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 Question from Nancy, a student: Use Newton's method to find the real root function, accurate to five decimal places f(x) = x^5+2x^2+3 How would I go about solving this? I used x[n] +1 = x[n] -(f(x[n]))/(f '(x[n])) So i got f(x)=x^5 +2x^2 +3 f '(x)= 5x^4 +4x Then I plugged it in and got x[n]=(x^5+2x^2+3)/(5x^4+4x) Now I'm stuck what do I do next?

Hi Nancy,

You have an error in the iteration step for Newton's Method, it should be

x[n +1] = x[n] -(f(x[n]))/(f '(x[n]))

or as I want to write it

$x_{n +1} = x_{n} -\frac{f(x_n)}{f '(x_n)}$

$x_{n +1} = x_{n} -\frac{x_n^5 +2x_n^2 +3}{5x_n^4 +4x_n}$

To get started with Newton's Method you need to select an initial value $x_0.$ Newton's Method works best if the starting value is close to the root you seeking. You might just think, why not just start with $x_0 = 0.$ Unfortunately $f'(0) = 0$ and $f'(x_0)$ is in the denominator so that won't work.

Let's be a little more careful. Looking at the function $f(x) = x^5 +2x^2 +3$ you can see that if $x$ is positive so is $f(x).$ $f(0)$ is also positive so any root must be negative. $f(-1)$ is also positive but $f(-2)$ is negative so there is a root between $-1$ and $-2.$ Hence it seems like starting with $x_0 = -1$ or $x_0 = -2$ would be good choices but again $f'(-1) = 0$ so the remaining choice is $x_0 = -2.$

The iterative step with $x_0 = -2$ gives

\begin{eqnarray*}
x_{1} &=& x_{0} -\frac{x_0^5 +2x_0^2 +3}{5x_0^4 +4x_0}\\
&=& -2 - \frac{-21}{72}\\
&=& -1.7083333
\end{eqnarray*}

Now find $x_2$ using the iterative step

$x_{2} = x_{1} -\frac{x_1^5 +2x_1^2 +3}{5x_1^4 +4x_1}$

Penny

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