Find the Taylor polynomial of order 3 generated by f at a.
-$$f ( x ) = \frac { 1 } { x + 3 } , a = 1$$
A) $P _ { 3 } ( x ) = \frac { 1 } { 2 } - \frac { x - 1 } { 4 } + \frac { ( x - 1 ) ^ { 2 } } { 8 } - \frac { ( x - 1 ) ^ { 3 } } { 16 }$
B) $P _ { 3 } ( x ) = \frac { 1 } { 4 } - \frac { x + 1 } { 16 } + \frac { ( x + 1 ) ^ { 2 } } { 64 } - \frac { ( x + 1 ) ^ { 3 } } { 256 }$
C) $\mathrm { P } _ { 3 } ( \mathrm { x } ) = \frac { 1 } { 4 } - \frac { \mathrm { x } - 1 } { 16 } + \frac { ( \mathrm { x } - 1 ) ^ { 2 } } { 64 } - \frac { ( \mathrm { x } - 1 ) ^ { 3 } } { 256 }$
D) $\mathrm { P } _ { 3 } ( \mathrm { x } ) = \frac { 1 } { 2 } - \frac { \mathrm { x } + 1 } { 4 } + \frac { ( \mathrm { x } + 1 ) ^ { 2 } } { 8 } - \frac { ( \mathrm { x } + 1 ) ^ { 3 } } { 16 }$

Question

Quizplus · Accepted Answer

The Taylor series formula for a function $f(x)$ centered at $a$ is:

$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$

Here, we have $f(x) = \frac{1}{x+3}$ and $a=1$. First, we need to compute the first few derivatives of $f$:

$f(x) = \frac{1}{x+3}$

$f'(x) = -\frac{1}{(x+3)^2}$

$f''(x) = \frac{2}{(x+3)^3}$

$f'''(x) = -\frac{6}{(x+3)^4}$

Next, we evaluate these derivatives at $x=a=1$:

$f(1) = \frac{1}{4}$

$f'(1) = -\frac{1}{16}$

$f''(1) = \frac{1}{32}$

$f'''(1) = -\frac{3}{256}$

Finally, we plug these values into the Taylor series formula and simplify:

$f(x) = \frac{1}{4} - \frac{1}{16}(x-1) + \frac{1}{2 \cdot 32}(x-1)^2 - \frac{3}{3! \cdot 256}(x-1)^3$

$f(x) = \frac{1}{4} - \frac{1}{16}(x-1) + \frac{1}{64}(x-1)^2 - \frac{1}{256}(x-1)^3$

Thus, the answer is choice C.

Find the Taylor Polynomial of Order 3 Generated by F f(x)=1x+3,a=1f ( x ) = \frac { 1 } { x + 3 } , a = 1f(x)=x+31​,a=1

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