Answer
a. $-1$
b. $0$
c. $1$
d. Results do not converge.
Work Step by Step
a. Given $f(x)=4x^4-4x^2$, we have $f'(x)=16x^3-8x$. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with starting points $x_0=-2$ and $x_0=-0.8$, we can obtain the zero shown in the table as $-1$ (two tables horizontally aligned).
b. Repeating the above procedure with starting points $x_0=-0.5$ and $x_0=0.25$, we can obtain the zero shown in the table as $0$
c. Repeating the above procedure with starting points $x_0=0.8$ and $x_0=2$, we can obtain the zero shown in the table as $1$
d. Repeating the above procedure with starting points $x_0=-\frac{\sqrt {21}}{7}$ and $x_0=\frac{\sqrt {21}}{7}$, we can observe that the $x_i$ values oscillate between positive and negative numbers, indicating that the results do not converge.
Please note that you may observe a result converging to zero, but this is due to the approximations during calculations. It can be proved that the results oscillate if the exact radical forms of $x_0$ are used.
$x_{1}=x_0-\frac{f(x_0)}{f'(x_0)}=x_0-\frac{x_0(x_0^2-1)}{4x_0^2-2}=-\frac{\sqrt {21}}{7}-\frac{(-\frac{\sqrt {21}}{7})((-\frac{\sqrt {21}}{7})^2-1)}{4(-\frac{\sqrt {21}}{7})^2-2}=\frac{\sqrt {21}}{7}=-x_0$ and $x_2=-x_1$, etc.
