$\text { Find the third Taylor polynomial for } f ( x ) = \frac { 7 } { x } \text {, expanded about } c = 1 \text {. }$
A) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) - 14 ( x - 1 ) ^ { 2 } - 7 ( x - 1 ) ^ { 3 }$
B) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) + 7 ( x - 1 ) ^ { 2 } - 7 ( x - 1 ) ^ { 3 }$
C) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) + 14 ( x - 1 ) ^ { 2 } - 14 ( x - 1 ) ^ { 3 }$
D) $P _ { 3 } ( x ) = 7 - 7 ( x + 1 ) + 7 ( x + 1 ) ^ { 2 } - 7 ( x + 1 ) ^ { 3 }$
E) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) - 42 ( x - 1 ) ^ { 2 } - 7 ( x - 1 ) ^ { 3 }$

Question

$\text { Find the third Taylor polynomial for } f ( x ) = \frac { 7 } { x } \text {, expanded about } c = 1 \text {. }$ 
A) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) - 14 ( x - 1 ) ^ { 2 } - 7 ( x - 1 ) ^ { 3 }$
B) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) + 7 ( x - 1 ) ^ { 2 } - 7 ( x - 1 ) ^ { 3 }$
C) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) + 14 ( x - 1 ) ^ { 2 } - 14 ( x - 1 ) ^ { 3 }$
D) $P _ { 3 } ( x ) = 7 - 7 ( x + 1 ) + 7 ( x + 1 ) ^ { 2 } - 7 ( x + 1 ) ^ { 3 }$
E) $P _ { 3 } ( x ) = 7 - 7 ( x - 1 ) - 42 ( x - 1 ) ^ { 2 } - 7 ( x - 1 ) ^ { 3 }$

Quizplus · Accepted Answer

The third Taylor polynomial for $f(x) = \frac{7}{x}$ about $c = 1$ is found by calculating the first three derivatives of $f(x)$, evaluating them at $x = 1$, and then using the formula for the Taylor series. The derivatives of $f(x) = \frac{7}{x}$ are $f'(x) = -\frac{7}{x^2}$, $f''(x) = \frac{14}{x^3}$, and $f'''(x) = -\frac{42}{x^4}$. Evaluating these at $x = 1$ gives $f(1) = 7$, $f'(1) = -7$, $f''(1) = 14$, and $f'''(1) = -42$. The Taylor polynomial is thus $P_3(x) = 7 - 7(x - 1) + 7(x - 1)^2 - 7(x - 1)^3$.

Find the third Taylor polynomial for f(x)=7x, expanded about c=1. \text { Find the third Taylor polynomial for } f ( x ) = \frac { 7 } { x } \text {, expanded about } c = 1 \text {. } Find the third Taylor polynomial for f(x)=x7​, expanded about c=1.

$\text { Find the third Taylor polynomial for } f ( x ) = \frac { 7 } { x } \text {, expanded about } c = 1 \text {. }$