Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (j)/Solution 1

To find the second degree Taylor polynomial, we use the equation

$T_2(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2}(x-a{)}^2$

We know that a =4 and that ƒ(x)=√x. We can easily compute the derivatives to get

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{1}{2}{x}^{-1/2}\\ f^{″}(x)& =-\frac{1}{4x^{3/2}}\end{array}$

Plugging this all in to the Taylor equation, we get:

$\begin{array}{rl}{T}_2(x)& =f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2}(x-a{)}^2\\ & =f(4)+f^{\prime }(4)(x-4)+\frac{f^{″}(4)}{2}(x-4{)}^2\\ & =\sqrt{4}+\frac{1}{2}{4}^{-1/2}(x-4)+\frac{-\frac{1}{4\times 4^{3/2}}}{2}(x-4{)}^2\\ & =2+\frac{1}{4}(x-4)-\frac{1}{64}(x-4{)}^2\\ & =2+\frac{x-4}{4}-\frac{(x-4{)}^2}{64}\\ \end{array}$