## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.4 Taylor Polynomials - Exercises - Page 493: 29

#### Answer

$x \in[1,2.9]$

#### Work Step by Step

Given $$f(x) =\sqrt{x}=x^{1/2},\ \ \ a=1$$ Since \begin{array}{rlrl} f(x) & =x^{1 / 2} & & f(1)=1 \\ f^{\prime}(x) & =\frac{1}{2} x^{-1 / 2} & & f^{\prime}(1)=\frac{1}{2} \\ f^{\prime \prime}(x) & =-\frac{1}{4} x^{-3 / 2} & f^{\prime \prime}(1) & =-\frac{1}{4} \\ f^{\prime \prime \prime}(x) & =\frac{3}{8} x^{-5 / 2} & f^{\prime \prime \prime}(1) & =\frac{3}{8} \end{array} Then \begin{aligned} T_{3}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\ &= 1+\frac{1}{2}(x-1)-\frac{1}{4 \cdot 2 !}(x-1)^{2}+\frac{3}{8 \cdot 3 !}(x-1)^{3}\\ &=1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^{2}+\frac{1}{16}(x-1)^{3} \end{aligned} Since $$|f(x)-T_3(x)|= |\sqrt{x}-1-\frac{1}{2}(x-1)+\frac{1}{8}(x-1)^{2}-\frac{1}{16}(x-1)^{3}|$$ From the following graph, we have for $x \in[1,2.9]$ that the error does not exceed $0.25 .$ The error at $x=3$ appears to just exceed $0.25$

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