Chapter 4, Problem 73RE

To determine

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

  $f\left(x\right)=8x^4-4x^3-2x-1$

Answer to Problem 73RE

The approximate zero value to two decimal places of $f\left(x\right)=8x^4-4x^3-2x-1$ is $0.93$

Explanation of Solution

Given:

  $f\left(x\right)=8x^4-4x^3-2x-1$

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 $\left[0,1\right]$ is used for the function $f\left(x\right)=8x^4-4x^3-2x-1$ as:

  $\begin{array}{l}f\left(0\right)=8{\left(0\right)}^4-4{\left(0\right)}^3-2\left(0\right)-1=-1\\ f\left(1\right)=8{\left(1\right)}^4-4{\left(1\right)}^3-2\left(1\right)-1=1\end{array}$

Here,

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

As $f\left(0\right)$ and $f\left(1\right)$ are opposite sign, it follows the intermediate value theorem that function $f$ has a zero between $0$ and $1$ .

Now, need to divide the given interval $\left[0,1\right]$ into equal $10$ subintervals as:

  $\left[0,0.1\right],\left[0.1,0.2\right],\left[0.2,0.3\right],\left[0.3,0.4\right],\left[0.4,0.5\right],\left[0.5,0.6\right],\left[0.6,0.7\right],\left[0.7,0.8\right],\left[0.8,0.9\right],\left[0.9,1\right]$

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

  $\begin{array}{l}f\left(0.1\right)=8{\left(0.1\right)}^4-4{\left(0.1\right)}^3-2\left(0.1\right)-1=-1.2032\\ f\left(0.2\right)=8{\left(0.2\right)}^4-4{\left(0.2\right)}^3-2\left(0.2\right)-1=-1.4192\\ f\left(0.3\right)=8{\left(0.3\right)}^4-4{\left(0.3\right)}^3-2\left(0.3\right)-1=-1.6432\\ f\left(0.4\right)=8{\left(0.4\right)}^4-4{\left(0.4\right)}^3-2\left(0.4\right)-1=-1.8512\\ f\left(0.5\right)=8{\left(0.5\right)}^4-4{\left(0.5\right)}^3-2\left(0.5\right)-1=-2\\ f\left(0.6\right)=8{\left(0.6\right)}^4-4{\left(0.6\right)}^3-2\left(0.6\right)-1=-2.0272\\ f\left(0.7\right)=8{\left(0.7\right)}^4-4{\left(0.7\right)}^3-2\left(0.7\right)-1=1.8512\\ f\left(0.8\right)=8{\left(0.8\right)}^4-4{\left(0.8\right)}^3-2\left(0.8\right)-1=-1.3712\\ f\left(0.9\right)=8{\left(0.9\right)}^4-4{\left(0.9\right)}^3-2\left(0.9\right)-1=-0.4672\\ f\left(1\right)=8{\left(1\right)}^4-4{\left(1\right)}^3-2\left(1\right)-1=1\end{array}$

So, the zero is observed between $0.9$ and $1$ .

Now, Now, need to divide the given interval $\left[0.9,1\right]$ into equal $10$ subintervals as:

  $\begin{array}{l}\left[0.9,0.91\right],\left[0.91,0.92\right],\left[0.92,0.93\right],\left[0.93,0.94\right],\left[0.94,0.95\right],\left[0.95,0.96\right],\\ \left[0.96,0.97\right],\left[0.97,0.98\right],\left[0.98,0.99\right],\left[0.99,1\right]\end{array}$

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

  $\begin{array}{l}f\left(0.91\right)=8{\left(0.91\right)}^4-4{\left(0.91\right)}^3-2\left(0.91\right)-1=-0.348\\ f\left(0.92\right)=8{\left(0.92\right)}^4-4{\left(0.92\right)}^3-2\left(0.92\right)-1=-0.224\\ f\left(0.93\right)=8{\left(0.93\right)}^4-4{\left(0.93\right)}^3-2\left(0.93\right)-1=-0.093\\ f\left(0.94\right)=8{\left(0.94\right)}^4-4{\left(0.94\right)}^3-2\left(0.94\right)-1=-0.044\end{array}$

So, the zero lies between $0.93$ and $0.94$ .

Therefore, the correct to two decimal places, the zero is $0.93$ .

Conclusion:

Hence, the correct to two decimal places, the zero is $0.93$ .