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

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

$T_{2}(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}$

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

${\begin{aligned}f'(x)&={\frac {1}{2}}x^{-1/2}\\f''(x)&=-{\frac {1}{4x^{3/2}}}\end{aligned}}$

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

$\displaystyle {\begin{aligned}T_{2}(x)&=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}\\&=f(4)+f'(4)(x-4)+{\frac {f''(4)}{2}}(x-4)^{2}\\&={\sqrt {4}}+{\frac {1}{2}}4^{-1/2}(x-4)+{\frac {-{\frac {1}{4\times 4^{3/2}}}}{2}}(x-4)^{2}\\&=2+{\frac {1}{4}}(x-4)-{\frac {1}{64}}(x-4)^{2}\\&=2+{\frac {x-4}{4}}-{\frac {(x-4)^{2}}{64}}\\\end{aligned}}$