Answer
Left hand $x_2\approx-1.65$
Right hand $x_2\approx1.17$.
Work Step by Step
Step 1. Given $f(x)=x^4+x-3$, we have $f'(x)=4x^3+1$. Starting with $x_0=-1$ for the left hand, we can use Newton’s method $x_{n+1}=x_n-\frac{f(x)}{f'(x)}$ to estimate the solution up to $x_2$ as shown in the table. The result gives $x_2\approx-1.65$ for this case.
Step 2. Repeat the above step for the right hand with $x_0=1$; we can get the result $x_2\approx1.17$ for this case as shown in the table.
