Chapter 4.5, Problem 66E

a.

To determine

Determine the coefficients $b_0,b_1\text{ and }{b}_2.$

Answer to Problem 66E

Coefficients: -

  $\begin{array}{l}{b}_0=f(a),\\ b_1=f\text{'}(a),\\ b_2=\frac{f\text{'}\text{'}(a)}{2}.\end{array}$

Quadratic approximation at $x=a$ :

  $Q(x)=f(a)+f\text{'}(a)\left(x-a\right)+\frac{1}{2}f\text{'}\text{'}(a){\left(x-a\right)}^2$ .

Explanation of Solution

Given:

Quadratic approximation: - $Q(x)=b_0+b_1\left(x-a\right)+b_2{\left(x-a\right)}^2$ to $f(x)$ at $x=a$ .

Properties: -

1. $Q(a)=f(a)$,
2. $Q\text{'}(a)=f\text{'}(a)$,
3. $Q\text{'}\text{'}(a)=f\text{'}\text{'}(a)$ .

Calculation:

Use property (i) $Q(a)=f(a)$ ,

  $\begin{array}{c}Q(a)=f(a)\\ \text{Put }x=a,\\ b_0+b_1\left(a-a\right)+b_2{\left(a-a\right)}^2=f(a)\\ b_0+0+0=f(a)\\ b_0=f(a)\end{array}$

Use property (ii) $Q\text{'}(a)=f\text{'}(a)$ ,

But, given that

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

First derivative of $Q(x)$:

  $\begin{array}{l}Q\text{'}(x)=0+b_1+2b_2\left(x-a\right)\\ Q\text{'}(x)=b_1+2b_2\left(x-a\right)\end{array}$

So,

  $\begin{array}{c}Q\text{'}(a)=f\text{'}(a)\\ \text{Put }x=a,\\ b_1+2b_2\left(a-a\right)=f\text{'}(a)\\ b_1+0=f\text{'}(a)\\ b_1=f\text{'}(a)\end{array}$

Use property (iii) $Q\text{'}\text{'}(a)=f\text{'}\text{'}(a)$ ,

Second derivative of $Q(x)$:

  $\begin{array}{l}Q\text{'}(x)=b_1+2b_2\left(x-a\right)\\ Q\text{'}\text{'}(x)=0+2b_2\\ Q\text{'}\text{'}(x)=2b_2\end{array}$

So,

  $\begin{array}{c}Q\text{'}\text{'}(a)=f\text{'}\text{'}(a)\\ \text{Put }x=a,\\ 2b_2=f\text{'}\text{'}(a)\\ b_2=\frac{1}{2}\left(f\text{'}\text{'}(a)\right)\end{array}$

Conclusion:

The coefficients are

  $\begin{array}{c}{b}_0=f(a),\\ b_1=f\text{'}(a),\\ b_2=\frac{f\text{'}\text{'}(a)}{2}.\end{array}$

Also, Quadratic approximation at $x=a$ :-

  $Q(x)=f(a)+f\text{'}(a)\left(x-a\right)+\frac{1}{2}f\text{'}\text{'}(a){\left(x-a\right)}^2$ .

b.

To determine

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

Answer to Problem 66E

Quadratic approximation at $x=0$ :

  $Q(x)=1+x+x^2$

Explanation of Solution

Given:

From part (a):-

Quadratic approximation: $Q(x)=f(a)+f\text{'}(a)\left(x-a\right)+\frac{1}{2}f\text{'}\text{'}(a){\left(x-a\right)}^2$ to $f(x)$ at $x=a$ .

Calculation:

Derivatives of $f(x)$ :-

  $\begin{array}{c}f(x)=\frac{1}{\left(1-x\right)}\\ f\text{'}(x)=\frac{1}{{\left(1-x\right)}^2}\\ f\text{'}\text{'}(x)=\frac{2}{{\left(1-x\right)}^3}\end{array}$

Quadratic approximation at $x=0$ :

  $\begin{array}{c}Q(x)=f(a)+f\text{'}(a)\left(x-a\right)+\frac{1}{2}f\text{'}\text{'}(a){\left(x-a\right)}^2\\ =\left(\frac{1}{1-a}\right)+\frac{1}{{\left(1-a\right)}^2}\left(x-a\right)+\frac{1}{2}\left(\frac{2}{{\left(1-a\right)}^3}\right){\left(x-a\right)}^2\\ =\left(\frac{1}{1-a}\right)+\frac{\left(x-a\right)}{{\left(1-a\right)}^2}+\frac{1}{{\left(1-a\right)}^3}{\left(x-a\right)}^2\\ \text{Put }a=0,\\ Q(x)=\left(\frac{1}{1-0}\right)+\frac{\left(x-0\right)}{{\left(1-0\right)}^2}+\frac{1}{{\left(1-0\right)}^3}{\left(x-0\right)}^2\\ Q(x)=1+x+x^2\end{array}$ .

Conclusion:

Quadratic approximation at $x=0$ :

  $Q(x)=1+x+x^2$

c.

