T_2(x)=f(4)+f'(4)(x−4)+\dfrac{f''(4)}{2!}(x−4)^2
f(x)=x^{\frac{1}{2}}⇒f(4)=2
f'(x) = \dfrac{1}{2} x^{\frac{-1}{2}} \Rightarrow f'(4) = \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{1}{4}
f''(x) = -\dfrac{1}{4} x^{\frac{-3}{2}} \Rightarrow f''(4) = -\dfrac{1}{4}\cdot\dfrac{1}{8} = -\dfrac{1}{32}
따라서 T_2(x) = 2 + \dfrac{1}{4}(x - 4) + \dfrac{\dfrac{-1}{32}}{2}(x - 4)^2 = 2 + \dfrac{1}{4}(x - 4) - \dfrac{1}{64}(x - 4)^2이다.