Chapter 4, Problem 72RE

To determine

To approximate: the zero correct to two decimal places.

Answer to Problem 72RE

The zero is between 1.33 and 1.34 and so correct to two decimal places, the zero is $1.34$ .

Explanation of Solution

Given:

  $f(x)=2x^3-x^2-3$

Calculation:

Given that the polynomial $2x^3-x^2-3$ has exactly one positive real root. Let the zeros is between 1 and 2. Divide the interval [1,2] into 10 equal subintervals:

  $\begin{array}{l}\left[1,1.1\right] \left[\begin{array}{l}1.1,1.2\hfill \end{array}\right]\left[\begin{array}{l}1.2,1.3\hfill \end{array}\right]\left[\begin{array}{l}1.3,1.4\hfill \end{array}\right]\left[\begin{array}{l}1.4,1.5\hfill \end{array}\right]\\ \left[1.5,1.6\right] \left[1.6,1.7\right] [1.7,1.8][1.8,1.9] \left[1.9,2\right]\end{array}$

Now find the value of $f$ at each endpoint until the intermediate the Intermediate value theorem applies.

  $\begin{array}{l}f(1)=-2f(1.1)=-1.548f(1.2)=-0.984\\ f(1.3)=-0.296f(1.4)=0.528\end{array}$

We conclude that the zero is between 1.3 and 1.4.

Now again divide $\left[1.3,1.4\right]$ into 10 equal subintervals and proceed to evaluate $f$ at each endpoints.

  $\begin{array}{l}f(1.30)=-0.296f(1.31)=-0.2199f(1.32)=-0.1425\\ f(1.33)=-0.0636f(1.34)=0.0166\end{array}$

Therefore,the zero is between 1.33 and 1.34 and so correct to two decimal places, the zero is $1.34$

Conclusion:

Therefore, the zero is between 1.33 and 1.34 and so correct to two decimal places, the zero is $1.34$ .