Find the third degree Taylor polynomial centered at $c = 1$ for the function. $$f ( x ) = \sqrt { x }$$
A) $1 + \frac { 1 } { 2 } ( x + 1 ) - \frac { 1 } { 8 } ( x - 1 ) ^ { 2 } + \frac { 1 } { 16 } ( x + 1 ) ^ { 3 }$
B) $1 + \frac { 1 } { 2 } ( x - 1 ) + \frac { 1 } { 8 } ( x - 1 ) ^ { 2 } - \frac { 1 } { 16 } ( x - 1 ) ^ { 3 }$
C) $1 + \frac { 1 } { 2 } ( x - 1 ) - \frac { 1 } { 8 } ( x - 1 ) ^ { 2 } + \frac { 1 } { 16 } ( x - 1 ) ^ { 3 }$
D) $1 + \frac { 1 } { 2 } ( x - 1 ) - \frac { 1 } { 8 } ( x - 1 ) ^ { 2 } - \frac { 1 } { 16 } ( x - 1 ) ^ { 3 }$
E) $1 + \frac { 1 } { 2 } ( x + 1 ) + \frac { 1 } { 8 } ( x - 1 ) ^ { 2 } - \frac { 1 } { 16 } ( x + 1 ) ^ { 3 }$

Question

Quizplus · Accepted Answer

The third degree Taylor polynomial for a function $f(x)$ centered at $c$ is given by $f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3$. For $f(x) = \sqrt{x}$ centered at $c=1$, we calculate the derivatives and evaluate them at $c=1$, resulting in the coefficients found in option C.

Find the Third Degree Taylor Polynomial Centered At c=1c = 1c=1

Find the Third Degree Taylor Polynomial Centered At $c = 1$