find-the-second-order-taylor-polynomial-for-fxsin1xfxsin-1x-centered-at-a12afrac

Question

Find the second-order Taylor polynomial for f(x)=sin⁡−1xf(x)=\$$sin^{-1}x $$centered at a=12a=\frac12.

Accepted Answer

π6+233(x−12)+239(x−12)2\frac{\pi}{6}+\frac{2\sqrt3}{3}\left(x-\frac12\right)+\frac{2\sqrt3}{9}\left(x-\frac12\$$right)^2$$

Answer

π3+233(x−12)+239(x−12)2\frac{\pi}{3}+\frac{2\sqrt3}{3}\left(x-\frac12\right)+\frac{2\sqrt3}{9}\left(x-\frac12\$$right)^2$$

Answer

π6+233(x+12)+239(x+12)2\frac{\pi}{6}+\frac{2\sqrt3}{3}\left(x+\frac12\right)+\frac{2\sqrt3}{9}\left(x+\frac12\$$right)^2$$

Answer

π6−233(x−12)−239(x−12)2\frac{\pi}{6}-\frac{2\sqrt3}{3}\left(x-\frac12\right)-\frac{2\sqrt3}{9}\left(x-\frac12\$$right)^2$$

Find the second-order Taylor polynomial for f(x)=sin−1xf(x)=\(\sin\)^{-1}x centered at a=12a=\(\frac\)12.