Chapter 5.5, Problem 4E

a.

To determine

To find the linearization $L\left(x\right)\text{of }f\left(x\right)\text{ at }x=0$

Answer to Problem 4E

The linearization of the function is $L\left(x\right)=x$

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\ln \left(x+1\right),x=0$

Since $f\left(0\right)=\ln \left(1+0\right)=0$ the point of tangency is $\left(0,0\right)$ .the slope of the tangent line is $f^{\prime }\left(0\right)=\frac{1}{1+0}=1$ thus,

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

b.

To determine

To check the accuracy of approximation $L\left(0+0.1\right)\approx f\left(0+0.1\right)$

Answer to Problem 4E

The approximation error is $\left|0.1-0.0953\right|\approx 0.004=4\times {10}^{-3}$ , the error is less than ${10}^{-2}$

Explanation of Solution

Given information:

The given value of a is 0

To approximate $\ln \left(\left(0+0.1\right)+1\right)$ , we use $x=0+0.1$ :

  $\ln \left(\left(0+0.1\right)+1\right)=f\left(0+0.1\right)\approx L\left(0+0.1\right)=0+0.1=0.1$

The calculator gives $\ln \left(\left(0+0.1\right)+1\right)=0.0953$ , so the approximation error is

  $\left|0.1-0.0953\right|\approx 0.004=4\times {10}^{-3}$ , the error is less than ${10}^{-2}$