{2}=1.1-\frac{f(1.1)}{f'(1.1)}\approx1.0843283016$ $x_{3}=1.0843283016-\frac{f(1.0843283016)}{f'(1.0843283016)}\approx 1.0842246211$ $x_{4}\approx 1.0842246163 $and $x_{5}\approx 1.0842246163$ Hence, root for the equation $f(x)=4e^{-x^{2}}sinx-x^{2}+x-1$ is $x= 0.21916368, 1.08422462$
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