Find the Taylor polynomial of order 3 generated by f at a.
-$$f ( x ) = \ln ( x + 1 ) , a = 6$$
A) $P _ { 3 } ( x ) = \ln 5 + \frac { x - 6 } { 5 } + \frac { ( x - 6 ) ^ { 2 } } { 50 } + \frac { ( x - 6 ) ^ { 3 } } { 375 }$
B) $P _ { 3 } ( x ) = \ln 7 - \frac { x - 6 } { 7 } + \frac { ( x - 6 ) ^ { 2 } } { 98 } - \frac { ( x - 6 ) ^ { 3 } } { 1029 }$
C) $P _ { 3 } ( x ) = \ln 5 - \frac { x - 6 } { 5 } + \frac { ( x - 6 ) ^ { 2 } } { 50 } - \frac { ( x - 6 ) ^ { 3 } } { 375 }$
D) $\mathrm { P } _ { 3 } ( \mathrm { x } ) = \ln 7 + \frac { \mathrm { x } - 6 } { 7 } - \frac { ( \mathrm { x } - 6 ) ^ { 2 } } { 98 } + \frac { ( \mathrm { x } - 6 ) ^ { 3 } } { 1029 }$

Question

Quizplus · Accepted Answer

The Taylor polynomial of order 3 for $f(x) = \ln(x + 1)$ at $a = 6$ is found by evaluating the function and its derivatives at $x = 6$, then using the formula for the Taylor series. The correct coefficients match those in option D, considering the derivatives of $f(x) = \ln(x + 1)$ and evaluating them at $x = 6$.

Find the Taylor Polynomial of Order 3 Generated by F f(x)=ln⁡(x+1),a=6f ( x ) = \ln ( x + 1 ) , a = 6f(x)=ln(x+1),a=6

Find the Taylor Polynomial of Order 3 Generated by F $f ( x ) = \ln ( x + 1 ) , a = 6$