The Taylor series for a function $f$ about $x = 4$ is given by $$\sum _ { n = 1 } ^ { \infty } \frac { ( x - 4 ) ^ { n + 1 } } { ( n + 1 ) 3 ^ { n } } = \frac { ( x - 4 ) ^ { 2 } } { 2 \cdot 3 } + \frac { ( x - 4 ) ^ { 3 } } { 3 \cdot 3 ^ { 2 } } + \frac { ( x - 4 ) ^ { 4 } } { 4 \cdot 3 ^ { 3 } } + \cdots + \frac { ( x - 4 ) ^ { n + 1 } } { ( n + 1 ) 3 ^ { n } } + \cdots$$ and converges to $f ( x )$ on its interval of convergence. A. Using the ratio test, find the interval of convergence of the Taylor series for $f$ about $x = 4$. Justify your answer. B. Find the first three nonzero terms and the general term of the Taylor series for $f ^ { \prime }$, the derivative of $f$, about $x = 4$. C. The Taylor series for $f ^ { \prime }$ described in part B is a geometric series. For all $x$ in the interval of convergence of the Taylor series for $f ^ { \prime }$, show that $f ^ { \prime } ( x ) = \frac { x - 4 } { 7 - x }$. D. It is known that the radius of convergence of the Taylor series for $f$ about $x = 4$ is the same as the radius of convergence of the Taylor series for $f ^ { \prime }$ about $x = 4$. Does the Taylor series for $f ^ { \prime }$ described in part B converge to $f ^ { \prime } ( x ) = \frac { x - 4 } { 7 - x }$ at $x = 8$ ? Give a reason for your answer.
The Taylor series for a function $f$ about $x = 4$ is given by
$$\sum _ { n = 1 } ^ { \infty } \frac { ( x - 4 ) ^ { n + 1 } } { ( n + 1 ) 3 ^ { n } } = \frac { ( x - 4 ) ^ { 2 } } { 2 \cdot 3 } + \frac { ( x - 4 ) ^ { 3 } } { 3 \cdot 3 ^ { 2 } } + \frac { ( x - 4 ) ^ { 4 } } { 4 \cdot 3 ^ { 3 } } + \cdots + \frac { ( x - 4 ) ^ { n + 1 } } { ( n + 1 ) 3 ^ { n } } + \cdots$$
and converges to $f ( x )$ on its interval of convergence.
A. Using the ratio test, find the interval of convergence of the Taylor series for $f$ about $x = 4$. Justify your answer.
B. Find the first three nonzero terms and the general term of the Taylor series for $f ^ { \prime }$, the derivative of $f$, about $x = 4$.
C. The Taylor series for $f ^ { \prime }$ described in part B is a geometric series. For all $x$ in the interval of convergence of the Taylor series for $f ^ { \prime }$, show that $f ^ { \prime } ( x ) = \frac { x - 4 } { 7 - x }$.
D. It is known that the radius of convergence of the Taylor series for $f$ about $x = 4$ is the same as the radius of convergence of the Taylor series for $f ^ { \prime }$ about $x = 4$. Does the Taylor series for $f ^ { \prime }$ described in part B converge to $f ^ { \prime } ( x ) = \frac { x - 4 } { 7 - x }$ at $x = 8$ ? Give a reason for your answer.