This is a little bit in advance, but I wanted to let everyone know that my servers will be undergoing some maintenance on May 17 and May 18 during 8:00 AM CST until 2:00 PM CST. Hopefully the only inconvenience will be the occasional “lost/broken” connection that should be fixed by simply reloading the page. Outside of that the maintenance should (fingers crossed) be pretty much “invisible” to everyone.

Paul

May 6, 2021

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### 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.
- Plot the function on the interval \(\left[ {0,9} \right]\).
- 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?
- 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?
- 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?
- 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?