Chapter 4, Problem 74RE

To determine

To calculate: Given polynomial function.

Answer to Problem 74RE

Correct to two decimal places the zero is $-0.26$ .

Explanation of Solution

Given information:

  $f\left(x\right)=3x^4+4x^3-8x-2$

Calculation:

As it is given that,

  $f\left(x\right)=3x^4+4x^3-8x-2$

Now evaluate $f$ at $0$ and $1$ ,

  $\begin{array}{l}f\left(0\right)=2{\left(0\right)}^3-{\left(0\right)}^2-3\\ =-3\\ f\left(-1\right)=3{\left(-1\right)}^4+4{\left(-1\right)}^3-8\left(-1\right)-2\\ =5\end{array}$

Since $f\left(0\right)<0$ and $f\left(1\right)>0$ and $f\left(0\right)$ and $f\left(1\right)$ are of opposite signs. It follows the intermediate value theorem that $f$ has a zero between $0$ and $1$ .

Devide the interval $\left[0,-1\right]$ into $10$ equal sub intervals $\left[0,-0.1\right],\left[-0.1,-0.2\right],\left[-0.2,-0.3\right],\left[-0.3,-0.4\right],\left[-0.4,-0.5\right],\left[-0.5,-0.6\right],\left[-0.6,-0.7\right],\left[-0.7,-0.8\right],\left[-0.8,-0.9\right],\left[-0.9,1\right]$ now, find out the value of $f$ at each endpoint until the intermediate value theorem applies.

  $\begin{array}{l}f\left(-0.1\right)=3{\left(-0.1\right)}^4+4{\left(-0.1\right)}^3-8\left(-0.1\right)-2=-1.2037\\ f\left(-0.2\right)=3{\left(-0.2\right)}^4+4{\left(-0.2\right)}^3-8\left(-0.2\right)-2=-0.4272\\ f\left(-0.3\right)=3{\left(-0.3\right)}^4+4{\left(-0.3\right)}^3-8\left(-0.3\right)-2=0.3163\\ f\left(-0.4\right)=3{\left(-0.4\right)}^4+4{\left(-0.4\right)}^3-8\left(-0.4\right)-2=1.0208\\ f\left(-0.5\right)=3{\left(-0.5\right)}^4+4{\left(-0.5\right)}^3-8\left(-0.5\right)-2=0.1875\\ f\left(-0.6\right)=3{\left(-0.6\right)}^4+4{\left(-0.6\right)}^3-8\left(-0.6\right)-2=2.3248\\ f\left(-0.7\right)=3{\left(-0.7\right)}^4+4{\left(-0.7\right)}^3-8\left(-0.7\right)-2=4.9483\\ f\left(-0.8\right)=3{\left(-0.8\right)}^4+4{\left(-0.8\right)}^3-8\left(-0.8\right)-2=3.5808\\ f\left(-0.9\right)=3{\left(-0.9\right)}^4+4{\left(-0.9\right)}^3-8\left(-0.9\right)-2=4.2523\\ f\left(1\right)=3{\left(1\right)}^4+4{\left(1\right)}^3-8\left(1\right)-2=-3\end{array}$

The zero is between $\left(-0.1\right)$ to $\left(1\right)$ .

Divide the interval $\left[-0.21,-0.29\right]$ into $10$ equal sub intervals now, find out the value of $f$ at each endpoint until the intermediate value theorem applies.

  $\begin{array}{l}f\left(-0.21\right)=3{\left(-0.21\right)}^4+4{\left(-0.21\right)}^3-8\left(-0.21\right)-2=-0.3512\\ f\left(-0.22\right)=3{\left(-0.22\right)}^4+4{\left(-0.22\right)}^3-8\left(-0.22\right)-2=-0.27556\\ f\left(-0.23\right)=3{\left(-0.23\right)}^4+4{\left(-0.23\right)}^3-8\left(-0.23\right)-2=-0.20027\\ f\left(-0.24\right)=3{\left(-0.24\right)}^4+4{\left(-0.24\right)}^3-8\left(-0.24\right)-2=-1.030046\\ f\left(-0.25\right)=3{\left(-0.25\right)}^4+4{\left(-0.25\right)}^3-8\left(-0.25\right)-2=-0.05078\\ f\left(-0.26\right)=3{\left(-0.26\right)}^4+4{\left(-0.26\right)}^3-8\left(-0.26\right)-2=0.02340\\ f\left(-0.27\right)=3{\left(-0.27\right)}^4+4{\left(-0.27\right)}^3-8\left(-0.27\right)-2=0.097211\\ f\left(-0.28\right)=3{\left(-0.28\right)}^4+4{\left(-0.28\right)}^3-8\left(-0.28\right)-2=0.17063\\ f\left(-0.29\right)=3{\left(-0.29\right)}^4+4{\left(-0.29\right)}^3-8\left(-0.29\right)-2=0.24366\end{array}$

We can conclude that zero is lies between $-0.26$ to $-0.29$ .

Correct to two decimal places the zero is $-0.26$ .