Find the Taylor polynomial of order 3 generated by f at a.
-$$f ( x ) = \frac { 1 } { x + 4 } , a = 0$$
A) $P _ { 3 } ( x ) = \frac { x } { 4 } + \frac { x ^ { 2 } } { 16 } + \frac { x ^ { 3 } } { 64 } + \frac { x ^ { 4 } } { 256 }$
B) $P _ { 3 } ( x ) = \frac { 1 } { 4 } + \frac { x } { 16 } + \frac { x ^ { 2 } } { 64 } + \frac { x ^ { 3 } } { 256 }$
C) $P _ { 3 } ( x ) = \frac { x } { 4 } - \frac { x ^ { 2 } } { 16 } + \frac { x ^ { 3 } } { 64 } - \frac { x ^ { 4 } } { 256 }$
D) $P _ { 3 } ( x ) = \frac { 1 } { 4 } - \frac { x } { 16 } + \frac { x ^ { 2 } } { 64 } - \frac { x ^ { 3 } } { 256 }$

Question

Quizplus · Accepted Answer

The Taylor polynomial of order 3 for $f(x) = \frac{1}{x+4}$ at $a = 0$ involves calculating the first three derivatives of $f(x)$, evaluating them at $a = 0$, and then using the formula for the Taylor series. The correct polynomial, $P_3(x) = \frac{1}{4} - \frac{x}{16} + \frac{x^2}{64} - \frac{x^3}{256}$, reflects the derivatives of $f(x)$ at $a = 0$ and follows the pattern of alternating signs due to the negative powers of $x+4$ in the derivatives.

Find the Taylor Polynomial of Order 3 Generated by F f(x)=1x+4,a=0f ( x ) = \frac { 1 } { x + 4 } , a = 0f(x)=x+41​,a=0

Find the Taylor Polynomial of Order 3 Generated by F $f ( x ) = \frac { 1 } { x + 4 } , a = 0$