Chapter 3, Problem 75RE

(a)

To determine

To find: The linearization of the given function and approximate the value of $\sqrt[3]{1.03}$ .

Explanation of Solution

Given: the function is $f\left(x\right)=\sqrt[3]{1+3x}$ .

Differentiate the function with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{1}{3}{\left(1+3x\right)}^{-2/3}\times 3\\ ={\left(1+3x\right)}^{-2/3}\\ f^{\prime }\left(0\right)=1\end{array}$

The linear approximation at $a=0$ is,

  $\begin{array}{c}L\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)\left(x-0\right)\\ =1+\left(x-0\right)\\ =1+x\end{array}$

Thus, ${\left(1+3x\right)}^{1/3}=1+x$

Now,

  $1.03=1+\left(3\times 0.1\right)$

Thus,

  $\begin{array}{c}{\left(1+\left(3\times 0.1\right)\right)}^{1/3}=1+.01\\ =1.01\end{array}$

Therefore, $\sqrt[3]{1.03}=1.01$

(b)

To determine

To find: The value of $x$ for which the linear approximation given in part (a) is accurate to within $0.1$ .

Explanation of Solution

Given: the function is $f\left(x\right)=\sqrt[3]{1+3x}$ .

The accuracy of $0.1$ is required.

So choose $x$ such that,

  ${\left(1+3x\right)}^{1/3}-0.1<1+x<{\left(1+3x\right)}^{1/3}+0.1$

The graph of $y={\left(1+3x\right)}^{1/3}-0.1$ , $y=1+x$ and $y={\left(1+3x\right)}^{1/3}+0.1$ is shown below.

  

Figure (1)

It is visible from the graph the linear approximation intersects the upper curve at $x\approx -0.232,0.398$ .

Therefore, for the accuracy it is required to have $-0.23<x<0.40$ .