Find the third degree Taylor polynomial centered at c = 4 for the function. $f ( x ) = \sqrt { x }$
A) $2 + \frac { 1 } { 4 } ( x - 4 ) - \frac { 1 } { 64 } ( x - 4 ) ^ { 2 } + \frac { 1 } { 512 } ( x - 4 ) ^ { 3 }$
B) $2 - \frac { 1 } { 4 } ( x - 4 ) + \frac { 1 } { 64 } ( x - 4 ) ^ { 2 } - \frac { 1 } { 512 } ( x - 4 ) ^ { 3 }$
C) $2 - \frac { 1 } { 4 } ( x - 4 ) - \frac { 1 } { 64 } ( x - 4 ) ^ { 2 } - \frac { 1 } { 512 } ( x - 4 ) ^ { 3 }$
D) $2 + \frac { 1 } { 4 } ( x + 4 ) - \frac { 1 } { 64 } ( x + 4 ) ^ { 2 } + \frac { 1 } { 512 } ( x + 4 ) ^ { 3 }$
E) $2 - \frac { 1 } { 4 } ( x + 4 ) + \frac { 1 } { 64 } ( x + 4 ) ^ { 2 } - \frac { 1 } { 512 } ( x + 4 ) ^ { 3 }$

Question

Quizplus · Accepted Answer

The Taylor series for $f(x) = \sqrt{x}$ centered at $c=4$ is given by$$f(x) = \sqrt{4} + \frac{1}{2\sqrt{4}}(x-4) - \frac{1}{8(4)^{3/2}}(x-4)^2 + \frac{1}{16(4)^{5/2}}(x-4)^3 + \cdots$$Simplifying, we get$$f(x) = 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2 + \frac{1}{512}(x-4)^3 + \cdots$$Therefore, the third degree Taylor polynomial for $f(x) = \sqrt{x}$ centered at $c=4$ is$$P_3(x) = 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2 + \frac{1}{512}(x-4)^3$$

Find the Third Degree Taylor Polynomial Centered at C = f(x)=xf ( x ) = \sqrt { x }f(x)=x​

Find the Third Degree Taylor Polynomial Centered at C = $f ( x ) = \sqrt { x }$