Chapter 4.5, Problem 62E

To determine

To choose: The correct option for the percentage error.

Answer to Problem 62E

The correct option is (A).

Explanation of Solution

Given information: $y=\sqrt[3]{x}$at $x=64$ which are used to approximate $\sqrt[3]{66}$ .

Concept used: According to the Linear Approximation Formula, if $f$ is differentiable at $x=a$ , then $L\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$ .

Calculation:

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

Differentiate the given function with respect to $x$ .

  $f^{\prime }\left(x\right)=\frac{1}{3{\left(x\right)}^{\frac{2}{3}}}$

Now, substitute 64 for $x$ into the obtained derivative and simplify it.

  $\begin{array}{c}{f}^{\prime }\left(64\right)=\frac{1}{3{\left(64\right)}^{\frac{2}{3}}}\\ =\frac{1}{48}\end{array}$

Substitute 64 for $a$ into the Linear Approximation Formula, $L\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$ , and simplify it.

  $L\left(x\right)=f\left(64\right)+f^{\prime }\left(64\right)\left(x-64\right)$

Now,

  $\begin{array}{c}L\left(x\right)={\left(64\right)}^{\frac{1}{3}}+\frac{1}{48}\left(x-64\right)\\ =\frac{8}{3}+\frac{x}{48}\end{array}$

Substitute 66 for $x$ into the obtained equation to approximate the value of $\sqrt[3]{66}$ .

  $\begin{array}{c}L\left(66\right)=\frac{8}{3}+\frac{66}{48}\\ \approx 4.0417\end{array}$

Now, determine the required percentage error as shown below.

  $\begin{array}{c}\%\text{ error}=\left[\frac{\sqrt[3]{66}-4.0417}{\sqrt[3]{66}}\right]\times 100\%\\ =0.01\%\end{array}$

This implies that option (A) is correct.