To determine

Graph the function $f(x)$ and quadratic approximation at $x=0.$

Explanation of Solution

Given:

Function: $f(x)=\frac{1}{\left(1-x\right)}$

Quadratic approximation at $x=0$ :

  $Q(x)=1+x+x^2$

Calculation:

Table for values: -

iteration	$x\text{'}s\text{ value}$	$f(x)=\frac{1}{\left(1-x\right)}$	$Q(x)=1+x+x^2$
1.	-2	$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$	3
2.	-1	$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$	1
3.	0	1	1
4.	2	-1	7
5.	3	$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$-2$}\right.$	13

Use graphing calculator to graph,

  

Here, $X-axis$ denotes values of $x$ .

  $Y-axis$ denotes values of $y$ .

Using graph, it can be seen that both the graph have same value at $x=0$ .

So, using the Quadratic approximation the approximate values of function in the neighborhood of point, can be calculated.

Conclusion:

Hence, the coefficients are

  $\begin{array}{c}{b}_0=f(a),\\ b_1=f\text{'}(a),\\ b_2=\frac{f\text{'}\text{'}(a)}{2}.\end{array}$

Quadratic approximation at $x=a$ :-

  $Q(x)=f(a)+f\text{'}(a)\left(x-a\right)+\frac{1}{2}f\text{'}\text{'}(a){\left(x-a\right)}^2$ .

d.

To determine

Find the quadratic approximation to $g(x)=\frac{1}{x}$ at $x=1.$

Answer to Problem 66E

Quadratic approximation at $x=1$ :

  $Q(x)=x+{\left(x-1\right)}^2$

Explanation of Solution

Given:

From part (a):

Quadratic approximation: $Q(x)=f(a)+f\text{'}(a)\left(x-a\right)+\frac{1}{2}f\text{'}\text{'}(a){\left(x-a\right)}^2$ to $f(x)$ at $x=a$ .

Calculation:

Derivatives of$g(x)$:

  $\begin{array}{c}g(x)=\frac{1}{x}\\ g\text{'}(x)=\frac{-1}{x^2}\\ g\text{'}\text{'}(x)=\frac{2}{x^3}\end{array}$

Quadratic approximation at $x=1$ :-

  $\begin{array}{c}Q(x)=g(a)+g\text{'}(a)\left(x-a\right)+\frac{1}{2}g\text{'}\text{'}(a){\left(x-a\right)}^2\\ =\left(\frac{1}{a}\right)-\frac{1}{a^2}\left(x-a\right)+\frac{1}{2}\left(\frac{2}{a^3}\right){\left(x-a\right)}^2\\ =\frac{1}{a}-\frac{1}{a^2}\left(x-a\right)+\frac{1}{a^3}{\left(x-a\right)}^2\end{array}$

  $\begin{array}{c}\text{Put }a=1,\\ Q(x)=\left(\frac{1}{1}\right)-\frac{1}{1^2}\left(x-1\right)+\left(\frac{1}{1^3}\right){\left(x-1\right)}^2\\ Q(x)=1+\left(x-1\right)+{\left(x-1\right)}^2\\ Q(x)=1-1+x+{\left(x-1\right)}^2\\ Q(x)=x+{\left(x-1\right)}^2\end{array}$

Graph of $g(x)$and $Q(x)=x+{\left(x-1\right)}^2$:-

Use graphing calculator to graph the graph,

  

Using graph, it can be seen that both the graph have same value at $x=1$ .

So, using the Quadratic approximation the approximate values of function in the neighborhood of point, can be calculated

Conclusion:

Quadratic approximation at $x=1$ :-

  $Q(x)=x+{\left(x-1\right)}^2$

e.

To determine

Find the quadratic approximation to $h(x)=\sqrt{1+x}\text{ at }x=0.$

Answer to Problem 66E

Quadratic approximation at $x=0$ :

  $Q(x)=1-\frac{1}{2}x-\frac{1}{8}{x}^2$

Explanation of Solution

Given:

From part (a):-

Quadratic approximation: $Q(x)=f(a)+f\text{'}(a)\left(x-a\right)+\frac{1}{2}f\text{'}\text{'}(a){\left(x-a\right)}^2$ to $f(x)$ at $x=a$ .

Calculation:

Derivatives of $h(x)$ :

  $\begin{array}{l}h(x)=\sqrt{1+x}\\ h\text{'}(x)=\frac{1}{2\sqrt{1+x}}\\ h\text{'}\text{'}(x)=\frac{-1}{4{\left(1+x\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$

Quadratic approximation at $x=0$ :

  $\begin{array}{c}Q(x)=h(a)+h\text{'}(a)\left(x-a\right)+\frac{1}{2}h\text{'}\text{'}(a){\left(x-a\right)}^2\\ =\left(\sqrt{1+a}\right)-\frac{1}{2\sqrt{1+a}}\left(x-a\right)+\frac{1}{2}\left(\frac{-1}{4{\left(1+a\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right){\left(x-a\right)}^2\\ =\left(\sqrt{1+a}\right)-\frac{1}{2\sqrt{1+a}}\left(x-a\right)-\frac{1}{8{\left(1+a\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left(x-a\right)}^2\end{array}$ .

