Calculus with Applications (10th Edition)

$x=-1.1319$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[-2,-1]$ $x=2.3742$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[2,3]$
We are given $2x^4-2x^3-3x^2-5x-8=0$ which is also $f(x)=2x^4-2x^3-3x^2-5x-8$ $f'(x)=8x^3-6x^2-6x-5$ Check that $f(-1)\lt0$ and $f(-2)\gt0$. Since $f(-1)$ and $f(-2)$ have opposite signs, there is a solution for the equation in the interval $[-2,-1]$ As an initial guess, let $c_1=-2$ $c_2$ can be as follows: $c_2=c_1-\frac{f(c_1)}{f'(c_1)}=-2-\frac{38}{-81}\approx-1.53086$ In the same way, $c_3=-1.2513$ $c_4=-1.45776$ $c_5=-1.1322$ $c_6=-1.1319$ $x=-1.1319$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[-2,-1]$ Check that $f(2)\lt0$ and $f(3)\gt0$. Since $f(2)$ and $f(3)$ have opposite signs, there is a solution for the equation in the interval $[2,3]$ As an initial guess, let $c_1=2$ $c_2$ can be as follows: $c_2=c_1-\frac{f(c_1)}{f'(c_1)}=2-\frac{-14}{3}\approx 2.6087$ In the same way, $c_3=2.414$ $c_4=2.3756$ $c_5=2.3742$ $x=2.3742$ is a reasonably accurate solution of $f(x)=2x^4-2x^3-3x^2-5x-8$ in the interval $[2,3]$