## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.4 Taylor Polynomials - Exercises - Page 492: 13

#### Answer

$$T_{2}(x) =(x-1)-\frac{3}{2}(x-1)^{2}$$ $$T_{3}(x) =(x-1)-\frac{3}{2}(x-1)^{2}+\frac{11}{6}(x-1)^{3}$$

#### Work Step by Step

Given $$f(x)=\frac{\ln x}{x}, \quad a=1$$ Since \begin{aligned} f(x) &=\frac{\ln x}{x} & & f(a)=0 \\ f^{\prime}(x) &=\frac{1-\ln x}{x^{2}} & & f(a)=1 \\ f^{\prime \prime}(x) &=\frac{-3+2 \ln x}{x^{3}} & f(a) &=-3 \\ f^{\prime \prime \prime}(x) &=\frac{11-6 \ln x}{x^{4}} & f(a) &=11 \end{aligned} then \begin{aligned} T_{2}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}=0+1(x-1)+\frac{-3}{2 !}(x-1)^{2} \\ &=(x-1)-\frac{3}{2}(x-1)^{2} \end{aligned} and \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} \\ &=0+1(x-1)+\frac{-3}{2 !}(x-1)^{2}+\frac{11}{3 !}(x-1)^{3} \\ &=(x-1)-\frac{3}{2}(x-1)^{2}+\frac{11}{6}(x-1)^{3} \end{aligned}

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