Chapter 5.5, Problem 66E

a.

To determine

To calculate the coefficients $b_0,b_1,\text{ and }{b}_2$

Answer to Problem 66E

  $f\left(a\right)=b_0,f^{\prime }\left(a\right)=b_1,\text{ and }{b}_2=\frac{f^{″}\left(x\right)}{2}$

Explanation of Solution

Given information:

Quadratic approximation to $f\left(x\right)\text{ at }x=a$ is

  $Q\left(x\right)=b_0+b_1\left(x-a\right)+b_2{\left(x-a\right)}^2$

With the properties

  $\begin{array}{l}\text{i}\text{. }Q\left(a\right)=f\left(a\right),\\ \text{ii}\text{. }{Q}^{\prime }\left(a\right)=f^{\prime }\left(a\right),\\ \text{iii}\text{. }{Q}^{″}\left(a\right)=f^{″}\left(a\right).\end{array}$

Calculation :

Write the function and get the values at $x=a$

  $\begin{array}{l}Q\left(x\right)=b_0+b_1\left(x-a\right)+b_2{\left(x-a\right)}^2\\ Q\left(a\right)=b_0+b_1\left(a-a\right)+b_2{\left(a-a\right)}^2\\ Q\left(a\right)=b_0\\ f\left(a\right)=b_0\left[Q\left(a\right)=f\left(a\right)\right]\\ Q^{\prime }\left(x\right)=b_1+2b_2\left(x-a\right)\\ Q^{\prime }\left(a\right)=b_1+2b_2\left(a-a\right)\\ Q^{\prime }\left(a\right)=b_1\\ f^{\prime }\left(a\right)=b_1\left[Q^{\prime }\left(a\right)=f^{\prime }\left(a\right)\right]\\ Q^{″}\left(x\right)=2b_2\\ f^{″}\left(x\right)=2b_2\left[Q^{″}\left(a\right)=f^{″}\left(a\right)\right]\\ b_2=\frac{f^{″}\left(x\right)}{2}\end{array}$

Therefore,

  $f\left(a\right)=b_0,f^{\prime }\left(a\right)=b_1,\text{ and }{b}_2=\frac{f^{″}\left(x\right)}{2}$

b.

To determine

To find the quadratic approximation to $f\left(x\right)=\frac{1}{\left(1-x\right)}\text{ at }\left(x=0\right)$

Answer to Problem 66E

The quadratic approximation is $Q\left(x\right)=1+x+x^2$

Explanation of Solution

Given information:

The given statement is that find the quadratic approximation to $f\left(x\right)=\frac{1}{\left(1-x\right)}\text{ at }\left(x=0\right)$

Calculation :

  $\begin{array}{l}f\left(0\right)=1\\ f^{\prime }\left(x\right)={\left(1-x\right)}^{-2}\\ f^{\prime }\left(0\right)={\left(1-0\right)}^{-2}\\ f^{\prime }\left(0\right)=1\\ f^{″}\left(x\right)=2{\left(1-x\right)}^{-3}\\ f^{″}\left(0\right)=2{\left(1-0\right)}^{-3}\\ f^{″}\left(0\right)=2\end{array}$

Substitute the values in quadratic approximation equation

  $\begin{array}{l}Q\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+\frac{f^{″}\left(0\right)}{2}{x}^2\\ Q\left(x\right)=1+1\cdot x+\frac{2x^2}{2}\\ Q\left(x\right)=1+x+x^2\end{array}$

Therefore,

The quadratic approximation is $Q\left(x\right)=1+x+x^2$

c.

To determine

To graph: $f\left(x\right)=\frac{1}{\left(1-x\right)}$ and its quadratic approximation $\text{at }\left(x=0\right)$ .

Explanation of Solution

Given information:

ZOOM IN on the two graphs at point $\left(0,1\right)$ and comment what you see.

Graph:

The graph of $f\left(x\right)=\frac{1}{\left(1-x\right)}$ and its quadratic approximation $\text{at }\left(x=0\right)$ .

  

The image of ZOOM IN at point $\left(0,1\right)$

  

Interpretation:

At the point $\left(0,1\right)$ the two graphs are coincide.

d.

To determine

To find the quadratic approximation to $f\left(x\right)=\frac{1}{x}\text{ at }\left(x=1\right)$ and graph g and its quadratic approximation together.

Answer to Problem 66E

The quadratic approximation is $Q\left(x\right)=1-\left(x-1\right)+{\left(x-1\right)}^2$

The function and its approximation behave similar around $x=1$ , although their global behaviour is different.

Explanation of Solution

Given information:

The given statement is that find the quadratic approximation to $f\left(x\right)=\frac{1}{x}\text{ at }\left(x=1\right)$ and graph g and its quadratic approximation together.

