Chapter 5.5, Problem 12E

To determine

To calculate:

The approximate root of $\sqrt[3]{26}$ by use of linearization

Answer to Problem 12E

The approximate value of the root is $2.96\overline{2}$

Explanation of Solution

Given information:

Calculation:

Let $fx=\sqrt[3]{x}$ .The closest perfect cube to 26 is 27, so center the linearization at $x=27$ . The tangent line at (27,3) has a slope

  $\begin{array}{l}f\text{'}\left(27\right)=\frac{1}{3}{\left(27\right)}^{\frac{-2}{3}}\\ f\text{'}\left(27\right)=\frac{1}{3}\times \frac{1}{{\left(\sqrt[3]{27}\right)}^2}\\ f\text{'}\left(27\right)=\frac{1}{3}\times \frac{1}{{\left(3\right)}^2}\\ f\text{'}\left(27\right)=\frac{1}{3}\times \frac{1}{9}\\ f\text{'}\left(27\right)=\frac{1}{27}\end{array}$

So, $\sqrt[3]{26}$

  $\approx L\left(126\right)$

  $\begin{array}{l}=3+\frac{1}{27}\left(26-27\right)\\ =3+\frac{1}{27}\left(-1\right)\\ =3-\frac{1}{27}=\frac{81-1}{27}=\frac{80}{27}=2.96\overline{2}\end{array}$

Therefore, the approximate value of the root is $2.96\overline{2}$