Calculus I - Newton's Method (Assignment Problems)

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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 \({x_{\,0}}\).

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

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.

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

\(2{x^3} + 5{x^2} - 10x - 4 = 0\)
\({x^4} + 4{x^3} - 54{x^2} - 92x + 105 = 0\)
\(\displaystyle \frac{3}{2} - {{\bf{e}}^{ - {x^{\,2}}}} = \cos \left( x \right)\)
\(\ln \left( x \right) = 2\cos \left( x \right)\)
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?
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.
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?