Chapter 5.6, Problem 4QQ

a.

To determine

To show: the computation that leads to your conclusion.

Answer to Problem 4QQ

The value of $\sqrt{26}$ is 5.1.

Explanation of Solution

Given information:

Approximate $\sqrt{26}$ by using the linearization of $y=\sqrt{x}$ at the point $\left(25,5\right)$ .

Calculation :

We have to analyze to decide where to centre the approximation.

Let, $f\left(y\right)=\sqrt{x}$

The closest perfect square to 26 is 25, so we centre the linearization at $x=25$ . The tangent line at $\left(25,5\right)$ has slope.

Therefore,

  $f\left(25\right)={\left(25\right)}^{\frac{1}{2}}$

Differentiate,

  $\begin{array}{l}f\text{'}\left(25\right)=\frac{1}{2}{\left(25\right)}^{\frac{1}{2}-1}\\ f\text{'}\left(25\right)=\frac{1}{2}{\left(25\right)}^{-\frac{1}{2}}\\ f\text{'}\left(25\right)=\frac{1}{2\sqrt{25}}\\ f\text{'}\left(25\right)=\frac{1}{10}\end{array}$

So, find value of derivative at $x=25$ .

  $\begin{array}{l}y-5=\frac{1}{10}\left(x-25\right)\\ \sqrt{26-5}=\frac{1}{10}\left(26-25\right)\left[\begin{array}{l}\text{When, }y=\sqrt{26}\\ \text{then, }x=26\end{array}\right]\\ \sqrt{26-5}=\frac{1}{10}\left(26-25\right)+5\\ \sqrt{26}=\frac{1}{10}+5\\ \sqrt{26}=0.1+5\\ \sqrt{26}=5.1\end{array}$

Hence, the value of $\sqrt{26}$ is 5.1.

b.

To determine

To show: the computation that leads to your conclusion.

Answer to Problem 4QQ

The computation is 5.1.

Explanation of Solution

Given information:

Approximate $\sqrt{26}$ by using a first guess of 5 and one iteration of Newton’s method to approximate the zero of $x^2-26$ .

Calculation :

We have to analyze to decide where to centre the approximation.

Therefore,

  $\begin{array}{l}f\left(x\right)=x^2-26\\ f\left(5\right)=5^2-26\left[\text{Since }x=5\right]\\ f\left(5\right)=25-26\\ f\left(5\right)=-1\end{array}$

Differentiate $f\left(x\right)$

  $\begin{array}{l}f\text{'}\left(x\right)=2x\\ f\text{'}\left(5\right)=2\left(5\right)\left[\text{Since }x=5\right]\\ f\text{'}\left(5\right)=10\end{array}$

The first iteration using Newton’s method is $x_1=x_0-\frac{f\left(x_0\right)}{f\text{'}\left(x_0\right)}$ .

Since the first guess is $x_0=5$ ,

Substitute $x_0=5$ , $f\left(5\right)=-1$ and $f\text{'}\left(5\right)=10$ .

  $\begin{array}{l}{x}_1=x_0-\frac{f\left(x_0\right)}{f\text{'}\left(x_0\right)}\\ x_1=5-\frac{f\left(5\right)}{f\text{'}\left(5\right)}\\ x_1=5-\frac{\left(-1\right)}{\left(10\right)}\\ x_1=5-\left(-0.1\right)\\ x_1=5+0.1\\ x_1=5.1\end{array}$

Hence, the computation is 5.1.

c.

To determine

To show: the computation that leads to your conclusion.

Answer to Problem 4QQ

The value of $\sqrt{26}$ is 2.963.

Explanation of Solution

Given information:

Approximate $\sqrt[3]{26}$ by using an appropriate linearization.

Calculation :

We have to analyze to decide where to centre the approximation.

Let, $f\left(x\right)=\sqrt[3]{x}$

The closest perfect cube to 26 is 27, so we centre the linearization at $x=27$ . The tangent line at $\left(27,3\right)$ has slope.

Therefore,

  $f\left(27\right)=\sqrt[3]{27}={\left(27\right)}^{\frac{1}{3}}$

Differentiate,

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

So, find value of derivative at $x=27$ .

  $\begin{array}{l}f\left(x\right)-3=\frac{1}{27}\left(x-27\right)\\ \sqrt{26-3}=\frac{1}{27}\left(26-27\right)\left[\begin{array}{l}\text{When, }y=\sqrt{26}\\ \text{then, }x=26\end{array}\right]\\ \sqrt{26-3}=\frac{1}{27}\left(26-27\right)+3\\ \sqrt{26}=-\frac{1}{27}+3\\ \sqrt{26}=-0.037+3\\ \sqrt{26}=0.963\end{array}$

Hence, the value of $\sqrt{26}$ is 2.963.