Question 7.6.1: Use formula (7.6.4) to approximate the function f (x) = √25 ......
Use formula (7.6.4) to approximate the function
f(x)=f(0)+f^{\prime}(0)x+{\frac{1}{2}}f^{\prime\prime}(z)x^{2} (7.6.4)
f(x)=\sqrt{25+x}=(25+x)^{1/2}Use the result to estimate \sqrt{25.01}, with a bound on the absolute value of the remainder.
To apply (7.6.4), we differentiate to obtain
f^{\prime}(x)=\frac{1}{2}(25+x)^{-1/2},~~\mathrm{and}~f^{\prime\prime}(x)=\frac{1}{2}\left(-\frac{1}{2}\right)(25+x)^{-3/2}Then f (0) = 5, whereas f^{\prime}(0)=1/2\cdot1/5=1/10\mathrm{~and~}f^{\prime\prime}(z)=-(1/4)\mathrm{~(25+z)^{-3/2}}. So by (7.6.4), there exists z between 0 and x such that
\sqrt{25+x}=5+\frac{1}{10}x+\frac{1}{2}\left(-\frac{1}{4}\right)(25+z)^{-3/2}x^{2}=5+\frac{1}{10}x-\frac{1}{8}(25+z)^{-3/2}x^{2} (∗)
In order to estimate \sqrt{25.01}, we write 25.01 = 25 + 0.01 and use (∗). If x = 0.01, then z lies between 0 and 0.01, so 25 + z > 25. Then (25 + z)^{−3/2} < (25)^{−3/2} = 1/125, so the absolute value of the remainder is
|R_{2}(0.01)|=\left|-\frac{1}{8}(25+z)^{-3/2}\left(\frac{1}{100}\right)^{2}\right|\lt \frac{1}{80000}\cdot\frac{1}{125}=\,10^{-7}We conclude that \sqrt{25.01 }≈ 5 + 1/10 · 1/100 = 5.001, with an error less than 10^{−7}.