Chapter 4.5, Problem 9E

a.

To determine

To find: To find Linearization of $f(x)={(1-x)}^6$ using the linearization $L(x)=1+kx$.

Answer to Problem 9E

The Linearization of $f(x)={(1-x)}^6$ is $L(x)=1-6x$ .

Explanation of Solution

Given: The function $f(x)={(1-x)}^6$ using the linearization $L(x)=1+kx$.Formula used: $L(x)=f(a)+f\text{'}(a)(x-a)$

Calculation${(1-x)}^6={(1+(-x))}^6$

By using the linearization $L(x)=1+kx$

  $\begin{array}{l}L(x)=1+6(-x)\\ L(x)=1-6x\end{array}$

Conclusion:

The approximation is near to True value.

b.

To determine

To find: To find Linearisation of $f(x)=\frac{2}{(1-x)}$ using the linearisation $L(x)=1+kx$.

Answer to Problem 9E

Linearisation of $f(x)=\frac{2}{(1-x)}$ using the linearisation $L(x)=1+kx$ is $L(x)=2+x$ .

Explanation of Solution

Given: Linearisation of $f(x)=\frac{2}{(1-x)}$ using the linearisation $L(x)=1+kx$.

Formula used: $L(x)=f(a)+f\text{'}(a)(x-a)$

Calculation$\frac{2}{(1-x)}=2{(1-x)}^{-0.5}$

By using the linearization $L(x)=1+kx$

  $\begin{array}{l}L(x)=2(1+(-x)(-0.5))\\ L(x)=2(1+0.5x)\\ L(x)=2+x\end{array}$

Conclusion:

Linearisation of $f(x)=\frac{2}{(1-x)}$ using the linearisation $L(x)=1+kx$ is $L(x)=2+x$ .

c.

To determine

To find: To find Linearisation of $f(x)=\frac{1}{\sqrt{1+x}}$ using the linearisation $L(x)=1+kx$.

Answer to Problem 9E

Linearisation of $f(x)=\frac{1}{\sqrt{1+x}}$ using the linearisation $L(x)=1+kx$ is $L(x)=1-\frac{1}{2}x$ .

Explanation of Solution

Given: The function $f(x)=\frac{1}{\sqrt{1+x}}$ using the linearisation $L(x)=1+kx$.Formula used: $L(x)=f(a)+f\text{'}(a)(x-a)$

Calculation$f(x)=1{(1+x)}^{-\frac{1}{2}}$

By using the linearization $L(x)=1+kx$

  $\begin{array}{l}L(x)=1-\frac{1}{2}x\\ L(x)=\frac{2-x}{2}\end{array}$

Conclusion:

Linearisation of $f(x)=\frac{1}{\sqrt{1+x}}$ using the linearisation $L(x)=1+kx$ is $L(x)=1-\frac{1}{2}x$ .