Question

consider the following function.
$f(x)=x^{4 / 5}, quad a=1, quad n=3, quad 0.7 leq x leq 1.3$
(a) approximate $f$ by a taylor polynomial with degree $n$ at the number $a$.
$t_{3}(x)=$
(b) use taylors inequality to estimate the accuracy of the approximation $f(x) approx t_{n}(x)$ when $x$ lies in the given interval. (round your answer to eight decimal places.)
$left|r_{3}(x)
ight| leq$
(c) check your result in part (b) by graphing $left|r_{n}(x)
ight|$.

Explanation:

Step1: Find the derivatives of \(f(x)=x^{4/5}\)

The formula for the Taylor polynomial \(T_n(x)\) of a function \(f(x)\) about \(a\) is \(T_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\).
First, \(f(x)=x^{4/5}\), so \(f(1)=1\).
The first - derivative \(f^{\prime}(x)=\frac{4}{5}x^{-1/5}\), and \(f^{\prime}(1)=\frac{4}{5}\).
The second - derivative \(f^{\prime\prime}(x)=-\frac{4}{25}x^{-6/5}\), and \(f^{\prime\prime}(1)=-\frac{4}{25}\).
The third - derivative \(f^{(3)}(x)=\frac{24}{125}x^{-11/5}\), and \(f^{(3)}(1)=\frac{24}{125}\).

Step2: Construct the Taylor polynomial \(T_3(x)\)

Using the Taylor polynomial formula \(T_3(x)=f(1)+f^{\prime}(1)(x - 1)+\frac{f^{\prime\prime}(1)}{2!}(x - 1)^2+\frac{f^{(3)}(1)}{3!}(x - 1)^3\)

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Step3: Use Taylor's inequality for the remainder \(R_3(x)\)

The \(n + 1\)th derivative \(f^{(4)}(x)=-\frac{264}{625}x^{-16/5}\).
For \(x\in[0.7,1.3]\), \(\vert f^{(4)}(x)\vert=\frac{264}{625}\vert x^{-16/5}\vert\). The maximum value of \(\vert x^{-16/5}\vert\) on the interval \([0.7,1.3]\) occurs at \(x = 0.7\).
\(\vert f^{(4)}(x)\vert\leq M=\frac{264}{625}(0.7)^{-16/5}\)
By Taylor's inequality \(R_n(x)=\frac{f^{(n + 1)}(z)}{(n+1)!}(x - a)^{n + 1}\), here \(n = 3\), \(a = 1\)
\(\vert R_3(x)\vert=\frac{\vert f^{(4)}(z)\vert}{4!}\vert x - 1\vert^{4}\leq\frac{\frac{264}{625}(0.7)^{-16/5}}{24}\vert x - 1\vert^{4}\)
Since \(\vert x-1\vert\leq0.3\) (because \(x\in[0.7,1.3]\))

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Answer:

\(T_3(x)=1+\frac{4}{5}(x - 1)-\frac{2}{25}(x - 1)^2+\frac{4}{125}(x - 1)^3\) (or the expanded form \(\frac{4}{125}x^{3}-\frac{22}{125}x^{2}+\frac{112}{125}x-\frac{37}{125}\)), \(\vert R_3(x)\vert\leq0.00002\)