  $\begin{array}{c}\text{Put }a=0,\\ Q(x)=\left(\sqrt{1+0}\right)-\frac{1}{2\sqrt{1+0}}\left(x-0\right)-\frac{1}{8{\left(1+0\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left(x-0\right)}^2\\ =1-\frac{1}{2}x-\frac{1}{8}{x}^2\\ \\ Q(x)=1-\frac{1}{2}x-\frac{1}{8}{x}^2\end{array}$

Graph of $h\left(x\right)\text{ and }Q(x)=1-\frac{1}{2}x-\frac{1}{8}{x}^2$:

Use graphing calculator to graph the graph,

  

Using graph, it can be seen that both the graph have same value at $x=0$ .

So, using the Quadratic approximation the approximate values of function in the neighborhood of point, can be calculated.

Conclusion:

Quadratic approximation at $x=0$ :-

  $Q(x)=1-\frac{1}{2}x-\frac{1}{8}{x}^2$

f.

To determine

Find the linearization of $f(x),g(x)\text{ and }h(x).$

Answer to Problem 66E

Linearization of $f(x)=1-x.$

Linearization of $g(x)=x.$

Linearization of $h(x)=1-\frac{x}{2}.$

Explanation of Solution

Given:

  $\begin{array}{l}f(x)=\frac{1}{\left(1-x\right)},\\ g(x)=\frac{1}{x},\\ h\left(x\right)=\sqrt{1+x}.\end{array}$

Concept used:

Linearization: If $f$ is differentiable at $x=a$ , then the equation of the tangent line,

  $L(x)=f(a)+f\text{'}(a)(x-a),$ defines the linearization of $f$ at $x=a$ .

Calculation:

Linearization of function $f(x)=\frac{1}{\left(1-x\right)}$at $x=0$:

Derivatives of $f(x)$ :-

  $\begin{array}{c}f(x)=\frac{1}{\left(1-x\right)}\\ f\text{'}(x)=\frac{1}{{\left(1-x\right)}^2}\end{array}$

Quadratic approximation at $x=0$ :

  $\begin{array}{c}L(x)=f(a)+f\text{'}(a)\left(x-a\right)\\ =\left(\frac{1}{1-a}\right)+\frac{1}{{\left(1-a\right)}^2}\left(x-a\right)\\ =\left(\frac{1}{1-a}\right)+\frac{\left(x-a\right)}{{\left(1-a\right)}^2}\end{array}$

  $\begin{array}{c}\text{Put }a=0,\\ L(x)=\left(\frac{1}{1-0}\right)+\frac{\left(x-0\right)}{{\left(1-0\right)}^2}\\ L(x)=1+\frac{x}{1}\\ L(x)=1+x\end{array}$

So, linearization of $f(x)=1+x$ .

Linearization of function $g(x)=\frac{1}{x}\text{ at }x=1$:

Derivatives of $g(x)$ :-

  $\begin{array}{l}g(x)=\frac{1}{x}\\ g\text{'}(x)=\frac{-1}{x^2}\end{array}$

Quadratic approximation at $x=1$ :-

  $\begin{array}{c}L(x)=g(a)+g\text{'}(a)\left(x-a\right)\\ =\left(\frac{1}{a}\right)-\frac{1}{a^2}\left(x-a\right)\\ =\frac{1}{a}-\frac{1}{a^2}\left(x-a\right)\\ \\ \text{Put }a=1,\\ L(x)=\left(\frac{1}{1}\right)-\frac{1}{1^2}\left(x-1\right)\\ L(x)=1+\left(x-1\right)\\ L(x)=1-1+x\\ L(x)=x\end{array}$

Linearization of $g(x)=x.$

Linearization of function $h(x)=\sqrt{1+x}$at $x=0$:

Derivatives of $h(x)$ :-

  $\begin{array}{c}h(x)=\sqrt{1+x}\\ h\text{'}(x)=\frac{1}{2\sqrt{1+x}}\end{array}$

Quadratic approximation at $x=0$ :-

  $\begin{array}{c}L(x)=h(a)+h\text{'}(a)\left(x-a\right)\\ =\left(\sqrt{1+a}\right)-\frac{1}{2\sqrt{1+a}}\left(x-a\right)\\ =\left(\sqrt{1+a}\right)-\frac{1}{2\sqrt{1+a}}\left(x-a\right)\\ \\ \text{Put }a=0,\\ L(x)=\left(\sqrt{1+0}\right)-\frac{1}{2\sqrt{1+0}}\left(x-0\right)\\ =1-\frac{1}{2}x\\ \\ L(x)=1-\frac{1}{2}x\end{array}$

So, linearization of $h(x)=1-\frac{x}{2}.$

Conclusion:

Linearization of $f(x)=1-x.$

Linearization of $g(x)=x.$

Linearization of $h(x)=1-\frac{x}{2}.$