Calculation :

  $\begin{array}{l}g\left(1\right)=1\\ g^{\prime }\left(x\right)=-{\left(x\right)}^{-2}\\ g^{\prime }\left(1\right)=-1\\ {g^{\prime }}^{\prime }\left(x\right)=2x^{-3}\\ {g^{\prime }}^{\prime }\left(1\right)=2\end{array}$

Substitute the values in quadratic approximation equation

  $\begin{array}{l}Q\left(x\right)=g\left(1\right)+g^{\prime }\left(1\right)\left(x-1\right)+\frac{{g^{\prime }}^{\prime }\left(1\right)}{2}{\left(x-1\right)}^2\\ Q\left(x\right)=1-\left(x-1\right)+{\left(x-1\right)}^2\\ \end{array}$

Therefore,

The quadratic approximation is $Q\left(x\right)=1-\left(x-1\right)+{\left(x-1\right)}^2$

The graph of g and its quadratic approximation together.

  

The function and its approximation behave similar around $x=1$ , although their global behaviour is different.

e.

To determine

To find the quadratic approximation to $h\left(x\right)=\sqrt{1+x}\text{at }\left(x=0\right)$ and graph h and its quadratic approximation together.

Answer to Problem 66E

The quadratic approximation is $Q\left(x\right)=1+\frac{x}{2}-\frac{x^2}{8}$

The function and its approximation behave similar around $x=0$ , although their global behaviour is different.

Explanation of Solution

Given information:

The given statement is that find the quadratic approximation to $h\left(x\right)=\sqrt{1+x}\text{at }\left(x=0\right)$ and graph h and its quadratic approximation together.

Calculation :

  $\begin{array}{l}h\left(0\right)=1\\ h^{\prime }\left(x\right)=\frac{1}{2{\left(x+1\right)}^{\frac{1}{2}}}\\ h^{\prime }\left(0\right)=\frac{1}{2}\\ {h^{\prime }}^{\prime }\left(x\right)=\frac{1}{4{\left(x+1\right)}^{\frac{3}{2}}}\\ {h^{\prime }}^{\prime }\left(0\right)=-\frac{1}{2}\end{array}$

Substitute the values in quadratic approximation equation

  $\begin{array}{l}Q\left(x\right)=h\left(0\right)+h^{\prime }\left(0\right)\left(x-0\right)+\frac{h^{″}\left(1\right)}{2}{\left(x-0\right)}^2\\ Q\left(x\right)=1+\frac{x}{2}-\frac{x^2}{8}\end{array}$

Therefore,

The quadratic approximation is $Q\left(x\right)=1+\frac{x}{2}-\frac{x^2}{8}$

The graph of h and its quadratic approximation together.

  

The function and its approximation behave similar around $x=0$ , although their global behaviour is different.

f.

To determine

To write the linearization of f , g , and h at the respective points.

Answer to Problem 66E

Linearization of f , g , and h are $L\left(x\right)=1+x$ , $L\left(x\right)=-x+2$ , and $L\left(x\right)=1+\frac{x}{2}$ respectively.

Explanation of Solution

Given information:

Functions are given in parts (b), (d), and (e)

Formula used:

Linearization.

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

Calculation :

For f

  $\begin{array}{l}f\left(0\right)=1\\ f^{\prime }\left(x\right)={\left(1-x\right)}^{-2}\\ f^{\prime }\left(0\right)={\left(1-0\right)}^{-2}\\ f^{\prime }\left(0\right)=1\end{array}$

Linearization.

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

For g

  $\begin{array}{l}g\left(1\right)=1\\ g^{\prime }\left(x\right)=-{\left(x\right)}^{-2}\\ g^{\prime }\left(1\right)=-1\end{array}$

Linearization.

  $\begin{array}{l}L\left(x\right)=g\left(a\right)+g^{\prime }\left(a\right)\left(x-a\right)\\ L\left(x\right)=g\left(1\right)+g^{\prime }\left(1\right)\left(x-1\right)\\ L\left(x\right)=1-1\left(x-1\right)\\ L\left(x\right)=-x+2\end{array}$

For h

  $\begin{array}{l}h\left(0\right)=1\\ h^{\prime }\left(x\right)=\frac{1}{2{\left(x+1\right)}^{\frac{1}{2}}}\\ h^{\prime }\left(0\right)=\frac{1}{2}\end{array}$

Linearization.

  $\begin{array}{l}L\left(x\right)=h\left(a\right)+h^{\prime }\left(a\right)\left(x-a\right)\\ L\left(x\right)=h\left(0\right)+h^{\prime }\left(0\right)\left(x-0\right)\\ L\left(x\right)=1+\frac{1}{2}\left(x\right)\\ L\left(x\right)=1+\frac{x}{2}\end{array}$

Therefore,

Linearization of f , g , and h are $L\left(x\right)=1+x$ , $L\left(x\right)=-x+2$ , and $L\left(x\right)=1+\frac{x}{2}$ respectively.