Chapter 11.11, Problem 1PT

To determine

The quadratic approximation for $f\left(x\right)=x^6$ at a = 1.

Answer to Problem 1PT

Quadratic approximation for $f\left(x\right)=x^6$ at a = 1 is $\underset{\_}{\text{d) }{T}_2\left(x\right)=1+6\left(x-1\right)+15{\left(x-1\right)}^2}$.

Explanation of Solution

Result used:

The Taylor polynomial of degree n for f about a is, $T_n\left(x\right)=f\left(a\right)+\frac{f^{\prime }\left(a\right)}{1!}\left(x-a\right)+\frac{f^{″}\left(a\right)}{2!}{\left(x-a\right)}^2+\dots +\frac{f^n\left(a\right)}{n!}{\left(x-a\right)}^n$. $T_n\left(x\right)$ may be used to approximate $f\left(x\right)$ for any given x in the interval of convergence for the Taylor series for f about a.

Calculation:

The given function is $f\left(x\right)=x^6$.

To find the quadratic approximation, find $T_2\left(x\right)$ where a = 1.

The Taylor polynomial of degree 2 for $f\left(x\right)=x^6$ about a = 1 is,

$T_2\left(x\right)=f\left(1\right)+\frac{f^{\prime }\left(1\right)}{1!}\left(x-1\right)+\frac{f^{″}\left(1\right)}{2!}{\left(x-1\right)}^2$

Obtain the values of $f\left(1\right),f^{\prime }\left(1\right)\text{and }{f}^{″}\left(1\right)$ as follows.

$\begin{array}{l}f\left(x\right)=x^6;f\left(1\right)=1\\ f^{\prime }\left(x\right)=6x^5;f^{\prime }\left(1\right)=6\\ f^{″}\left(x\right)=30x^4;f^{″}\left(1\right)=30\end{array}$

Substitute these values in $T_2\left(x\right)=f\left(1\right)+\frac{f^{\prime }\left(1\right)}{1!}\left(x-1\right)+\frac{f^{″}\left(1\right)}{2!}{\left(x-1\right)}^2$. Then,

$\begin{array}{c}{T}_2\left(x\right)=1+\frac{6}{1}\left(x-1\right)+\frac{30}{2}{\left(x-1\right)}^2\\ =1+6\left(x-1\right)+15{\left(x-1\right)}^2\end{array}$

Therefore, the quadratic approximation is, $\underset{\_}{\text{d) }{T}_2\left(x\right)=1+6\left(x-1\right)+15{\left(x-1\right)}^2}$.