Chapter 4, Problem 71RE

To determine

The correct and approximate zero value to two decimal places of the following polynomial function is to be estimated:

  $f\left(x\right)=x^3-x-2$

Answer to Problem 71RE

The approximate zero value to two decimal places of $f\left(x\right)=x^3-x-2$ is $1.52$

Explanation of Solution

Given:

  $f\left(x\right)=x^3-x-2$

Concept Used:

Intermediate value theorem:

The polynomial function $f$ has interval $\left[a,b\right]$ . There is at least one real zero for the given polynomial function if $a<b$ and $f\left(a\right)$ and $f\left(b\right)$ are of opposite sign.

Calculation:

The interval $\left[1,2\right]$ is used for the function $f\left(x\right)=x^3-x-2$ as:

  $\begin{array}{l}f\left(1\right)={\left(1\right)}^3-\left(1\right)-2=-2\\ f\left(2\right)={\left(2\right)}^3-\left(2\right)-2=4\end{array}$

Here,

  $\begin{array}{l}f\left(1\right)=-2,f\left(1\right)<0\\ f\left(2\right)=4,f\left(2\right)>0\end{array}$

As $f\left(1\right)$ and $f\left(2\right)$ are opposite sign, it follows the intermediate value theorem that function $f$ has a zero between $1$ and $2$ .

According to the intermediate value theorem that function $f$ has a zero between $1$ and $2$ , the equation will have at least one root. The given function is continuous at each point in the given interval.

Now, need to divide the given interval $\left[1,2\right]$ into equal $10$ subintervals as:

  $\left[1,1.1\right],\left[1.1,1.2\right],\left[1.2,1.3\right],\left[1.3,1.4\right],\left[1.4,1.5\right],\left[1.5,1.6\right],\left[1.6,1.7\right],\left[1.7,1.8\right],\left[1.8,1.9\right],\left[1.9,2\right]$

Using intermediate value theorem, the value of $f$ at each endpoint is calculated as:

  $\begin{array}{l}f\left(1\right)={\left(1\right)}^3-\left(1\right)-2=-2\\ f\left(1.1\right)={\left(1.1\right)}^3-\left(1.1\right)-2=-1.769\\ f\left(1.2\right)={\left(1.2\right)}^3-\left(1.2\right)-2=-1.472\\ f\left(1.3\right)={\left(1.3\right)}^3-\left(1.3\right)-2=-1.103\\ f\left(1.4\right)={\left(1.4\right)}^3-\left(1.4\right)-2=-0.656\\ f\left(1.5\right)={\left(1.5\right)}^3-\left(1.5\right)-2=-0.125\\ f\left(1.6\right)={\left(1.6\right)}^3-\left(1.6\right)-2=0.496\end{array}$

So, the zero is observed between $1.5$ and $1.6$ .

Now, Now, need to divide the given interval $\left[1.5,1.6\right]$ into equal $10$ subintervals as:

  $\begin{array}{l}\left[1.5,1.51\right],\left[1.51,1.52\right],\left[1.52,1.53\right],\left[1.53,1.54\right],\left[1.54,1.55\right],\left[1.55,1.56\right],\left[1.56,1.57\right],\left[1.57,1.58\right],\\ \left[1.58,1.59\right],\left[1.59,1.6\right]\end{array}$

Using intermediate value theorem, the value of $f$ at each endpoint is calculated as:

  $\begin{array}{l}f\left(1.51\right)={\left(1.51\right)}^3-\left(1.51\right)-2=-0.067\\ f\left(1.52\right)={\left(1.52\right)}^3-\left(1.52\right)-2=-0.0082\\ f\left(1.53\right)={\left(1.53\right)}^3-\left(1.53\right)-2=0.0516\end{array}$

So, the zero lies between $1.52$ and $1.53$ .

Therefore, the correct to two decimal places, the zero is $1.52$ .

Conclusion:

Hence, the correct to two decimal places, the zero is $1.52$ .