Find the fourth degree Taylor polynomial centered at $c = 7$ for the function. $f ( x ) = \ln x$
A) $\ln 7 - \frac { 1 } { 7 } ( x - 7 ) - \frac { 1 } { 98 } ( x - 7 ) ^ { 2 } - \frac { 1 } { 1029 } ( x - 7 ) ^ { 3 } - \frac { 1 } { 9604 } ( x - 7 ) ^ { 4 }$
B) $\ln 7 + 7 ( x - 7 ) - 98 ( x - 7 ) ^ { 2 } + 1029 ( x - 7 ) ^ { 3 } - 9604 ( x - 7 ) ^ { 4 }$
C) $\ln 7 + \frac { 1 } { 7 } ( x - 7 ) - \frac { 1 } { 98 } ( x - 7 ) ^ { 2 } + \frac { 1 } { 1029 } ( x - 7 ) ^ { 3 } - \frac { 1 } { 9604 } ( x - 7 ) ^ { 4 }$
D) $\ln 7 - 7 ( x - 7 ) + 98 ( x - 7 ) ^ { 2 } - 1029 ( x - 7 ) ^ { 3 } + 9604 ( x - 7 ) ^ { 4 }$
E) $\ln 7 - \frac { 1 } { 7 } ( x - 7 ) + \frac { 1 } { 98 } ( x - 7 ) ^ { 2 } - \frac { 1 } { 1029 } ( x - 7 ) ^ { 3 } + \frac { 1 } { 9604 } ( x - 7 ) ^ { 4 }$

Question

Quizplus · Accepted Answer

We have the formula for the nth degree Taylor polynomial of f(x) centered at c as:

$P_n(x) = f(c) + \dfrac{f'(c)}{1!}(x-c) + \dfrac{f''(c)}{2!}(x-c)^2 + \dots + \dfrac{f^{(n)}(c)}{n!}(x-c)^n$

Here, we have:

$f(x) = \ln x \implies f'(x) = \dfrac{1}{x}, f''(x) = -\dfrac{1}{x^2}, f'''(x) = \dfrac{2}{x^3}, f^{(4)}(x) = -\dfrac{6}{x^4}$

And, we need to find the fourth degree Taylor polynomial centered at c = 7:

\begin{align*}
P_4(x) &= f(7) + \dfrac{f'(7)}{1!}(x-7) + \dfrac{f''(7)}{2!}(x-7)^2 + \dfrac{f'''(7)}{3!}(x-7)^3 + \dfrac{f^{(4)}(7)}{4!}(x-7)^4 \
&= \ln 7 + \dfrac{1}{7}(x-7) - \dfrac{1}{2\cdot 49}(x-7)^2 + \dfrac{2}{3\cdot 343}(x-7)^3 - \dfrac{6}{4!\cdot 2401}(x-7)^4 \
&= \ln 7 + \dfrac{1}{7}(x-7) - \dfrac{1}{98}(x-7)^2 + \dfrac{1}{1029}(x-7)^3 - \dfrac{1}{9604}(x-7)^4 \
\end{align*}

Therefore, our answer is choice C.

Find the Fourth Degree Taylor Polynomial Centered At c=7c = 7c=7

Find the Fourth Degree Taylor Polynomial Centered At $c = 7$