Chapter 9.2, Problem 23E

(a)

To determine

To find: The third order Taylor polynomial for $f$ at $x=1$ and the value at $f\left(1.2\right)$ .

Answer to Problem 23E

The third order Taylor polynomial is $f\left(x\right)=4-\left(x-1\right)+\frac{3}{2}{\left(x-1\right)}^2+\frac{1}{3}{\left(x-1\right)}^3$ .

The value of $f\left(1.2\right)$ is $f\left(1.2\right)=3.863$ .

Explanation of Solution

Given:

The value of function at $x=1$ is $f\left(1\right)=4$ .

The value of first derivative of function at $x=1$ is $f\text{'}\left(1\right)=-1$ .

The value of second derivative of function at $x=1$ is $f\text{'}\text{'}\left(1\right)=3$ .

The value of third derivative of function at $x=1$ is $f\text{'}\text{'}\text{'}\left(1\right)=2$ .

Concept used:

Taylor polynomial is given by:

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(a\right)}{n!}}{\left(x-a\right)}^n\\ =f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)+\frac{f\text{'}\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2+\frac{f\text{'}\text{'}\text{'}\left(a\right)}{3!}{\left(x-a\right)}^3+\mathrm{............}\end{array}$

Calculation:

Put $a=1$ in general expression of Taylor polynomial.

  $\begin{array}{c}f\left(x\right)=f\left(1\right)+f\text{'}\left(1\right)\left(x-1\right)+\frac{f\text{'}\text{'}\left(1\right)}{2!}{\left(x-1\right)}^2+\frac{f\text{'}\text{'}\text{'}\left(1\right)}{3!}{\left(x-1\right)}^3\\ =4+\left(-1\right)\left(x-1\right)+\frac{3}{2!}{\left(x-1\right)}^2+\frac{2}{3!}{\left(x-1\right)}^3\\ =4-\left(x-1\right)+\frac{3}{2\times 1}{\left(x-1\right)}^2+\frac{2}{3\times 2\times 1}{\left(x-1\right)}^3\\ =4-\left(x-1\right)+\frac{3}{2}{\left(x-1\right)}^2+\frac{1}{3}{\left(x-1\right)}^3\mathrm{....................}\left(1\right)\end{array}$

The value of $f\left(1.2\right)$ is obtained if $x=1.2$ in equation $\left(1\right)$ is:

  $\begin{array}{c}f\left(1.2\right)=4-\left(1.2-1\right)+\frac{3}{2}{\left(1.2-1\right)}^2+\frac{1}{3}{\left(1.2-1\right)}^3\\ =4-0.2+\frac{3}{2}{\left(0.2\right)}^2+\frac{1}{3}{\left(0.2\right)}^3\\ =4-0.2+\frac{3}{2}\times 0.04+\frac{1}{3}\times 0.008\\ =3.863\end{array}$

Conclusion:

The third order Taylor polynomial is $f\left(x\right)=4-\left(x-1\right)+\frac{3}{2}{\left(x-1\right)}^2+\frac{1}{3}{\left(x-1\right)}^3$ .

The value of $f\left(1.2\right)$ is $f\left(1.2\right)=3.863$ .

(b)

To determine

The second order Taylor polynomial for derivative of $f$ at $x=1$ and find the value at $f\text{'}\left(0.2\right)$ .

Answer to Problem 23E

The second order Taylor polynomial for derivative of $f$ at $x=1$ is $f\text{'}\left(x\right)={\left(x-1\right)}^2+3\left(x-1\right)-1$ .

The value of $f\text{'}\left(1.2\right)$ is $f\text{'}\left(1.2\right)=-1.56$ .

Explanation of Solution

Given:

The third order Taylor polynomial is $f\left(x\right)=\frac{1}{6}\left(2x^3-3x^2-18x+37\right)$

The derivative of Taylor polynomial is given by:

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(4-\left(x-1\right)+\frac{3}{2}{\left(x-1\right)}^2+\frac{1}{3}{\left(x-1\right)}^3\right)\\ f\text{'}\left(x\right)=\left[\frac{d}{dx}\left(4\right)-\frac{d}{dx}\left(x-1\right)+\frac{3}{2}\times \frac{d}{dx}{\left(x-1\right)}^2+\frac{1}{3}\times \frac{d}{dx}{\left(x-1\right)}^3\right]\\ f\text{'}\left(x\right)=0-1+\frac{3}{2}\times 2\left(x-1\right)\left(1\right)+\frac{1}{3}\times 3{\left(x-1\right)}^2\left(1\right)\\ f\text{'}\left(x\right)={\left(x-1\right)}^2+3\left(x-1\right)-1\mathrm{................}\left(2\right)\end{array}$

Put the value $x=1.2$ in the equation $\left(2\right)$ .

  $\begin{array}{c}f\text{'}\left(1.2\right)={\left(1.2-1\right)}^2+3\left(1.2-1\right)-1\\ ={\left(0.2\right)}^2-3\left(0.2\right)-1\\ =0.04-0.6-1\\ =-1.56\end{array}$

Conclusion:

The second order Taylor polynomial for derivative of $f$ at $x=1$ is $f\text{'}\left(x\right)={\left(x-1\right)}^2+3\left(x-1\right)-1$ .

The value of $f\text{'}\left(1.2\right)$ is $f\text{'}\left(1.2\right)=-1.56$ .