Chapter 4.5, Problem 86AYU

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

$3x^3-10x+9=0;-3\le r\le -2$

To determine

The approximate solution $r$ correct to two decimal places of the equation $3x^3-10x+9=0$ in the interval of $-3\le r\le -2$

Answer to Problem 86AYU

Solution:

The equation $3x^3-10x+9=0$ has a solution $r$, in the interval of $-3\le r\le -2$ correct to two decimal places is $-2.18$.

Explanation of Solution

Given Information:

$3x^3-10x+9=0$; $-3\le r\le -2$

The solution $r$ of the equation $3x^3-10x+9=0$ in the interval of $-3\le r\le -2$ is obtained as follows:

Step 1: First, divide the interval $-3\le r\le -2$ into $10$ equal subintervals.

The intervals are $\left[-3,-2.9\right]$, $\left[-2.9,-2.8\right]$, $\left[-2.8,-2.7\right]$, $\left[-2.7,-2.6\right]$, $\left[-2.6,-2.5\right]$,

$\left[-2.5,-2.4\right]$, $\left[-2.4,-2.3\right]$, $\left[-2.3,-2.2\right]$, $\left[-2.2,-2.1\right]$ and $\left[-2.1,-2\right]$

Step 2: Now, find the value of the function $f\left(x\right)=3x^3-10x+9$ at each end point until the sign of $f\left(x\right)$ changes.

$f\left(-2.9\right)=3{\left(-2.9\right)}^3-10\left(2.9\right)+9=-35.167$

$f\left(-2.8\right)=3{\left(-2.8\right)}^3-10\left(2.8\right)+9=-28.856$

$f\left(-2.7\right)=3{\left(-2.7\right)}^3-10\left(2.7\right)+9=-23.049$

$f\left(-2.6\right)=3{\left(-2.6\right)}^3-10\left(2.6\right)+9=-17.728$

$f\left(-2.5\right)=3{\left(-2.5\right)}^3-10\left(2.5\right)+9=-12.875$

$f\left(-2.4\right)=3{\left(-2.4\right)}^3-10\left(2.4\right)+9=-8.472$

$f\left(-2.3\right)=3{\left(-2.3\right)}^3-10\left(2.3\right)+9=-4.501$

$f\left(-2.2\right)=3{\left(-2.2\right)}^3-10\left(2.2\right)+9=-0.944$

$f\left(-2.1\right)=3{\left(-2.1\right)}^3-10\left(2.1\right)+9=2.217$

By using an Intermediate Value Theorem, the zero lies between $-2.2$ and $-2.1$.

Step 3: Now, divide the interval $\left[-2.2,-2.1\right]$ into $10$ equal subintervals.

The intervals are $\left[-2.2,-2.19\right]$, $\left[-2.19,-2.18\right]$, $\left[-2.18,-2.17\right]$, $\left[-2.17,-2.16\right]$, $\left[-2.16,-2.15\right]$,

$\left[-2.15,-2.14\right]$, $\left[-2.14,-2.13\right]$, $\left[-2.13,-2.12\right]$, $\left[-2.12,-2.11\right]$ and $\left[-2.11,-2.1\right]$

Step 4: Now, find the value of the function $f\left(x\right)=3x^3-10x+9$ at each end point until the sign of $f\left(x\right)$ changes.

$f\left(-2.19\right)=3{\left(-2.19\right)}^3-10\left(2.19\right)+9=-0.6104$

$f\left(-2.18\right)=3{\left(-2.18\right)}^3-10\left(2.18\right)+9=-0.2807$

$f\left(-2.17\right)=3{\left(-2.17\right)}^3-10\left(2.17\right)+9=0.0451$

Thus, by using an Intermediate Value Theorem, the zero lies between $-2.18$ and $-2.17$

Therefore, the equation $3x^3-10x+9=0$ has a solution $r$ in the interval of $-3\le r\le -2$ correct to two decimal places is $-2.18$