Chapter 3.9, Problem 9E

To determine

To find: The Linearization of the given function and the approximate the numbers $\sqrt{0.9}$ and $\sqrt{0.99}$ .

Answer to Problem 9E

The Linearization of the given function is $-\frac{1}{2}x+1$ , the approximate value $\sqrt{0.9}$ is $0.95$ , and the approximate value of $\sqrt{0.99}$ is $0.995$ .

Explanation of Solution

Given information:

The given function is,

  $f\left(x\right)=\sqrt{1-x}$

The given value of $a$ is $0$ .

Concept used:

The Linearization of the function is calculated as,

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

The Linearization of the function is calculated as,

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

The approximation of the number $\sqrt{0.9}$ is calculated as,

  $\begin{array}{c}\sqrt{0.9}=\sqrt{1-0.1}\\ =-\frac{1}{2}\left(0.1\right)+1\\ =0.95\end{array}$

The approximation of the number $\sqrt{0.99}$ is calculated as,

  $\begin{array}{c}\sqrt{0.99}=\sqrt{1-0.01}\\ =-\frac{1}{2}\left(0.01\right)+1\\ =0.995\end{array}$

Therefore, The Linearization of the given function is $-\frac{1}{2}x+1$ , the approximate value $\sqrt{0.9}$ is $0.95$ , and the approximate value of $\sqrt{0.99}$ is $0.995$ .

Consider the graph of the function and the tangent line.