Chapter 4.5, Problem 84AYU

In Problems $85-88$, each equation has a solution $r$ in the interval indicated. Use the method of Example $10$ to approximate this solution correct to two decimal places

$x^4+8x^3-x^2+2=0;-1\le r\le 0$

To determine

The approximate solution $r$ correct to two decimal places of the equation $x^4+8x^3-x^2+2=0$ in the interval of $-1\le r\le 0$

Answer to Problem 84AYU

Solution:

The equation $x^4+8x^3-x^2+2=0$ has a solution $r$ in the interval of $-1\le r\le 0$ correct to two decimal places is $-0.61$

Explanation of Solution

Given Information:

$x^4+8x^3-x^2+2=0$; $-1\le r\le 0$.

The solution $r$ of the equation $x^4+8x^3-x^2+2=0$ in the interval $-1\le r\le 0$ is obtained as follows:

Step 1: First, divide the interval $-1\le r\le 0$ into $10$ equal sub-intervals.

The intervals are $\left[-1,-0.9\right]$, $\left[-0.9,-0.8\right]$, $\left[-0.8,-0.7\right]$, $\left[-0.7,-0.6\right]$, $\left[-0.6,-0.5\right]$,

$\left[-0.5,-0.4\right]$, $\left[-0.4,-0.3\right]$, $\left[-0.3,-0.2\right]$, $\left[-0.2,-0.1\right]$ and $\left[-0.1,0\right]$

Step 2: Now, find the value of the function $f\left(x\right)=x^4+8x^3-x^2+2$ at each end point until the sign of $f\left(x\right)$ changes.

$f\left(-0.9\right)={\left(-0.9\right)}^4+8{\left(-0.9\right)}^3-{\left(-0.9\right)}^2+2=-3.9859$

$f\left(-0.8\right)={\left(-0.8\right)}^4+8{\left(-0.8\right)}^3-{\left(-0.8\right)}^2+2=-2.3264$

$f\left(-0.7\right)={\left(-0.7\right)}^4+8{\left(-0.7\right)}^3-{\left(-0.7\right)}^2+2=-0.9939$

$f\left(-0.6\right)={\left(-0.6\right)}^4+8{\left(-0.6\right)}^3-{\left(-0.6\right)}^2+2=0.0416$

By using an Intermediate Value Theorem, the zero lies between $-0.7$ and $-0.6$

Step 3: Now, divide the interval $\left[-0.7,-0.6\right]$ into $10$ equal subintervals.

The intervals are $\left[-0.7,-0.69\right]$, $\left[-0.69,-0.68\right]$, $\left[-0.68,-0.67\right]$, $\left[-0.67,-0.66\right]$, $\left[-0.66,-0.65\right]$, $\left[-0.65,-0.64\right]$, $\left[-0.64,-0.63\right]$, $\left[-0.63,-0.62\right]$, $\left[-0.62,-0.61\right]$ and $\left[-0.61,-0.60\right]$

Step 4: Now, find the value of the function $f\left(x\right)=x^4+8x^3-x^2+2$ at each end point until the sign of $f\left(x\right)$ changes.

$f\left(-0.69\right)={\left(-0.69\right)}^4+8{\left(-0.69\right)}^3-{\left(-0.69\right)}^2+2=-0.8775$

$f\left(-0.68\right)={\left(-0.68\right)}^4+8{\left(-0.68\right)}^3-{\left(-0.68\right)}^2+2=-0.7640$

$f\left(-0.67\right)={\left(-0.67\right)}^4+8{\left(-0.67\right)}^3-{\left(-0.67\right)}^2+2=-0.6535$

$f\left(-0.66\right)={\left(-0.66\right)}^4+8{\left(-0.66\right)}^3-{\left(-0.66\right)}^2+2=-0.5458$

$f\left(-0.65\right)={\left(-0.65\right)}^4+8{\left(-0.65\right)}^3-{\left(-0.65\right)}^2+2=-0.4410$

$f\left(-0.64\right)={\left(-0.64\right)}^4+8{\left(-0.64\right)}^3-{\left(-0.64\right)}^2+2=-0.3390$

$f\left(-0.63\right)={\left(-0.63\right)}^4+8{\left(-0.63\right)}^3-{\left(-0.63\right)}^2+2=-0.2397$

$f\left(-0.62\right)={\left(-0.62\right)}^4+8{\left(-0.62\right)}^3-{\left(-0.62\right)}^2+2=-0.1433$

$f\left(-0.61\right)={\left(-0.61\right)}^4+8{\left(-0.61\right)}^3-{\left(-0.61\right)}^2+2=-0.0495$

$f\left(-0.6\right)={\left(-0.6\right)}^4+8{\left(-0.6\right)}^3-{\left(-0.6\right)}^2+2=0.0416$

Thus, by using an Intermediate Value Theorem, the zero lies between $-0.61$ and $-0.6$

Therefore, the equation $x^4+8x^3-x^2+2=0$ has a solution $r$ in the interval of $-1\le r\le 0$, correct to decimal places is $-0.61$.