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### Section 4-13 : Newton's Method

For problems 1 – 3 use Newton’s Method to determine $${x_{\,2}}$$ for the given function and given value of $${x_{\,0}}$$.

1. $$f\left( x \right) = 7{x^3} - 8x + 4$$, $${x_{\,0}} = - 1$$
2. $$f\left( x \right) = \cos \left( {3x} \right) - \sin \left( x \right)$$, $${x_{\,0}} = 0$$
3. $$f\left( x \right) = 7 - {{\bf{e}}^{2x - 3}}$$, $${x_0} = 5$$

For problems 4 – 8 use Newton’s Method to find the root of the given equation, accurate to six decimal places, that lies in the given interval.

1. $${x^5} = 6$$ in $$\left[ {1,2} \right]$$
2. $$2{x^3} - 9{x^2} + 17x + 20 = 0$$ in $$\left[ { - 1,1} \right]$$
3. $$3 - 12x - 4{x^3} - 3{x^4} = 0$$ in $$\left[ { - 3, - 1} \right]$$
4. $${{\bf{e}}^x} = 4\cos \left( x \right)$$ in $$\left[ { - 1,1} \right]$$
5. $${x^2} = {{\bf{e}}^{2 - {x^{\,2}}}}$$in $$\left[ {0,2} \right]$$

For problems 9 – 12 use Newton’s Method to find all the roots of the given equation accurate to six decimal places.

1. $$2{x^3} + 5{x^2} - 10x - 4 = 0$$
2. $${x^4} + 4{x^3} - 54{x^2} - 92x + 105 = 0$$
3. $$\displaystyle \frac{3}{2} - {{\bf{e}}^{ - {x^{\,2}}}} = \cos \left( x \right)$$
4. $$\ln \left( x \right) = 2\cos \left( x \right)$$
5. Suppose that we want to find the root to $${x^3} - 7{x^2} + 8x - 3 = 0$$. Is it possible to use $${x_{\,0}} = 4$$ as the initial point? What can you conclude about using Newton’s Method to approximate roots from this example?
6. Use the function $$f\left( x \right) = {\cos ^2}\left( x \right) - \sin \left( x \right)$$ for this problem.
1. Plot the function on the interval $$\left[ {0,9} \right]$$.
2. Use $${x_{\,0}} = 4$$ to find one of the roots of this function to six decimal places. Did you get the root you expected to?
3. Use $${x_{\,0}} = 5$$ to find one of the roots of this function to six decimal places. Did you get the root you expected to?
4. Use $${x_{\,0}} = 6$$ to find one of the roots of this function to six decimal places. Did you get the root you expected to?
5. What can you conclude about choosing values of $${x_{\,0}}$$ to find roots of equations using Newton’s Method.
7. Use $${x_{\,0}} = 0$$ to find one of the roots of $$2{x^5} - 7{x^3} + 3x - 1 = 0$$ accurate to six decimal places. Did we choose a good value of $${x_{\,0}}$$ for this problem?