Show that $x^3 −2x−17 = 0$ has only one real root and determine its correct value to 2 decimals using Newton's method.

Question

What did I do

From the graph I know that the root exists and is approximately 2.83. So I used 2 as $x_0$.

graph

$f(x) = x^3 − 2x − 17 = 0$

$f'(x) = 3x^2 − 2 = 0$

$x_{k+1} = x_k - \frac{f(x)}{f'(x)} $

$x_{1} = x_0 - \frac{ x_0^3 − 2x_0 − 17 }{3x_0^2 − 2} = 2.3 $

$e = \lvert{\frac{(x_1-x_0)}{x_1}}\rvert=0.1304$

$x_{2} = 2.3 - \frac{ 2.3^3 − 2*2.3 − 17 }{3*2.3^2 − 2} = 2.9801 $

$e = \lvert{\frac{(x_2-x_1)}{x_2}}\rvert= 0.2281$

$x_3 = 2.8378$

$e = 0.0501$

$x_4 = 2.8298$

$e = 0.0028 $

So the approximate root is $x_4 = 2.8298$

Is what I did correct ?

Is there a way to show that $f(x) = x^3 − 2x − 17 = 0$ has only one root without using the graph?

Thanks for any help.

Answer 1

Since $f'(x)=3x^2-2$, $f$ is strictly decreasing on $\left[-\sqrt{\frac23},\sqrt{\frac23}\right]$ and strictly increasing on $\left(-\infty,-\sqrt{\frac23}\,\right]$ and on $\left[\sqrt{\frac23},\infty\right)$. But$$f\left(-\sqrt{\frac23}\right)=\frac43\sqrt{\frac23}-17<0\quad\text{and}\quad f\left(\sqrt{\frac23}\right)=-\frac43\sqrt{\frac23}-17<0.$$Therefore, and since $\lim_{x\to\pm\infty}f(x)=\pm\infty$, $f$ has one and only one zero, which belongs to $\left[\sqrt{\frac23},\infty\right)$.

Answer 2

The elementary way to prove it is to divide x^3-2x-17 by x-2.8298, giving a quadratic expression. In finding the roots of that expression, you'll find a negative number for the familiar b^2-4ac under the radical. You will have found the two complex roots.