Answer
$-1.306562965,-0.5411961 ,0.5411961,1.306562965$
Work Step by Step
Step 1. Given $f(x)=2x^4-4x^2+1$, we have $f'(x)=8x^3-8x$.
Step 2. Graph the function and estimate the starting points for using the Newton’s method as $x_0=-1.3, -0.5, 0.5, 1.3$.
Step 3. Use Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with the above starting points; we can obtain the four solutions shown in the table as $-1.306562965,-0.5411961 ,0.5411961,1.306562965$
